text:
\section{Introduction}

The purpose of this note is to show that the metrics constructed in \cite{as} have finite energy as conjectured by Donaldson in \cite{d2}. Namely we prove the following proposition:
\begin{thm}\label{main}
Let $X$ be a strictly unbounded K\"ahler toric orbifold of real dimension $4$ endowed with one of the scalar- flat K\"ahler toric metric constructed in \cite{as}. The Calabi functional
\[
\int_X|\mathcal{R}|^2,
\]
where $\mathcal{R}$ is the Riemann curvature tensor, is finite.
\end{thm}
An unbounded toric $4$-manifold is roughly a non-compact symplectic $4$-manifold with a $\mathbb{T}^2$ action which is effective and Hamiltonian. Such a manifold is said to be strictly unbounded if the moment map image has non-parallel unbounded edges. The simplest examples of such a manifolds are $\mathbb{R}^4$ and all its blow-ups. We will give a precise definition ahead.

The asymptotic behavior of the metrics in our main theorem was investigated in \cite{as} in order to prove that the constructed metrics were complete. The results there were not sufficient to show that the Calabi functional of these metrics is finite as conjectured by Donaldson in \cite{d2}. In this note, we use the explicit construction of \cite{as} to prove Theorem \ref{main}.  In fact our proof of theorem \ref{main} is essentially direct and we basically proceed by actually calculating
\[
\int_X|\mathcal{R}|^2.
\]
In section \ref{example} we give an example where we carry out the calculation for the the Calabi functional explicitly. The general case is not much harder and is reduced to determining the integral of a rational function of a single real variable. Given any specific example of a strictly unbounded K\"ahler toric orbifold of real dimension $4$ endowed with one of the scalar- flat K\"ahler toric metric from \cite{as} we can carry out the computation of its Calabi functional by integrating a rational function of a single real variable. The degree of the polynomials in the denominator and numerator increases with the number of edges of the moment map image so the calculation becomes more complicated, but can be carried out.

There are still interesting open questions about the asymptotic behavior of the metrics from \cite{as}. For example, for parameters $\nu=(\alpha,\beta)$ for which $\alpha/\beta$ is irrational it is unclear if the corresponding metrics fall into the ALF category. This is a question we plan to address in subsequent work.

Also we conjecture that any scalar-flat K\"ahler toric metric on the strictly unbounded toric 4-manifold $X$ is a so-called Donaldson generalized Taub-NUT metric corresponding to an admissible parameter $\nu \in \mathbb{R}^2$ as constructed in \cite{as}. If this is true, the above proposition shows that the Calabi functional of any scalar- flat K\"ahler toric metric on a strictly unbounded $4$-dimensional orbifold is finite.

We will assume that the reader is somewhat familiar with the construction in \cite{as} but we will quickly give some background on unbounded toric 4-orbifolds and scalar-flat K\"ahler toric metrics on them.

The paper is organized as follows. We first give a very brief review of the results in \cite{as}. Then in section \ref{chern_classes} we explain how to calculate
\[
\int_X|\mathcal{R}|^2,
\]
in terms of
\[
\int_X c_1^2 \quad \text{and} \, \int_X c_2.
\]
This is probably well understood by most readers but we recall it here for the sake of completeness. We describe a set of convenient coordinates called isothermal coordinates which were also used in \cite{as} and explain how to express the Hermitian metric on the tangent bundle corresponding to the scalar-flat K\"ahler metric in these coordinates. In sections \ref{c_1} and \ref{c_2} we show that by using isothermal coordinates our calculations reduce to the calculation of a pair of integrals on $\mathbb{H}$. We also show that the involved integrals are finite. These sections are the heart of the paper. Section \ref{c_2} has rather cumbersome calculations but section \ref{c_1} already gives a flavor of what is involved in the proof of our main theorem and yields relatively simple results. Section \ref{example} particularizes to the case when the underlying  toric manifold is the minimal resolution of a cyclic singularity. Calculations simplify in this setting making it possible to carry out the integrals whose finiteness was shown in sections \ref{c_1} and \ref{c_2}.

\vspace{.5cm}
\noindent \textbf{Acknowledgments:} Most of this work was carried out in 2010 while I was visiting Imperial College. I would like to take the opportunity to thank the Mathematics department there for its hospitality. In particular I would like to thank Simon Donaldson,  Andr\'e Neves and Richard Thomas. I would also like to thank Claude Lebrun for his encouragement and for reviving my own interest in these calculations.

\section{background}
For a more leisurely treatment of unbounded toric manifolds the reader should refer to \cite{as}. We begin by recalling the definition of a K\"ahler toric orbifold
\begin{defn} \label{def:torb}
A K\"ahler toric $4$-orbifold is a connected $4$-dimensional
K\"ahler orbifold $(X,\omega,J)$ equipped with an effective, Hamiltonian,
holomorphic action of the standard (real) $2$-torus $\mathbb{T}^2 = \mathbb{R}^2/2\pi\mathbb{Z}^2$, such that
the corresponding moment map $\phi : X \to \mathbb{R}^2$,  is proper onto its convex image $P=\phi(X)\subset \mathbb{R}^2$.
\end{defn}
We will actually work with a subclass of non-compact K\"ahler toric orbifolds so we make the following definition
\begin{defn}
A K\"ahler toric $4$-orbifold is said to be \emph{unbounded} if its
moment polygon is unbounded and has a finite number of vertices.

A K\"ahler toric $4$-orbifold is said to be \emph{strictly unbounded} if it is unbounded and its moment polygon has non-parallel unbounded edges.
\end{defn}

We showed in \cite{as} that any unbounded K\"ahler toric $4$-manifolds is biholomorphic to a finite blow up of minimal resolutions of cyclic singularities. In the same way unbounded K\"ahler toric $4$-orbifolds are a finite blow up of $\mathbb{C}^2/\mathbb{Z}^k$ for some integer $k$ where  $\mathbb{C}^2/\mathbb{Z}^k$ has an isolated singularity at the origin.

Let us fix some notation. As we have seen in \cite{as}, if $X$ is a strictly unbounded toric $4$-dimensional orbifold, the image of $\phi$ is a convex unbounded polygonal region in $\mathbb{R}^2$ which we will denote by $P$. The region $P$ has a finite number of edges labeled $1,\cdots,d$. Each of the edges has a primitive normal vector in $\mathbb{Z}^2$ $\nu_i, i=1,\cdots, d$ and $P$ is given by
\[
P=\{x\in \mathbb{R}^2: x\cdot\nu_i\geq \lambda_i, i=1,\cdots,d\}
\]
for some real numbers $\lambda_i$. There are exactly two unbounded edges on $P$. The first edge normal to $\nu_1$ and the last edge normal to $\nu_d$. The fact that $X$ is strictly unbounded ensures that these are not parallel. See \cite{as} for more details.  We recall the main results in \cite{as}
\begin{thm}[Abreu,Sena-Dias]
Let $X$ be a strictly unbounded toric 4-manifold with moment map image $P$
\[
P=\{x\in \mathbb{R}^2: x\cdot\nu_i\geq \lambda_i, i=1,\cdots,d\}.
\]
 Let $\nu = (\alpha,\beta)$ be a vector in $\mathbb{R}^2$ such that
\[
\det(\nu_1,\nu),\,\,\det(\nu_d,\nu)>0.
\]
Then $X$ admits a complete scalar-flat K\"ahler toric metric called Donaldson generalized Taub-NUT metric with parameter $\nu$.
\end{thm}
Theorem \ref{main} applies  to these metrics.

For later reference we recall the definition of symplectic coordinates on $X$. The reader who is not familiar with these coordinates can look at \cite{a1} or \cite{a2}. Alternatively there is some background in \cite{as}.

Let $X_0=\phi^{-1} (\text{int} (P))$. The torus $\mathbb{T}^2$ acts freely on $X_0$ and
\[
X_0\simeq\text{int} (P)\times \mathbb{T}^2.
\]
There are (symplectic) coordinates on $X_0$ that realize the above identification. We denote these by $(x_1,x_2,\theta_1,\theta_2)$ where $(x_1,x_2)$ take values in $P$ and $(\theta_1,\theta_2)$ are the angle coordinates on the torus $\mathbb{T}^2$. There is also a so-called symplectic potential associated to each K\"ahler toric metric on $X$. It is a smooth function on $\text{int} (P)$ which we will denote by $u$. It completely determines the metric. We set $\xi_1=u_{x_1}$ and $\xi_2=u_{x_2}$. As it turns out $(\xi_1,\xi_2,\theta_1,\theta_2)$ give complex coordinates on $X$.

Another useful set of coordinates on $X_0$ is given as follows. The toric K\"ahler metric determines a metric on the real locus of $X$. The metric induced on this real  2-dimensional manifold admits a set of isospectral coordinates which we call $(H,r)$ where
$$
r=(\Det(\Hess u))^{-\frac{1}{2}}
$$
These were described by Donaldson in \cite{d2} and used in \cite{as}. They play a very important role here as well. It turns out that they define a diffeomorphism between $\text{int} (P)$ and $\mathbb{H}=\{(H,r):r>0\}$ which we will denote by $\gamma$.

We start by showing that the finiteness of the first and second Chern classes of $X$ imply the finiteness of the Calabi functional.

\section{The Calabi functional and Chern classes}\label{chern_classes}
\subsection{Characteristic classes on $X$}
The K\"ahler metric determines a Hermitian connection $\theta$ on the complex fiber bundle $TX$. On a coordinate patch, using a local frame, $\theta$ can be interpreted as a $2$ by $2$ matrix of $1$-forms.  As in the compact case, such a connection has a curvature form which we denote by $\Theta$. It is given by
\[
\Theta=d\theta+\theta\wedge  \theta.
\]
Locally, the curvature $\Theta$ can be seen as a $2$ by $2$ matrix of $2$-forms using a local frame for $TX$. Changing frames on $TX$ changes the connection form to an adjoint matrix and therefore, using ad-invariant polynomials (i.e. polynomials that are invariant under the adjoint action of $U(2)$ on the space of $2$ by $2$ matrices), there are globally well defined $2$-forms to be extracted from the the curvature $\Theta$. These are called characteristic forms. See \cite{w} for a more precise definition.

We will use the following characteristic forms.
\begin{itemize}
\item $c_1=\Tr \Theta$, where $\Tr$ denotes the trace, is the first Chern class of $X$,
\item  $c_2=\det\Theta$, where $\det$ denotes the determinant, is the second Chern class of $X$,
\item $p=\Tr (\Theta \wedge \Theta)$ is the Pontryagin class of $X$.
\end{itemize}
On a $2$-dimensional complex manifold these are related because the space of ad-invariant polynomials of $2$ by $2$ matrices is $2$-dimensional. We will use the following very easy lemma.
\begin{lemma}\label{2dimcc}
The forms $c_1^2$ and $p$ determine the form $c_2$.
\end{lemma}
\begin{proof}
Using a local frame write the curvature form as
\[
\Theta=
\begin{pmatrix}
\Theta_{11}&\Theta_{12}\\
\Theta_{21}&\Theta_{22}
\end{pmatrix}.
\]
Then
\[
c_1^2=(\Theta_{11}+\Theta_{22})\wedge (\Theta_{11}+\Theta_{22})=\Theta_{11}^2+\Theta_{22}^2+2\Theta_{11}\wedge \Theta_{22},
\]
and
\[
p=\Theta_{11}^2+\Theta_{12}\wedge\Theta_{21}+\Theta_{21}\wedge\Theta_{12}+\Theta_{22}^2=
\Theta_{11}^2+\Theta_{22}^2+2\Theta_{12}\wedge\Theta_{21}.
\]
As for the $c_2$ it is given by
\[
c_2=\Theta_{11}\wedge \Theta_{22}-\Theta_{21}\wedge\Theta_{12},
\]
and therefore $c_1^2-p=2c_2$.
\end{proof}

\subsection{The Calabi functional and characteristic classes}
In fact we have introduced characteristic classes because our main result turns out to be equivalent to the integrability of the characteristic classes of $TX$. The following lemma holds.
\begin{lemma}
Let $X$ be a non-compact 4-dimensional scalar-flat K\"ahler manifold. Then if
\[
\int_X c_1^2, \,\,\, \int_X p
\]
are finite, $X$ has finite energy i.e.
\[
\int_X |\mathcal{R}|^2
\]
is finite.
\end{lemma}
\begin{proof}
On a K\"ahler manifold, following the notation in \cite{b}, we can write the Riemann curvature tensor norm as linear combination of $s^2$, $|\rho_0|^2$ and $|B_0|^2$ where
\begin{itemize}
\item $s$ is the scalar curvature,
\item $\rho_0$ is the primitive part of the Ricci tensor,
\item and $|B_0|^2$ is the primitive part of the so-called $B$ tensor (see \cite{b}, pag. 77).
\end{itemize}
Now consider the following Apte identities (see \cite{b}, pag. 80)
\begin{equation}
\frac{1}{2}\wedge^2 c_2=\frac{1}{8\pi^2}\left(\frac{s^2}{12}-|\rho_0|^2+|B_0|^2\right),
\end{equation}
\begin{equation}
\frac{1}{2}\wedge^2 c_1=\frac{1}{4\pi^2}\left(\frac{s^2}{8}-|\rho_0|^2\right).
\end{equation}
where $\wedge$ is contraction by $\omega$ i.e. the formal adjoint of wedging with $\omega$. Our metrics are all scalar-flat i.e. $s=0,$ so we see from the above formulas, that to show that $\int|\mathcal{R}|^2$ is finite it is enough to show that $\int c_2$ and $\int c_1^2$ are finite. Or, using the above lemma \ref{2dimcc}, we see that to show that the Calabi functional is finite, it is enough to see that  $\int c_1^2$ and  $\int p$ are finite.
 \end{proof}

We will work in the coordinates $(H,r)$ introduced in \cite{as}. These are isospectral coordinates for the metric on the real locus of $X$. We have
\[
r=(\det \Hess u)^{-\frac{1}{2}}.
\]
It is not hard to see that $r$ is harmonic for the restriction of the scalar-flat metric on $X$ to the real locus of $X$. We define $H$ to be its harmonic conjugate. In these coordinates the scalar-flat metric is given by specifying an $\mathbb{R}^2$ valued function $\xi(H,r)$ which axi-symmetric and harmonic when seen as function in $\mathbb{R}^3$. Here $H$ is the third coordinate in $\mathbb{R}^3$ and $r$ is the distance to $H$-axis. In \cite{as} we study a specific choice for $\xi$ given a polytope. Namely
\begin{displaymath}
\xi = \nu_1\log(r)+\frac{1}{2}\sum_{i=1}^{d-1}(\nu_{i+1}-\nu_{i})\log\left( H_i+\rho_i\right)+\nu H
\end{displaymath}
where
\begin{itemize}
\item the $\nu_i$ denote the normals to the facets of the moment polytope of $X$,
\item the vector $\nu$ is such such that
\[
\Det(\nu,\nu_1),\Det(\nu,\nu_d) \geq 0,
\]
\item Given real number $0<a_1<\cdots<a_{d-1}$ appropriately chosen in terms of the moment polytope, $H_i$ is simply $H+a_i$ for all $i=1, \cdots, d-1$
\item $\rho_i$ is simply $\sqrt{r^2+(H+a_i)^2}$ for all $i=1, \cdots, d-1$.
\end{itemize}
Given the vector $\xi$, the coordinates $(H,r)$ give a bijective proper identification of the moment polytope of $X$ with the upper half plane $\mathbb{H}$. See \cite{as} for more details.

We set up some notation that will be useful later on. Let $(\alpha_i,\beta_i)$ denote the components of the normal $\nu_i$ to the $i$th facet of $P$ where $i=1,\cdots,d$.  We have
\[
D\xi=\begin{pmatrix}
\sum_{i=1}^{d-1}\frac{(\alpha_{i+1}-\alpha_{i})}{2\rho_i}+\alpha&\frac{1}{r}\left({\alpha_1}+\frac{1}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{\rho_i}\right)(\alpha_{i+1}-\alpha_{i})\right)  \\
\sum_{i=1}^{d-1}\frac{(\beta_{i+1}-\beta_{i})}{2\rho_i}+\beta&\frac{1}{r}\left({\beta_1}+\frac{1}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{\rho_i}\right)(\beta_{i+1}-\beta_{i})\right)
\end{pmatrix}.
\]
Let
\begin{equation}\label{f}
f=\nu_1+\frac{1}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{\rho_i}\right)(\nu_{i+1}-\nu_{i})
\end{equation}
and
\begin{equation}\label{g}
g=\nu+\sum_{i=1}^{d-1}\frac{(\nu_{i+1}-\nu_{i})}{2\rho_i},
\end{equation}
denote the components of $g$ by $(g_a,g_b)$ and the components of $f$ by $(f_a,f_b)$ then
\begin{equation}\label{Dxi}
D\xi=\begin{pmatrix}
g_a&\frac{f_a}{r}  \\
g_b&\frac{f_b}{r}
\end{pmatrix}.
\end{equation}
Said in another way, the columns of $D\xi$ are precisely $g$ and $\frac{f}{r}$. This implies that $V=r\Det D\xi$ is given by
$$
V=\Det (g,f).
$$
We also have that
\begin{equation}\label{inversa_Dxi}
D\xi^{-1}=\frac{1}{V}\begin{pmatrix}
f_b&-{f_a}  \\
-rg_b&rg_a
\end{pmatrix}.
\end{equation}
Said in another way, the lines of $D\xi^{-1}$ are $f^\perp=(f_b,-f_a)$ and $-g^\perp=(-g_b,g_a)$.

Let $R$ be a positive large number and $B(R)$ be the pre-image via the moment map in $X$ of the ball of radius $R$ in $\mathbb{H}$ i.e.
\[
B(R)=\phi^{-1}\gamma^{-1}(\{(H,r)\in \mathbb{H}: \rho=\sqrt{r^2+H^2}<R\})
\]
where $\phi$ denotes the moment map. We are actually using $(H,r)$ to identify the moment polytope with $\mathbb{H}$. If we show that the quantities
\[
\int_{B(R)}c_1^2
\]
and
\[
\int_{B(R)}p
\]
are bounded as $R$ tends to infinity, this will finish our proof. We will often write $B(R)$ for both the set $\{(H,r)\in \mathbb{H}: \rho=\sqrt{r^2+H^2}<R\}$ and its pre-image via the moment map $\gamma\circ\phi$.

The boundary of $B(R)$ in $\mathbb{H}$ has $2$ distinct subsets that we will treat separately in calculations. Namely
$$
{\partial} B(R)=L_R \cup C_R
$$
where
\begin{itemize}
\item the set $L_R$ is given by $L_R=\{(H,0): -R<H<R\}$,
\item  the set $C_R$ is given by $C_R=\{(H,r): \sqrt{r^2+H^2}=R\}$.
\end{itemize}

\subsection{The Hermitian metric in the toric setting}
Consider a non-compact toric manifold $X$ endowed with a K\"ahler toric metric. This K\"ahler metric determines a Hermitian metric on $TX$ and we want to give an expression for this metric in a local frame using the symplectic potential.
\begin{lemma}\label{hermitian_metric}
Let $X$ be a toric manifold with a K\"ahler toric metric. Consider complex coordinates in the usual way in the pre-image via the moment map of the interior of the moment map image of $X$. In this frame, the Hermitian metric on $TX$ is simply $2\Hess u$ where $u$ is the symplectic potential associated to the K\"ahler toric metric. The induced connection is given by
\[
\theta=-\partial \Hess u \Hess^{-1} u.
\]
\end{lemma}
\begin{proof}
We will use isospectral coordinates $(H,r)$ on $X$ to prove this lemma. Let $P$ be the moment map image of $X$ in $\mathbb{R}^2$. Let $(\xi_1, \xi_2)$ denote coordinates on $\mathbb{R}^2$. Over the interior of $P$, the torus action on $X$ is free and $X$ admits complex coordinates $(z_1,z_2)$ given by
\[
z_a=\xi_a+i\theta_a
\]
for $a=1,2$.
In the above formula $\xi_a=u_{x_a}$ can be expressed as functions of the coordinates $(H,r)$ by the formulas in \cite{as} and $(\theta_1,\theta_2)$ refer to the angle coordinates on the tori above each point in $\text{int}(P)$. For $a=1,2$, we can write
\[
\frac{\partial}{\partial z_a}=\frac{\partial H}{\partial \xi_a}\frac{\partial}{\partial H}+\frac{\partial r}{\partial \xi_a}\frac{\partial}{\partial r}-i\frac{\partial}{\partial \theta_a}
\]
In the coordinates $(x_1,x_2,\theta_1,\theta_2)$ the metric is given by
\begin{equation}\nonumber
\begin{bmatrix}
\phantom{-}\Hess u & \vdots & 0\  \\
\hdotsfor{3} \\
\phantom{-}0 & \vdots & \Hess^{-1} u
\end{bmatrix},
\end{equation}
where $u$ is the symplectic potential of the metric. Since $(H,r)$ are isospectral coordinates for the restriction of the metric to the polytope, in the coordinates $(H,r,\theta_1,\theta_2)$ the metric is given by
\begin{equation} \label{metric_rH}
\begin{bmatrix}
\phantom{-}\begin{pmatrix}
 V&0\\
  0&V
  \end{pmatrix}
   & \vdots & 0\  \\
\hdotsfor{3} \\

\phantom{-}0 & \vdots & \Hess^{-1}u
\end{bmatrix}
\end{equation}
where $V=r \det D\xi$. Here, $D\xi$ denotes the matrix of the derivatives of $(\xi_1,\xi_2)$ as a function of $(H,r)$. The fiber bundle $TX$ is locally trivialized by $(\frac{\partial}{\partial z_1},\frac{\partial}{\partial z_2})$ and using expression (\ref{metric_rH}) for the metric, we can see that the Hermitian metric in this trivialization is given by
\[
\left( \frac{\partial}{\partial z_a}, \frac{\partial}{\partial z_b}\right)=V\left(\frac{\partial r}{\partial \xi_a}\frac{\partial r}{\partial \xi_b}+ \frac{\partial H}{\partial \xi_a}\frac{\partial H}{\partial \xi_b} \right)+ u^{ab},
\]
with $a,b \in \{1,2\}$. Here $u^{ab}$ denotes the entries of the matrix $\Hess^{-1} u$, as usual. The matrix of the Hermitian metric in the frame $(\frac{\partial}{\partial z_1},\frac{\partial}{\partial z_2})$ is given by
\[
V (D\xi^{-1})^tD\xi^{-1}+\Hess^{-1} u.
\]
Next we prove a lemma that will be extremely useful to us ahead.
\begin{lemma}\label{Hess}
Let $g$ be a K\"ahler toric metric on a symplectic toric four-fold $X$ with symplectic potential $u$ on its moment polytope $P$. Let $(H,r)\in \mathbb{H}$ be isothermal coordinates for $g$ and $\xi=(u_{x_1},u_{x_2})$. Then
\[
\Hess u(x)=\frac{D\xi D\xi^{t}}{V}(H,r)
\]
where $V=r\det{D\xi}$ and $(H,r)$ corresponds to $x=(x_1,x_2)$ under the identification $\mathbb{H}\simeq P$.
\end{lemma}
\begin{proof}
Let $\eta=(u_{x_1},u_{x_2})$ and $\mu:\mathbb{H}\rightarrow P$ be the moment map in $(H,r)$ coordinates. We have that $\eta=\xi\circ\mu^{-1}$ and
\[
\Hess u=D\eta=D\xi D\mu^{-1}
\]
Now as we have seen in \cite{as}
\[
D\mu=r\begin{pmatrix}
\xi_{2,r} & -\xi_{2,H}  \\
-\xi_{1,r} & \xi_{1,H}
\end{pmatrix},
\]
hence
\[
D\mu^{-1}=\frac{D\xi ^t}{r\det D\xi}
\]
and the result follows.
\end{proof}
From the above lemma it follows that the Hermitian metric is simply $2\Hess^{-1} u$. This implies that the Hermitian connection matrix $\theta$ in our frame is given by
\[
\theta=\partial \Hess^{-1} u \Hess u=-\partial \Hess u \Hess^{-1} u.
\]
as stated.
\end{proof}

\section{The first Chern class}\label{c_1}
In this section we start by showing that
\[
\int_X c_1^2
\]
is finite. The matrix of the Hermitian metric on $TX$ in a local frame over $X_0$ is given by $2\Hess^{-1} u$ (see lemma \ref{hermitian_metric}) and thus
\[
c_1=\frac{1}{2}\overline{\partial}\p \log \det \Hess^{-1} u.
\]
From \cite{d1} we know that
\[
r=(\det \Hess u)^{-\frac{1}{2}},
\]
therefore
\[
c_1=d{\partial} \log r.
\]
We can write
\[
\int_{B(R)}c_1^2=\int_{B(R)}d\left({\partial} \log r\wedge d {\partial}\log r\right).
\]
Using Stokes theorem this becomes
\[
\int_{B(R)}c_1^2=\int_{{\partial} B(R)}\frac{{\partial} r}{r}\wedge d\frac{{\partial} r}{r}=\int_{{\partial} B(R)}\frac{{\partial} r\wedge d{\partial} r}{r^2}.
\]
Now we need to calculate ${\partial} r$ using the complex structure associated to our K\"ahler metric. This calculation will be important to us ahead so we state it as a lemma.
\begin{lemma}\label{partial}
Let $X$ be a strictly unbounded toric manifold with a scalar-flat K\"ahler metric given via a symplectic potential $u$. Let $(H,r)$ be isospectral coordinates for the metric on the real locus of $X$ with
\[
r=(\det \Hess u)^{-\frac{1}{2}},
\]
and $(\theta_1,\theta_2)$ be angle coordinates on $\mathbb{T}^2$. Write $\xi=(u_{x_1},u_{x_2})$ where $x=(x_1,x_2)$ are the coordinates on $P$. Then
\[
\begin{pmatrix}
{\partial} H\\
{\partial} r
\end{pmatrix}=
\begin{pmatrix}
dH\\
dr
\end{pmatrix}+iD\xi^{-1}
\begin{pmatrix}
d\theta_1\\
d\theta_2
\end{pmatrix},
\]
where $D\xi$ denotes the matrix of derivatives of $\xi$ as a function of $(H,r)$.
\end{lemma}
\begin{proof}
We have
\[
\begin{pmatrix}
dH\\
dr
\end{pmatrix}=\frac{{\partial} (H,r)}{{\partial} x } \begin{pmatrix}
dx_1\\
dx_2
\end{pmatrix}
\]
where
\[
\frac{{\partial} (H,r)}{{\partial} x}=\begin{pmatrix}
\frac{{\partial} H}{{\partial} x_1}&\frac{{\partial} H}{{\partial} x_2}\\
\frac{{\partial} r}{{\partial} x_1}&\frac{{\partial} r}{{\partial} x_2}
\end{pmatrix}.
\]
In symplectic coordinates $(x_1,x_2, \theta_1,\theta_2)$ the complex structure can be written as
\[
\begin{bmatrix}
\phantom{-}
   0& \vdots & -\Hess^{-1}u\  \\
\hdotsfor{3} \\

\phantom{-}\Hess u & \vdots & 0
\end{bmatrix}
\]
so that
\[
J\begin{pmatrix}
dx_1\\
dx_2
\end{pmatrix}=-\Hess^{-1}u\begin{pmatrix}
d\theta_1\\
d\theta_2
\end{pmatrix}
\]
and
\[
J\begin{pmatrix}
dH\\
dr
\end{pmatrix}=\frac{{\partial} (H,r)}{{\partial} x }J \begin{pmatrix}
dx_1\\
dx_2
\end{pmatrix}=-\frac{{\partial} (H,r)}{{\partial} x } \Hess^{-1} u\begin{pmatrix}
d\theta_1\\
d\theta_2
\end{pmatrix}.
\]
We will use the following simple facts
\begin{enumerate}
\item
\[
\Hess u=\frac{D\xi D\xi^t}{V}
\]
\item
\[
D\gamma=\frac{{\partial} (H,r)}{{\partial} x }=\frac{D\xi^t}{V}.
\]
\end{enumerate}
The first is proved in lemma \ref{Hess}, the second comes up in the proof of lemma \ref{Hess}. We  have
\[
\frac{{\partial} (H,r)}{{\partial} x } \Hess^{-1} u=\frac{D\xi^t}{V}V(D\xi^{-1})^tD\xi^{-1}=D\xi^{-1}.
\]
and
\[
J\begin{pmatrix}
dH\\
dr
\end{pmatrix}=-D\xi^{-1}\begin{pmatrix}
d\theta_1\\
d\theta_2
\end{pmatrix}.
\]
Because
\[
\begin{pmatrix}
{\partial} H\\
{\partial} r
\end{pmatrix}=
\begin{pmatrix}
dH\\
dr
\end{pmatrix}-iJ
\begin{pmatrix}
dH\\
dr
\end{pmatrix},
\]
the result follows.
\end{proof}
The above lemma gives a more tractable expression for $\int_X c_1^2$. We have the following
\begin{lemma}\label{c1_as_det}
We have
$$
\int_Xc_1^2=\lim_{R\rightarrow \infty}\int_{\partial B(R)}\frac{\Det(dg,g)}{\Det^2(g,f)}.
$$
\end{lemma}
\begin{proof}
We start by showing that
$$
\int_Xc_1^2=\lim_{R\rightarrow \infty}\int_{\partial B(R)}\frac{adb-bda}{r^2},
$$
where $(a,b)$ is the last line of the matrix $D\xi^{-1}$. We know that $\int_Xc_1^2$ is given by
\[
\lim_{R\rightarrow \infty}\int_{{\partial} B(R)}\frac{{\partial} r \wedge d{\partial} r}{r^2}
\]
First note that from lemma \ref{partial}, if we denote the elements in the second line of the matrix $D\xi^{-1}$ by $a$ and $b$ we have
\[
{\partial} r=dr+i(ad\theta_1+bd\theta_2).
\]
It follows that
\[
d{\partial} r=i\left(da\wedge d\theta_1+db\wedge d\theta_2\right),
\
\]
and
\[
{\partial} r \wedge d{\partial} r=-(adb-bda)\wedge d \theta_1\wedge d\theta_2+idr\wedge (da\wedge d\theta_1+db\wedge d\theta_2).
\]
Since we are integrating over the pre-image of a curve in $P$, any multiple of the form $dr\wedge(da\wedge d\theta_1+db\wedge d\theta_2)$ integrates to zero and we see that
\[
\int_Xc_1^2=\lim_{R\rightarrow \infty}\int_{{\partial} B(R)}\frac{adb-bda}{r^2}.
\]
It follows from formula \ref{inversa_Dxi} that
$$
a=-\frac{rg_b}{V}, \quad b=\frac{rg_a}{V}
$$
so that
$$
adb-bda=\frac{r^2}{V^2}(-g_bdg_a+g_adg_b)-g_ag_bd\left(\frac{r}{V}\right)+g_ag_bd\left(\frac{r}{V}\right)
$$
hence
$$
\frac{adb-bda}{r^2}=\frac{\Det(dg,g)}{V^2}
$$
and the claim follows from the fact that $V=\Det(g,f)$.
\end{proof}
We are now in a position to prove the main result in this section, namely:
\begin{prop}\label{c1}
Let $X$ be an unbounded toric manifold with a scalar-flat toric K\"ahler metric. Then
\[
\int_X c_1^2
\]
is finite.
\end{prop}
Before we begin with proof we set up some notation.
\begin{nota}
We will write
$$O(0), \, O(-1),\, O(-2),\, O(-3), O(-4)$$
for functions on $\mathbb{H}$ which are bounded by
$$C,\,\frac{C}{\rho},\,\frac{C}{\rho^2},\,\frac{C}{\rho^3}, \, \frac{C}{\rho^4}$$
respectively.  Here $C$ is some constant.
\end{nota}

\begin{proof}
Given equations \ref{f} and \ref{g} for the vectors $f$ and $g$ we can give an explicit formula for the $1$-form
$$
\frac{\Det(dg,g)}{\Det(g,f)^2}.
$$
We have
$$
dg=-\sum_{i=1}^{d-1}\frac{rdr+H_idH}{\rho_i^3}(\nu_{i+1}-\nu_i)
$$
so that $\Det(dg,g)$ is given by
$$
\frac{1}{2}\sum_{i=1}^{d-1}\frac{rdr+H_idH}{\rho_i^3}\Det(\nu,\nu_{i+1}-\nu_i)+\frac{1}{4}\sum_{i,j=1}^{d-1}\frac{rdr+H_idH}{\rho_i^3\rho_j}\Det(\nu_{j+1}-\nu_j,\nu_{i+1}-\nu_i)
$$
and from the formulas \ref{f} and \ref{g} we can write $\Det(g,f)$ as
$$
\begin{aligned}
&\Det(\nu,\nu_1)+\frac{1}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{\rho_i}\right)\Det(\nu,\nu_{i+1}-\nu_{i})-\sum_{i=1}^{d-1}\frac{1}{2\rho_i}\Det(\nu_1,\nu_{i+1}-\nu_{i})\\
&+\frac{1}{4}\sum_{i,j=1}^{d-1}\frac{1}{\rho_i}\left(1-\frac{H_j}{\rho_j}\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j)\\
\end{aligned}
$$

Next we will analyze the behavior of $\Det(dg,g)/\Det(g,f)^2$ on $C_R$. We have that for any $i=1,\cdots d-1$
$$
\frac{1}{\rho}-\frac{1}{\rho_i}=O(-2).
$$
We also have that for any $i,j=1,\cdots d-1$
$$
 \frac{H}{\rho^3}-\frac{H_i}{\rho_i^3}=O(-3), \, \frac{r}{\rho^3}-\frac{r}{\rho_i^3}=O(-3)
$$
which implies that
$$
\Det(dg,g)=\frac{(rdr+HdH)\Det(\nu,\nu_{d}-\nu_1)}{2\rho^3}+O(-3), \, \text{if}\quad \nu \ne 0
$$
and
$$
\frac{H}{\rho^4}-\frac{H_i}{\rho_i^3\rho_j}=O(-4), \, \frac{r}{\rho^4}-\frac{r}{\rho_i^3\rho_j}=O(-4)
$$
together with the fact that
$$
\sum_{i,j=1}^{d-1}\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j)
$$
this implies that
$$
\Det(dg,g)=O(-4),  \, \text{if} \quad \nu=0.
$$
On $C_R$, $\rho$ is constant so the $1$ form $rdr+HdH$ vanishes and we conclude that
$$\Det(dg,g)=O(-3), \, \text{if}\quad \nu \ne 0$$
and
$$\Det(dg,g)=O(-4), \, \text{if}\quad \nu= 0.$$ Also $\Det(g,f)$ is bounded from below in the case where $\nu\ne0$ and bounded from below by a constant $C$ times $1/\rho$ in the case where $\nu=0$. More precisely we have
$$
\Det(g,f)=\Det(\nu,\nu_1)+\frac{1}{2}\left(1-\frac{H}{\rho}\right)\Det(\nu,\nu_d-\nu_1)+O(-1), \, \text{if}\quad \nu\ne 0
$$
and
$$
\Det(g,f)=\frac{\Det(\nu_d,\nu_1)}{\rho}+O(-2), \, \text{if}\quad \nu= 0.
$$
Hence we may conclude that in both cases $\Det(dg,g)/\Det(f,g)^2$  is $O(-2)$ on $C_R$ and therefore its integral on $C_R$ tends to zero as $R$ tends to infinity.

In the region $L_R$, $\Det(dg,g)/\Det(f,g)^2$ is given by
\begin{equation}\label{c1LR}
\frac{\left(\frac{1}{2}\sum\frac{H_i}{|H_i|^3}\Det(\nu,\nu_{i+1}-\nu_i)+\frac{1}{4}\sum\frac{H_i}{|H_i|^3|H_j|}\Det(\nu_{j+1}-\nu_j,\nu_{i+1}-\nu_i)\right)dH
}{V^2}
\end{equation}
where
$$
\begin{aligned}
V=&\Det(\nu,\nu_1)+\frac{1}{2}\sum_{i=1}^{d-1}\Det(\left(1-\frac{H_i}{|H_i|}\right)\nu-\frac{\nu_1}{2|H_i|},\nu_{i+1}-\nu_{i})\\
&+\frac{1}{4}\sum_{i,j=1}^{d-1}\frac{1}{|H_i|}\left(1-\frac{H_j}{|H_j|}\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j).\\
\end{aligned}
$$
This is a rational form on each of the intervals $]-a_{i+1},-a_i[$  as well as on the intervals $]-\infty,-a_{d-1}[$ and $]-a_1,+\infty[$ which we can in principle integrate explicitly. Note that is has the appropriate behavior both at infinity and at the boundaries of the intervals so that it is integrable and therefore the corresponding integral on $L_R$ is finite. The lemma follows from this.
\end{proof}

\section{The Pontryagin class}\label{c_2}
The aim of this section is to show that the integral of the Pontryagin class of a scalar-flat K\"ahler toric metric on an unbounded toric $4$-manifold is finite. As for the corresponding calculation for $c_1^2$ which we carried out in the previous section, this involves some explicit knowledge about the asymptotic behavior of such metrics which were written down in \cite{as}.

We start by using the coordinates $(H,r)$ to give a more tractable formula for $\int_X p$.
\begin{lemma}\label{lemmaT}
Let
$$
M_H={\partial}_H \left(\frac{D\xi D\xi^t}{V}\right) V(D\xi D\xi^t)^{-1}-\frac{V_H}{V}\id
$$
 and
 $$
 M_r={\partial}_r \left(\frac{D\xi D\xi^t}{V}\right) V(D\xi D\xi^t)^{-1}-\frac{V_r}{V}\id
 $$
respectively.
Let
\begin{equation}\label{T}
\begin{aligned}
T=&\frac{\Tr (M_H^2) \Det(df,f)-\Tr (rM_rM_H) (\det(dg,f)+\Det(df,g))+\Tr (r^2M_r^2) \Det(dg,g)}{V^2}\\
 &+\frac{\Tr (rM_rdM_H-M_Hd(rM_r))}{V}.
\end{aligned}
\end{equation}
Then $\int_X p$ is given by the limit as $R$ tends to $\infty$ of the integral of $T$ on ${\partial} B(R)$ i.e.
$$
\int_X p=\lim_{R\rightarrow \infty} \int_{{\partial} B(R)} T
 $$
 \end{lemma}
\begin{proof}
By definition we have $p=\Tr (\Theta\wedge \Theta)$ where $\Theta$ is the curvature of the connection induced on $TX$ by our scalar-flat toric K\"ahler metric. Now
\[
\Theta=d\theta+\theta\wedge \theta
\]
where the connection form $\theta$ is of type $(1,0)$. Because our manifold is of complex dimension $2$ we see that $\theta\wedge \theta\wedge \theta\wedge \theta$ is zero because it is of type $(4,0)$. As for $d\theta\wedge \theta\wedge\theta$, it is a sum of a form of type $(4,0)$ with a form of type $(3,1)$ both of which are zero. Hence
\[
p=\Tr \left((d\theta+\theta\wedge \theta)\wedge (d\theta+\theta\wedge \theta)\right)=\Tr (d\theta \wedge d\theta)=d\Tr (\theta\wedge d\theta).
\]
By applying Stokes theorem we see that
\[
\int_{B(R)}p=\int_{{\partial} B(R)} \Tr (\theta \wedge d\theta).
\]
We know that
\[
\theta={\partial} \Hess u \Hess^{-1}u.
\]
Let $M_H$ and $M_r$ denote the matrices ${\partial}_H \Hess u \Hess^{-1}u$ and ${\partial}_r \Hess u \Hess^{-1}u$ respectively. Because
$$
\Hess u =\frac{D\xi D\xi^t}{V}
$$
these matrices are also given by the formulas in the lemma. Namely
$$
M_H={\partial}_H \left(\frac{D\xi D\xi^t}{V}\right) V(D\xi D\xi^t)^{-1}-\frac{V_H}{V}\id
$$
 and
 $$
 M_r={\partial}_r \left(\frac{D\xi D\xi^t}{V}\right) V(D\xi D\xi^t)^{-1}-\frac{V_r}{V}\id.
 $$
Now
\[
\theta=M_H {\partial} H +M_r {\partial} r.
\]
We write this as
\[
\theta= \begin{pmatrix} M_H& M_r \end{pmatrix}\begin{pmatrix}{\partial} H\\ {\partial} r \end{pmatrix}.
\]
In lemma \ref{partial} we calculated ${\partial} H$ and ${\partial} r$. Replacing above we see that
\[
\theta= \begin{pmatrix} M_H& M_r \end{pmatrix}\begin{pmatrix}d H\\ d r \end{pmatrix}+i\begin{pmatrix} M_H& M_r \end{pmatrix}D\xi^{-1}\begin{pmatrix}d \theta_1\\ d \theta_2 \end{pmatrix}.
\]
Let $A$ and $B$ be matrices defined by
\[
\begin{pmatrix} A& B \end{pmatrix}=\begin{pmatrix} M_H& M_r \end{pmatrix}D\xi^{-1}.
\]
The matrices $A$ and $B$ are only function of $H$ and $r$, they do not depend on $\theta_1$ and $\theta_2$. We have
\[
\theta= M_HdH+ M_rdr +i(Ad \theta_1+B d \theta_2),
\]
and
\[
d\theta= dM_H\wedge dH+d M_r\wedge dr +i(dA\wedge d \theta_1+dB \wedge d \theta_2).
\]
By wedging these two together we see that
\[
\theta \wedge d\theta= -(Ad \theta_1+B d \theta_2)\wedge(dA\wedge d \theta_1+dB \wedge d \theta_2)+M
\]
where $M$ is a matrix of two forms which integrate to zero on ${\partial} B(R)$. In fact, the entries of the matrix $M$ are linear combinations of the forms $dH\wedge dr\wedge d \theta_1$ and $dH\wedge dr\wedge  d\theta_2$, which vanish on ${\partial} B(R)$. From this we see that
\[
\int_X p=\lim_{R\rightarrow \infty }\int_{{\partial} B(R)} \Tr (AdB-BdA)
\]
Now
$$
\begin{pmatrix} A& B \end{pmatrix}=\begin{pmatrix} M_H& M_r \end{pmatrix}D\xi^{-1}
$$
so that using formula \ref{inversa_Dxi} for the inverse of $D\xi$ it follows that
$$
 A=\frac{1}{V}\left({f_bM_H}-{g_b}rM_r \right)
$$
and
$$
 B=-\frac{1}{V}\left({f_aM_H}-{g_a}rM_r \right).
$$
By expanding out $\Tr (AdB-BdA)$ using the above expressions the result of the lemma follows in a straightforward manner.
\end{proof}
It is also possible to obtain expressions for the matrices $M_H$ and $M_r$ in terms of the functions $f$ and $g$. These expressions although slightly cumbersome will prove very important for our calculations. So we write them in a separate lemma.
\begin{lemma}
Let $M_H$ and $M_r$ be defined as in lemma \ref{lemmaT}. Then we have the following expression for $M_H$
\begin{equation}
\begin{aligned}\label{MH}
M_H=& \frac{\Det(g,f)g_H(f^\perp)^t+r^2\Det(g_H,g)g(g^\perp)^t}{V^2}\\
+&\frac{\Det(g_H,f)g(f^\perp)^t-\Det(g,f)f_H(g^\perp)^t}{V^2} \\
+& \frac{-\Det(g,f_H)f(g^\perp)^t+\frac{\Det(f_H,f)}{r^2}f(f^\perp)^t}{V^2}\\
-&\frac{\Det(g_H,f)+\Det(g,f_H)}{\Det (g,f)}\id,
\end{aligned}
\end{equation}
and a corresponding expression for $M_r$ namely
\begin{equation}\label{Mr}
\begin{aligned}
M_r=&\frac{ \Det(g,f)g_r(f^\perp)^t+r^2\Det(g_r,g)g(g^\perp)^t}{V^2}\\
&\frac{\Det(g_r,f)g(f^\perp)^t-\Det(g,f)f_r(g^\perp)^t}{V^2} \\
&+ \frac{-\Det(g,f_r)f(g^\perp)^t+\frac{\Det(f_r,f)}{r^2}f(f^\perp)^t}{V^2}\\
&+ \frac{2\Det(g,f)f(g^\perp)^t}{V^2r}\\
&-\frac{\Det(g_r,f)+\Det(g,f_r)}{\Det (g,f)}\id
\end{aligned}
\end{equation}
\end{lemma}
\begin{proof}
From formula \ref{Dxi} it is easy to see that
 $$
 D\xi D\xi^t=gg^t+\frac{ff^t}{r^2}
 $$
 and
 $$
 (D\xi D\xi^t)^{-1}=\frac{r^2}{V^2}\left(g^\perp(g^\perp)^t+\frac{f^\perp(f^\perp)^t}{r^2}\right)
 $$
where $f^\perp=(f_b,-f_a)$ and $g^\perp=(g_b,-g_a)$.  Also
$$
{\partial}_H (D\xi D\xi^t)=g_Hg^t+gg_H^t+\frac{f_Hf^t}{r^2}+\frac{ff_H^t}{r^2}.
$$
By noting that for any vectors $u$ and $v$,
$$
u^t v^{\perp}=u\cdot v^\perp =\Det(u,v)
$$
this gives an expression for ${\partial}_H (D\xi D\xi^t)(D\xi D\xi^t)^{-1}$.

We also have that $V=r\Det D\xi=\det(g,f)$ and $V_H=\Det(g_H,f)+\Det(g,f_H)$. Now
$$
M_H={\partial}_H (D\xi D\xi^t)(D\xi D\xi^t)^{-1}-\frac{{\partial}_H V}{V}\id.
$$
Substituting above we see that $M_H$ is given by
\begin{equation}
\begin{aligned}
& \frac{\Det(g,f)g_H(f^\perp)^t+r^2\Det(g_H,g)g(g^\perp)^t}{V^2}\\
+&\frac{\Det(g_H,f)g(f^\perp)^t-\Det(g,f)f_H(g^\perp)^t}{V^2} \\
+& \frac{-\Det(g,f_H)f(g^\perp)^t+\frac{\Det(f_H,f)}{r^2}f(f^\perp)^t}{V^2}\\
-&\frac{\Det(g_H,f)+\Det(g,f_H)}{\Det (g,f)}\id
\end{aligned}
\end{equation}
as claimed. We can do a similar calculation for $M_r$.
\end{proof}
It is easy to see from the above formulas that $M_H$ is actually smooth up to the boundary of $\mathbb{H}$ except at the points $(-a_i, 0)$. The matrix-valued function $M_r$ is not. To deal with this fact we will often work with $rM_r$ which is smooth.

We start by working out an example. This is meant to help the reader follow the general calculations later on but also, it will turn out to be important in the proof of our main result.
\begin{exa}\label{example2edges}
Let $X(\nu_1,\nu_d)$ be the toric orbifold whose polytope $P_0$ has exactly two edges which are unbounded edges with normals $\nu_1$ and $\nu_d$. Endow $X(\nu_1,\nu_d)$ with the ALE metric corresponding to the parameter $\nu=0$ from \cite{as}.  Then  we will show that
$$
\int_{X(\nu_1,\nu_d)}p=0,
$$
what is more
$$
\lim_{R \rightarrow +\infty}\int_{L_R}T=0, \, \lim_{R \rightarrow +\infty}\int_{C_R}T=0.
$$

Let $f$, $h$, $M_H$ and $M_r$ be defined by the expressions \ref {f}, \ref{g}, \ref{MH} and \ref{Mr} respectively. The formulas for $f$ and $g$ simplify in this context to give
$$
f=\nu_1+\frac{1}{2}\left(1-\frac{H}{\rho}\right)(\nu_{d}-\nu_{1})
$$
and
$$
g=\frac{(\nu_{d}-\nu_{1})}{2\rho}.
$$
 We have
$$
df=\frac{r(-rdH+Hdr)}{2\rho^3}(\nu_{d}-\nu_{1})=\frac{r(-rdH+Hdr)}{\rho^2}g,
$$
and
$$
dg=-\frac{HdH+rdr}{2\rho^3}(\nu_{d}-\nu_{1})=-\frac{HdH+rdr}{\rho^2}g.
$$
Or to write things in another way
$$
f_H=-\frac{r^2}{\rho^2}g, \, f_r=\frac{rH}{\rho^2}g, \, g_H=-\frac{H}{\rho^2}g, \, g_r=-\frac{r}{\rho^2}g.
$$
We have $$\Det(df,g),\Det(f_H,g),\Det(f_r,g)=0,$$ and also $$\Det(dg,g),\Det(g_H,g),\Det(g_r,g)=0.$$ Further
$$
V=\frac{\Det(\nu_d,\nu_1)}{2\rho}.
$$
Formula \ref{T} for $T$ simplifies to yield
\begin{equation}
\begin{aligned}
T=&\frac{\Tr (M_H^2) \Det(df,f)-\Tr (rM_rM_H) \det(dg,f)}{V^2}\\&
+\frac{\Tr (rM_rdM_H-M_Hd(rM_r))}{V}.
\end{aligned}
\end{equation}
Now
$$
\Det(df,f)=\frac{r(-rdH+Hdr)}{\rho^2}V,
$$
and
$$
\Det(dg,f)=-\frac{HdH+rdr}{\rho^2}V
$$
so that we can substitute in the above formula for $T$ to get
\begin{equation}\label{T0}
\begin{aligned}
T=&\frac{\Tr (M_H^2) r(-rdH+Hdr)+\Tr (rM_rM_H)(HdH+rdr) }{\rho^2V}\\
&\frac{d\Tr(M_HrM_r)}{V}-\frac{2\Tr (M_Hd(rM_r))}{V}
\end{aligned}
\end{equation}
or
\begin{equation}
\begin{aligned}
T=&\frac{\Tr (M_H^2) r(-rdH+Hdr)+\Tr (rM_rM_H)(HdH+rdr) }{\rho^2V}\\
&\frac{2\Tr(d(M_H)rM_r}{V}-\frac{\Tr d(M_HrM_r))}{V}.
\end{aligned}
\end{equation}
We have
$$
g_H(f^\perp)^t=\frac{-Hg (f^\perp)^t}{\rho^2},\, \, f_H(g^\perp)^t=\frac{-r^2g (g^\perp)^t}{\rho^2}, \, ,
$$
and
$$
\Det(g_H,f)=\frac{-HV}{\rho^2},\,\Det(f_H,f)=\frac{-r^2V}{\rho^2}
$$
The expression in equation \ref{MH}  simplifies accordingly to give
$$
\begin{aligned}
M_H=& \frac{-2Hg (f^\perp)^t+r^2g (g^\perp)^t-f(f^\perp)^t}{\rho^2V} +\frac{H}{\rho^2}\id,
\end{aligned}
$$
Similarly we have
$$
g_r(f^\perp)^t=\frac{-rg (f^\perp)^t}{\rho^2},\, \, f_r(g^\perp)^t=\frac{rHg (g^\perp)^t}{\rho^2}, \, ,
$$
and
$$
\Det(g_r,f)=\frac{-rV}{\rho^2},\,\Det(f_r,f)=\frac{rHV}{\rho^2}
$$
replacing in equation \ref{Mr} the expression for $M_r$ becomes
$$
\begin{aligned}
M_r=&-\frac{ 2rg (f^\perp)^t+rHg (g^\perp)^t}{\rho^2V}+ \frac{Hf(f^\perp)^t}{r\rho^2V}+ \frac{2f(g^\perp)^t}{Vr}+\frac{r}{\rho^2}\id
\end{aligned}
$$
so that
$$
rM_r=-\frac{ 2r^2g (f^\perp)^t+r^2Hg (g^\perp)^t-Hf(f^\perp)^t}{\rho^2V}+  \frac{2f(g^\perp)^t}{V}+\frac{r^2}{\rho^2}\id
$$
There are two separate calculations to do in order to calculate $\int_{{\partial} B(R)} T$ namely $\int_{C_R} T$ and $\int_{L_R} T$. We start with the later.

Assume that $r=0$. Replacing in equation \ref{T0} we see that
\begin{equation}\label{T0r=0}
T=\frac{\Tr (rM_rM_H)HdH}{\rho^2V}+\frac{d\Tr(M_HrM_r)}{V}-\frac{2\Tr (M_Hd(rM_r))}{V}.
\end{equation}
We can also simplify the expressions for $M_H$ and $M_r$ further so that we have
$$
M_H=- \frac{2Hg (f^\perp)^t+f(f^\perp)^t}{\rho^2V} +\frac{H}{\rho^2}\id,
$$
and
$$
rM_r=\frac{ Hf(f^\perp)^t}{\rho^2V}+  \frac{2f(g^\perp)^t}{V}.
$$
It follows that
$$
\Tr(M_HrM_r)=-\frac{2H}{\rho^2}
$$
and
$$
d(rM_r)=d\left(\frac{ H}{\rho^2V}\right)f(f^\perp)^t+  2d\left(\frac{1}{V}\right)f(g^\perp)^t-\frac{2Hf (g^\perp)^tdH}{\rho^2V}.
$$
But on $L_R$,
$$
d\left(\frac{1}{V}\right)=-\frac{V_HdH}{V^2}=-\frac{\Det(g_H,f)dH}{V^2}=\frac{HVdH}{\rho^2V^2}=\frac{HdH}{\rho^2V}
$$
so that
$$
 2d\left(\frac{1}{V}\right)f(g^\perp)^t-\frac{2Hf (g^\perp)^tdH}{\rho^2V}=0.
$$
and
$$
d(rM_r)=d\left(\frac{ H}{\rho^2V}\right)f(f^\perp)^t.
$$
Therefore we have that
$$
M_Hd(rM_r)=\frac{H}{\rho^2}d\left(\frac{ H}{\rho^2V}\right)f(f^\perp)^t
$$
and $M_Hd(rM_r)$ has zero trace. We can now return to equation \ref{T0r=0}. We have
$$
T=-\frac{2H^2dH}{\rho^4V}-\frac{2}{V}d\left(\frac{H}{\rho^2}\right)
$$
i.e.
$$
T=-\frac{2H^2dH}{\rho^4V}-\frac{2dH}{V\rho^2}+\frac{4H^2dH}{V\rho^4}
$$
which is zero when $r=0$. Hence $\int_{L_R}T$ is actually zero in this example.

Let us now turn our attention to $\int_{C_R}T$. Note that on $C_R$ we have $HdH+rdr=0$ and $dg=0$. Substituting in equation \ref{T0} we see that
\begin{equation}\label{T0R}
\begin{aligned}
T=&\frac{\Tr (M_H^2) r(-rdH+Hdr)}{\rho^2V}+\frac{2\Tr(d(M_H)rM_r)}{V}-\frac{\Tr d(M_HrM_r)}{V}.
\end{aligned}
\end{equation}
From the expression
$$
M_H= \frac{-2Hg (f^\perp)^t+r^2g (g^\perp)^t-f(f^\perp)^t}{\rho^2V} +\frac{H}{\rho^2}\id,
$$
we se that
$$
\Tr M_H^2=\frac{2(\rho^2+r^2)}{\rho^4}=\frac{2}{\rho^2}
$$
and also that
$$
dM_H= -\frac{2dHg (f^\perp)^t+2Hg(df^\perp)^t-2rdrg (g^\perp)^t+df(f^\perp)^t+f(df^\perp)^t}{\rho^2V} +\frac{dH}{\rho^2}\id,
$$
where we have used the fact on $C_R$, $\rho$ and therefore $V$ are constants. By using the fact that
$$
df=\frac{r}{\rho^2}(-rdH+Hdr)g
$$
it follows that
$$
\begin{aligned}
dM_H&= -\left(2\rho^2dH+{r}(-rdH+Hdr)\right)\frac{g(f^\perp)^t}{\rho^4V}-{r}(-rdH+Hdr)\frac{f(g^\perp)^t}{\rho^4V} \\
&-\left({2Hr}(-rdH+Hdr)-2r\rho^2dr \right)\frac{g(g^\perp)^t}{\rho^4V}+\frac{dH}{\rho^2}\id,
\end{aligned}
$$
We also have that
$$
rM_r=-\frac{ 2r^2g (f^\perp)^t+r^2Hg (g^\perp)^t-Hf(f^\perp)^t}{\rho^2V}+  \frac{2f(g^\perp)^t}{V}+\frac{r^2}{\rho^2}\id
$$
and by multiplying out we conclude that
$$
\Tr(rM_rdM_H)=-\frac{2Hd\rho}{\rho^3}
$$
and that
$$
\Tr(rM_rM_H)=-\frac{2H}{\rho^2}
$$
so that
$$
T=\frac{2}{\rho^2V}\left(\frac{ r(-rdH+Hdr)}{\rho^2V}+dH\right)
$$
which is zero on $C_R$. So, for the ALE metric on $X(\nu_1,\nu_d)$ we have
$$
\int_{X(\nu_1,\nu_d)}p=0.
$$
\end{exa}

We are now in a position to prove the main result of this section
\begin{prop}\label{finitep}
Let $X$ be a strictly unbounded toric surface with a scalar-flat toric K\"ahler metric from \cite{as} with parameter $\nu$. Then
\[
\int_X p
\]
is finite. In fact if we set $a_0=-\infty$ and $a_d=+\infty$,
$$
\int_X p= \sum_{i=1}^d\left[\frac{2V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}+4\int_{r=0}\frac{\Det(g_H,g)dH}{V^2}.
 $$
Here $V=\Det(g,f)$ where $g$ and $f$ are given by the formulas \ref{g} and \ref{f}.
\end{prop}
\begin{proof}
To do this it is enough to show that
$$
\lim_{R\rightarrow \infty} \int_{{\partial} B(R)} T
$$
is finite. As in the example there are two terms in the above integral which we treat separately. First we will study
$$
\int_{C_R} T .
$$
 To this end we study the asymptotic behavior of $M_H$ and $M_r$.

 Consider first the case when $\nu\ne 0$. It follows from the formulas \ref{f} and \ref{g} that $f$  and $g$ are bounded, $df$ is $O(-1)$ and $dg$ is $O(-2)$. When $\nu\ne 0$ this gives us the asymptotic behavior of $M_H$. Namely
$$
M_H=O_H(0)+O_H(-1)+O_H(-2)
$$
where $O_H(0),O_H(-1)$ and $O_H(-2)$ are $O(0)$, $O(-1)$ and $O(-2)$ respectively. They are given explicitly by
$$
O_H(0)=\frac{r^2\Det(g_H,g)}{V^2}g(g^\perp)^t;
$$
$$
O_H(-1)=-\frac{1}{V^2}\left( \Det(g,f)f_H(g^\perp)^t+\Det(g,f_H)f(g^\perp)^t \right)-\frac{\Det(g,f_H)}{\Det (g,f)}\id;
$$
$$
\begin{aligned}
O_H(-2)=\frac{1}{V^2}&\left(\Det(g,f)g_H(f^\perp)^t+\Det(g_H,f)g(f^\perp)^t+\frac{\Det(f_H,f)}{r^2}f(f^\perp)^t \right)\\
&-\frac{\Det(g_H,f)}{\Det (g,f)}\id.
\end{aligned}
$$
respectively. From the formula  \ref{Mr} we write
$$
M_r=O_r(0)+O_r(-1)+O_r(-2)
$$
where $O_r(0),O_r(-1)$ and $O_r(-2)$ are $O(0)$, $O(-1)$ and $O(-2)$ respectively. They are given explicitly by
$$
O_r(0)=\frac{r^2\Det(g_r,g)}{V^2}g(g^\perp)^t;
$$
$$
\begin{aligned}
O_r(-1)=-\frac{1}{V^2}&\left( \Det(g,f)f_r(g^\perp)^t+\Det(g,f_r)f(g^\perp)^t-\frac{2\Det(g,f)}{r}f(g^\perp)^t\right)\\
&-\frac{\Det(g,f_r)}{\Det (g,f)}\id;
\end{aligned}
$$
$$
\begin{aligned}
O_r(-2)=\frac{1}{V^2}&\left(\Det(g,f)g_r(f^\perp)^t+  \Det(g_r,f)g(f^\perp)^t+\frac{\Det(f_r,f)}{r^2}f(f^\perp)^t\right)\\
&-\frac{\Det(g_r,f)}{\Det (g,f)}\id
\end{aligned}
$$
respectively. From these calculation it follows that $\Tr (M_H^2)$ is bounded. But in fact:
\begin{itemize}
\item its highest order term is the trace of $O_H(0)^2$ which is zero because $$g(g^\perp)^tg(g^\perp)^t=0,$$
\item as for the term in $O(-1)$ of this trace it is given by the trace of $$2O_H(0)O_H(-1)$$ which is zero as well because the matrices
$$g(g^\perp)^tf_H(g^\perp)^t,\, g(g^\perp)^tf(g^\perp)^t,\,  g(g^\perp)$$
all have zero trace. In fact they are all multiples of $g(g^\perp)$.
\end{itemize}
From these two observations we conclude that $\Tr (M_H)^2$ is in fact $O(-2)$. Similarly $\Tr M_r^2=O(-2)$ and $\Tr (M_HM_r)=O(-2)$. It follows that
\begin{itemize}
\item $\Tr(M_H^2)\Det(df,f)$ is $O(-3)$,
\item $\Tr (r^2 M_r^2)\Det(dg,g)$ is $O(-2)$
\item $\Tr (rM_rM_H)(\det(dg,f)+\det (df,g))$ is $O(-2)$.
\end{itemize}
We can conclude that the integral of the sum of the above expressions on $C_R$ tends to zero as $R$ tends to $\infty$. We are left with studying
$$
\int_{C_R}\frac{\Tr(M_Hd(rM_r)-rM_rdM_H)}{V}.
$$
Let us start by considering $\Tr (M_rdM_H)$. This is in principle $O(-1)$. But again
\begin{itemize}
\item its highest order term which is $O_r(0)dO_H(0)$ has zero trace as it is given as a linear combination of the traces of the matrices $g(g^\perp)^tg(g^\perp)^t$, $g(g^\perp)^tdg(g^\perp)^t$ or $g(g^\perp)^tg(dg^\perp)^t$ all of which vanish because $g(g^\perp)^t$ has zero trace or $(g^\perp)^tg=0$,
\item the next term, namely $O_r(0)dO_H(-1)+O_r(-1)dO_H(0)$ is a priori of order $O(-2)$, but in fact by carrying out explicit calculations one sees that this term reduces to
$$
\begin{aligned}
&\frac{2r^2\Det(g,dg)}{V^3} \left( \Det(g_r,g)\Det(f_H,g)+\Det(f_r,g)\Det(g_H,g)\right)\\
&-\frac{2r^2\Det(g,dg)\Det(g_H,g)}{V^2r}
\end{aligned}
$$
which is $O(-3)$.
\end{itemize}
From this we conclude that $\Tr (rM_rdM_H)$ is $O(-2)$. Now as we have seen
$$
\Tr (M_HM_r)=O(-2),
$$
which implies that
$$
\Tr (M_HrM_r)=O(-1), \, \Tr (d M_HrM_r)=O(-2)
$$
therefore
$$
\Tr (d(rM_r)M_H)=d\Tr (M_HrM_r)-\Tr (rM_rdM_H)=O(-2).
$$
Because $V$ is bounded from below it follows that
$$
 \int_{ C_R} T
 $$
 has limit zero as $R$ tends to $+\infty$.

 Next we consider the case when $\nu=0$. From formulas \ref{f} and \ref{g} we see that in this case $f$ is bounded but $g=O(-1)$. On the other hand $V$ is $O(-1)$. Formulas \ref{MH} and \ref{Mr} still hold but now they imply a different asymptotic behavior for $M_H$ and $M_r$. Namely each of the terms in the sum is now $O(-1)$. Therefore $M_H=O(-1)$ and $rM_r$ is bounded. It follows that
 \begin{itemize}
\item $\Tr(M_H^2)\Det(df,f)$ is $O(-3)$,
\item $\Tr (r^2 M_r^2)\Det(dg,g)$ is $O(-3)$,
\item $\Tr (rM_rM_H)(\det(dg,f)+\det (df,g))$ is $O(-3)$,
\item $\Tr (rM_rdM_H-M_Hd(rM_r))\Det(g,f)$ is $O(-3)$.
\end{itemize}
Each of the terms will be multiplied by $1/V^2$ which is of order $\rho^2$ so that we see that the integral over $C_r$ is actually $O(-1)$ and therefore in this case there might be a contribution towards $\int_X p$.

Consider the polytope $P_0$ with only two unbounded edges with normals $\nu_1$ and $\nu_d$. This is the polytope of the toric orbifold $X(\nu_1,\nu_d)$ from example \ref{example2edges} for which the construction of \cite{as} also works. We can associate to it functions $f_0$ and $g_0$ via the formulas
$$
f_0=\nu_1+\frac{1}{2}\left(1-\frac{H}{\rho}\right)(\nu_{d}-\nu_{1})
$$
and
$$
g_0=\frac{(\nu_{d}-\nu_{1})}{2\rho}.
$$
These are as in \ref{f} and \ref{g} respectively for the case when there are only two normals and $\nu=0$. Let $M_H$ and $M_r$ be defined by $f_0$ and $g_0$ as in equations \ref{MH} and \ref{Mr} respectively and $T_0$ be given by equation \ref{T} accordingly. The point is that
$$
f-f_0=O(-1), \, g-g_0=O(-2)
$$
and this implies that $T$ differs from $T_0$ by a term in $O(-2)$. Thus
$$
\lim_{R \rightarrow 0} \int_{C_R} T-T_0=0
$$
We know from example \ref{example2edges} that
$$
\lim_{R \rightarrow 0} \int_{C_R} T_0=0,
$$
therefore
$$
\lim_{R \rightarrow 0} \int_{C_R} T=0.
$$

We will now analyze the term
 $$
\int_{L_R} T .
$$
When $r=0$ we have that $f$ is constant since
$$
\left(1-\frac{H_i}{\rho_i}\right)
$$
is locally constant for all $i=1,\dots d-1$. It is either equal to $0$ or $2$ so that $f_r,f_H,df$ are all zero on $L_R$. Now consider equation \ref{MH} for $M_H$. This simplifies slightly when $r=0$ to give
 \begin{equation}\label{MH0}
M_H= \frac{g_H(f^\perp)^t}{V}+\frac{\Det(g_H,f)g(f^\perp)^t}{V^2}+ \frac{{\Det(f_H,f)}f(f^\perp)^t}{r^2V^2}-\frac{\Det(g_H,f)}{V}\id\\
\end{equation}
We can also see that $g_r=0$ on $L_R$ and equation \ref{Mr} for $M_r$ also simplifies. We get
$$
rM_r= \frac{{\Det(f_r,f)}f(f^\perp)^t}{rV^2}+ \frac{2f(g^\perp)^t}{V}.
$$
But we can easily see from formulas \ref{f} and \ref{g}  that
$$
\frac{f_r}{r}=\frac{1}{2}\sum_{i=1}^{d-1} \frac{H_i}{\rho_i^3}(\nu_{i+1}-\nu_i)=-g_H
$$
so the above becomes
 \begin{equation}\label{Mr0}
rM_r=- \frac{{\Det(g_H,f)}f(f^\perp)^t}{V^2}+ \frac{2f(g^\perp)^t}{V}.
\end{equation}
Because $(f^\perp)^tf=0$, it follows from these two formulas that
$$
M_HrM_r=\frac{\Det(g_H,f)}{V}\left( \frac{{\Det(g_H,f)}f(f^\perp)^t}{V^2}- \frac{2f(g^\perp)^t}{V}\right)
$$
and thus
\begin{equation}\label{trMHMr}
\Tr(M_HrM_r)=\frac{2\Det(g_H,f)}{V},
\end{equation}
implying that on $L_R$
$$
d\Tr(M_HrM_r)=2\left(\frac{\Det(g_{HH},f)}{V}-\frac{\Det(g_{H},f)^2}{V^2}\right).
$$
One can also check using equation \ref{Mr0} that $\Tr (r^2M_r^2)=4$. Equation \ref{T} for $T$ in turn also simplifies and we see that $T$ is given by
$$
\begin{aligned}
&\frac{-\Tr (rM_rM_H) \det(dg,f)+\Tr (r^2M_r^2) \Det(dg,g)}{V^2}\\
&+\frac{d\Tr (rM_rM_H)}{V}-\frac{2\Tr (M_Hd(rM_r))}{V}
\end{aligned}
 $$
hence on $L_R$ we have
$$
\begin{aligned}
T=\frac{-2 \det(g_H,f)^2dH}{V^3}+\frac{4\Det(g_H,g)dH}{V^2}+\frac{d\Tr (rM_rM_H)}{V}-\frac{2\Tr (M_Hd(rM_r))}{V}.
\end{aligned}
 $$
Next we calculate $\Tr (M_Hd(rM_r))$. It follows from equation \ref{Mr0} that on $L_R$
$$
\frac{d(rM_r)}{dH}=-\frac{{\partial}}{{\partial} H}\left( \frac{{\Det(g_H,f)}}{V^2}\right) f(f^\perp)^t+ \frac{2f(g_H^\perp)^t}{V}-\frac{2\Det(g_H,f)f(g^\perp)^t}{V^2}.
$$
 and $\Tr(M_Hd(rM_r))$ is given by
 $$
 -\frac{\Det(g_H,f)}{V}\left(  \frac{2\Det(f,g_H)}{V}-\frac{2\Det(g_H,f)\Det(f,g)}{V^2}\right)=0
 $$
so that
$$
\begin{aligned}
T=\frac{2dH}{V^2}\left(\frac{-2 \det(g_H,f)^2}{V}+{2\Det(g_H,g)}{}+{\Det(g_{HH},f)}{}\right).
\end{aligned}
 $$
 or
 $$
\begin{aligned}
T=\frac{4\Det(g_H,g)dH}{V^2}+2\left(\frac{V_{HH}}{V^2}-\frac{2 V_H^2}{V^3}\right)dH.
\end{aligned}
 $$
Integrating
$$
\frac{-2 V_H^2}{V^3}
$$
by parts we see that
$$
\begin{aligned}
\int_a^b T_{|L_R}=\left[\frac{2V_H}{V^2}\right]_a^b+4\int_a^b\frac{\Det(g_H,g)dH}{V^2}.
\end{aligned}
 $$
This proves the formula in the statement of the proposition. To show that this formula implies finiteness of $\int_X p$ note that
$$
\begin{aligned}
V=&\det(\nu,\nu_1)+\sum_{i=1}^{d-1}\frac{\Det(\nu_{i+1}-\nu_i,\nu_1)}{2\rho_i}+\sum_{i=1}^{d-1}\frac{1}{2}\left(1-\frac{H_i}{\rho_i}\right)\Det(\nu,\nu_{i+1}-\nu_i)\\
&+\sum_{i,j=1}^{d-1}\frac{1}{4\rho_i}\left(1-\frac{H_j}{\rho_j}
\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j).
\end{aligned}
$$
Since $V_H=\Det(g_H,f)$ on $L_R$
$$
\begin{aligned}
V_H=&-\sum_{i=1}^{d-1}\frac{H_i\Det(\nu_{i+1}-\nu_i,\nu_1)}{2\rho_i^3}\\
&-\sum_{i,j=1}^{d-1}\frac{H_i}{4\rho_i^3}\left(1-\frac{H_j}{\rho_j}
\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j).
\end{aligned}
$$
We can also write down a formula for $\Det(g_H,g)$.
$$
\begin{aligned}
\Det(g_H,g)=&\sum_{i=1}^{d-1}\frac{H_i\Det(\nu,\nu_{i+1}-\nu_i)}{2\rho_i^3}\\
-&\sum_{i,j=1}^{d-1}\frac{H_i}{4\rho_i^3\rho_j}\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j).
\end{aligned}
$$
When $H$ is close to $-a_i$ we have that
$$
V= \frac{C_1}{|H_i|}+O(0),\quad V_H=\frac{C_2H_i}{|H_i|^3}+O(0), \quad \Det(g_H,g)=\frac{C_3H_i}{|H_i|^3}+O(0)
$$
where $O(0)$ denotes a bounded function of $H$ in a neighborhood of $-a_i$ and $C_1,C_2$ and $C_3$ are constants. This then implies that
$$
\frac{\Det(g_H,g)}{V^2}, \frac{V_H}{V^2}
$$
are bounded which in turn implies that $T$ is bounded close to $-a_i$. Similarly when $H$ tends to infinity it is easy to see that $T$ is $O\left(\frac{1}{H}\right)$ (i.e. $T$ is bounded by $C/H$ for some constant $C$ on $r=0$) both when $\nu\ne 0$ and $\nu=0$. We can thus conclude that
$$
 \int_{L_R} T
 $$
is bounded. This concludes the proof.
 \end{proof}
Before we end this section we sum up our results in the following proposition
\begin{prop}
Let $X$ be a strictly unbounded toric surface with a scalar-flat toric K\"ahler metric from \cite{as} with parameter $\nu$. Then
$$
\begin{aligned}
\int_X |\mathcal{R}|^2= \sum_{i=1}^d\left[\frac{2V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}+3\int_{\mathbb{R}}\frac{G}{V^2}
\end{aligned}
$$
where $a_0=-\infty$ and $a_d=+\infty$, with
$$
\begin{aligned}
V=&\Det(\nu,\nu_1)+\frac{1}{2}\sum\Det(\left(1-\frac{H_i}{|H_i|}\right)\nu-\frac{\nu_1}{2|H_i|},\nu_{i+1}-\nu_{i})\\
&+\frac{1}{4}\frac{1}{|H_i|}\left(1-\frac{H_j}{|H_j|}\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j),
\end{aligned}
$$
$$
V_H=\frac{1}{2}\sum\frac{H_i}{|H_i|^3}\Det(\nu_1,\nu_{i+1}-\nu_i)+\frac{1}{4}\sum\frac{H_j}{|H_j|^3}\left(1-\frac{H_i}{|H_i|}\right)\Det(\nu_{i+1}-\nu_i,\nu_{j+1}-\nu_j)
$$
and
$$
G=\frac{1}{2}\sum\frac{H_i}{|H_i|^3}\Det(\nu,\nu_{i+1}-\nu_i)+\frac{1}{4}\sum\frac{H_i}{|H_i|^3|H_j|}\Det(\nu_{j+1}-\nu_j,\nu_{i+1}-\nu_i)
$$
\end{prop}
\section{Explicit calculations}\label{example}

In this sections we use the above results to calculate $\int_X|\mathcal{R}|^2$ for an important case namely that of minimal resolutions of cyclic singularities endowed with scalar-flat metrics. In \cite{as} this case is also treated in more detail.

Gravitational instantons are complete hyperk\"ahler metrics on non-compact manifolds. These are always K\"ahler and Ricci-flat. Some of the metrics constructed in \cite{as} turn out to be Ricci-flat (but note that some are not). These corresponds to special choices of the parameter $\nu$ indexing the metrics.

In a recent preprint Atiyah and LeBrun derive a beautiful Gauss-Bonnet formula for metrics with edge cone singularities. As an application of their formula they give an explicit expression for the Calabi functional of gravitational instantons on spaces which are of the form $\mathbb{C}^2/\Gamma$, where $\Gamma$ is a finite group of $SU(2)$, in terms of $\Gamma$. The authors only consider Ricci-flat metrics whereas our calculation applies to a larger family of scalar-flat metrics. On the other hand they do not limit themselves to toric metrics as we do here. Also the methods appearing in \cite{al} are likely to have other applications.

Let $\Gamma=\Gamma_d$ is the finite subgroup of $SU(2)$, of order
$d-1\in\mathbb{N}$, generated by
\begin{displaymath}
\begin{pmatrix}
  e^{\frac{2i\pi}{d-1}}&0\\
 0& e^{\frac{2i\pi(d-2)}{d-1}}
\end{pmatrix}.
\end{displaymath}
Let $X$ be the minimal toric resolution of $\mathbb{C}^2/\Gamma$. Its moment polygon is
$SL(2,\mathbb{Z})$ equivalent to one with normals $\nu_1=(0,1)$, $\nu_2=(1,0)$, \dots,
$\nu_{d}=(d-1,-(d-2))$. Endow $X$ with a scalar-flat K\"ahler toric metric with parameter $\nu$ as constructed in \cite{as}. Such a metric exits and is complete as long as $\Det(\nu,\nu_1),\Det(\nu,\nu_d)>0$. This implies that
$$
\alpha>0, \quad \alpha-(\alpha+\beta)(d-1)>0.
$$

We start by explicitly calculating $\int_X c_1^2$ using formula  \ref{c1LR}  in this case. Namely we prove the following lemma
\begin{lemma}
Let $X$ be the minimal toric resolution of a cyclic singularity of order $d-1$. Consider a scalar-flat metric from \cite{as} corresponding to an admissible parameter $\nu=(\alpha,\beta)$ i.e.
$$
\alpha>0, \quad \alpha-(\alpha+\beta)(d-1)>0.
$$
Then
$$
\int_X c_1^2=(\alpha+\beta)\left(\frac{1}{\alpha}-\frac{1}{\alpha-(\alpha+\beta)(d-1)}\right)
$$
\end{lemma}
\begin{proof}
 As we have mentioned before, $X$ is the minimal toric resolution of $\mathbb{C}^2/\Gamma$. Its moment polygon is
$SL(2,\mathbb{Z})$ equivalent to one with normals $\nu_1=(0,1)$, $\nu_2=(1,0)$, \dots,
$\nu_{d}=(d-1,-(d-2))$. Endow $X$ with one of the scalar-flat metrics from \cite{as}. For all $i=1,\cdots d-1$, $\nu_{i+1}-\nu_i=(1,-1)$ and the form $\Det(dg,g)/\Det(f,g)^2$ simplifies a great deal as a consequence. More precisely it follows from the expression \ref{c1LR} for
$$
\frac{\Det(dg,g)}{\Det(f,g)^2}
$$
that we have
$$
\frac{\Det(dg,g)}{\Det(f,g)^2}=-\frac{\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}\frac{\sg{H_i}}{H_i^2} dH
}{
\left( \alpha-\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}(1-\frac{H_i}{|H_i|})+\sum_{i=1}^{d-1}\frac{\sg{H_i}}{2H_i} \right)^2},
$$
where $\sg{H_i}$ is $1$ if $H_i\geq 0$ and $-1$ otherwise. Now
$$
\sum_{i=1}^{d-1}\left(1-\frac{H_i}{|H_i|}\right)
$$
is constant on the intervals $]-a_{i+1},-a_i[$  as well as on the intervals $]-\infty,-a_{d-1}[$ and $]-a_1,+\infty[$ so that
$$
-\sum\frac{\sg{H_i}}{2H_i^2}
$$
is the derivative of
$$
V=\alpha-\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{|H_i|}\right)+\sum_{i=1}^{d-1}\frac{\sg{H_i}}{2H_i}
$$
and therefore
$$
\frac{\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}\frac{\sg{H_i}}{H_i^2}
}{
\left( \alpha-\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}(1-\frac{H_i}{|H_i|})+\sum_{i=1}^{d-1}\frac{\sg{H_i}}{2H_i} \right)^2}=-(\alpha+\beta)\frac{V_H}{V^2}.
$$
As we have seen before
$$
\int_X c_1^2=\lim_{R\rightarrow \infty}\int_{C_R}\frac{\Det(dg,g)}{\Det(g,f)^2}+\lim_{R\rightarrow \infty}\int_{L_R}\frac{\Det(dg,g)}{\Det(g,f)^2}
$$
and the first limit in the sum is zero. As for the second limit it is
$$
\int_\mathbb{R} (\alpha+\beta)\frac{V_H}{V^2}=-(\alpha+\beta)\sum_{i=1}^d\left[\frac{1}{V}\right]_{-a_i}^{-a_{i-1}},
$$
where we set $a_0=-\infty$ and $a_d=+\infty$. We have
$$
V=\alpha-\frac{\alpha+\beta}{2}\sum_{i=1}^{d-1}\left(1-\frac{H_i}{|H_i|}\right)+\sum_{i=1}^{d-1}\frac{\sg{H_i}}{2H_i}
$$
and near $-a_i$
$$
V=\frac{\sg{H_i}}{2H_i}+O(0)
$$
with $O(0)$ being a bounded function near $-a_i$. This shows that $1/V$ vanishes at  $-a_i$ for $i=1,\cdots d-1$. The limit of $V$ at $+\infty$ is $\alpha$ because
$$
H>>0 \implies 1-\frac{H_i}{|H_i|}=0, \,\forall i=1,\cdots d-1.
$$
Similarly
$$
H<<0 \implies 1-\frac{H_i}{|H_i|}=2, \,\forall i=1,\cdots d-1.
$$
so that the limit of $V$ at $-\infty$ is $\alpha-(\alpha+\beta)(d-1)$ and
$$
\int_X c_1^2=(\alpha+\beta)\left(\frac{1}{\alpha-(\alpha+\beta)(d-1)}-\frac{1}{\alpha}\right).
$$
Notice that when $\nu=0$
$$
g=\sum_{i=1}^{d-1}\frac{v}{\rho_i}
$$
where we set $v=(1,-1)$ and thus
$$
dg=\sum_{i=1}^{d-1}d\left(\frac{1}{\rho_i}\right)v
$$
so that $\Det(dg,g)=0$ and $\int_X c_1^2=0$.
\end{proof}
We can use similar calculations to prove the following lemma
\begin{lemma}\label{Ad}
Let $X$ be the minimal toric resolution of a cyclic singularity of order $d-1$. Consider a scalar-flat metric from \cite{as} corresponding to an admissible parameter $\nu=(\alpha,\beta)$ such that
$$
\alpha>0, \quad \alpha-(\alpha+\beta)(d-1)>0.
$$
Then when $\nu=0$
$$
\int_X p=8(d-1)-\frac{8}{d-1}
$$
and when $\nu\ne0$ and $\alpha\ne 0$ then
$$
\int_X p=8(d-1)+4(\alpha+\beta)\left(\frac{1}{\alpha-(\alpha+\beta)(d-1)}-\frac{1}{\alpha}\right)
$$
\end{lemma}
\begin{proof}
From proposition \ref{finitep} we know that
$$
\int_X p= \sum_{i=1}^d\left[\frac{2V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}+4\int_{r=0}\frac{\Det(g_H,g)dH}{V^2}.
$$
In our case as we have seen before
$$
\nu_{i+1}-\nu_i=(1,-1)=v, \forall i=1,\cdots d
$$
thus
$$
g=\nu+\sum_{i=1}^{d-1}\frac{v}{2\rho_i}
$$
and
$$
f=\nu_1+\sum_{i=1}^{d-1}\frac{1}{2}\left(1-\frac{H_i}{\rho_i}\right)v.
$$
We will start by calculating
$$
 \sum_{i=1}^d\left[\frac{V_H}{V^2}\right]_{-a_i}^{-a_{i-1}},
 $$
where $a_0=-\infty$ and $a_d=+\infty$. Now $V=\Det(g,f)$ and in this setting this can be simplified to yield
$$
\Det(g,f)=\Det(\nu,\nu_1)+\sum_{i=1}^{d-1}\frac{\Det(v,\nu_1)}{2\rho_i}+\sum_{i=1}^{d-1}\frac{\Det(\nu,v)}{2}\left(1-\frac{H_i}{\rho_i}\right)
$$
Now $\Det(\nu,\nu_1)=\alpha$ and $\Det(\nu,v)=-(\alpha+\beta)$ so that
$$
\Det(g,f)=\alpha+\sum_{i=1}^{d-1}\frac{1}{2\rho_i}-(\alpha+\beta)\sum_{i=1}^{d-1}\frac{1}{2}\left(1-\frac{H_i}{\rho_i}\right)
$$
On $L_R$ we also have $V_H=\Det(g_H,f)$ because $g_r=0$. Now
$$
g_H=-\sum_{i=1}^{d-1}\frac{H_i}{2\rho_i^3}v
$$
therefore
$$
\Det(g_H,f)=-\sum_{i=1}^{d-1}\frac{H_i}{2\rho_i^3}
$$
When $H$ is close to $-a_i$ we have
$$
V=\frac{1}{2|H_i|}+O(0), \quad \Det(g_H,f)=-\frac{H_i}{2|H_i|^3}+O(0),
$$
where $O(0)$ is a bounded function of $H$ defined in a neighborhood of $-a_i$, so that
$$
\lim_{H\rightarrow -a_i}\frac{V_H}{V^2}=-2\sg(H_i).
$$
The behavior of $\frac{V_H}{V^2}$ at infinity depends on weather $\nu=0$ or $\nu\ne 0$. Namely:
\begin{itemize}
\item when $\nu\ne 0$ at $+\infty$ we have
$$
\lim_{H\rightarrow +\infty}V=\alpha, \, V_H=-{}\frac{d-1}{2H^2}+O(0), \, \lim_{H\rightarrow +\infty}\frac{V_H}{V^2}=0;
$$
\item similarly when $\nu\ne 0$ and at $-\infty$
$$
\lim_{H\rightarrow +\infty}V=\alpha-(d-1)(\alpha+\beta), \, V_H={}\frac{d-1}{2H^2}+O(0),\, \lim_{H\rightarrow -\infty}\frac{V_H}{V^2}=0;
$$
\item When $\nu=0$ at $+\infty$
$$
V={}\frac{d-1}{2H}+O(0), \, V_H=-{}\frac{d-1}{2H^2}+O(0), \, \lim_{H\rightarrow \infty}\frac{V_H}{V^2}=-\frac{2}{d-1};
$$
\item Similarly when $\nu=0$ at $-\infty$
$$
V=-{}\frac{d-1}{2H}+O(0), \, V_H={}\frac{d-1}{2H^2}+O(0), \, \lim_{H\rightarrow -\infty}\frac{V_H}{V^2}=\frac{2}{d-1};
$$
\end{itemize}
Putting these together we see that for $i=2,\cdots d-1$
$$
 \left[\frac{V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}=4.
 $$
When $\nu=0$ we also have
$$
 \left[\frac{V_H}{V^2}\right]_{-a_1}^{+\infty}=2-\frac{2}{d-1}, \quad \left[\frac{V_H}{V^2}\right]_{-\infty}^{-a_{d-1}}=2-\frac{2}{d-1},
$$
so that
$$
\sum_{i=1}^d\left[\frac{V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}=4(d-1)-\frac{4}{d-1}.
$$
In the case $\nu\ne 0$ we have instead that
$$
 \left[\frac{V_H}{V^2}\right]_{-a_1}^{+\infty}=2, \quad \left[\frac{V_H}{V^2}\right]_{-\infty}^{-a_{d-1}}=2,
$$
so that
$$
\sum_{i=1}^d\left[\frac{V_H}{V^2}\right]_{-a_i}^{-a_{i-1}}=4(d-1).
$$
On $L_R$ we have
$$
\frac{\Det(dg,g)}{\Det(g,f)^2}=\frac{\Det(g_H,g)}{V^2}
$$
and we have already calculated
$$
\lim_{R\rightarrow +\infty} \int_{L_R}\frac{\Det(dg,g)}{\Det(g,f)^2}
$$
in the proof of the previous lemma. Namely
$$
\lim_{R\rightarrow +\infty}\int_{L_R}\frac{\Det(g_H,g)dH}{V^2}=(\alpha+\beta)\left(\frac{1}{\alpha-(\alpha+\beta)(d-1)}-\frac{1}{\alpha}\right)
$$
\end{proof}
Putting the above two lemmas together  it is very easy to prove the main result in this section.
\begin{prop}
Let $X$ be the minimal toric resolution of a cyclic singularity of order $d-1$. Consider a scalar-flat metric from \cite{as} corresponding to an admissible parameter $\nu=(\alpha,\beta)$ such that
$$
\alpha>0, \quad \alpha-(\alpha+\beta)(d-1)>0.
$$
Then when $\nu=0$
$$
\int_X |\mathcal{R}|^2=8(d-1)-\frac{8}{d-1}.
$$
When $\nu\ne 0$
$$
\int_X |\mathcal{R}|^2=8(d-1)+3(\alpha+\beta)\left(\frac{1}{\alpha-(\alpha+\beta)(d-1)}-\frac{1}{\alpha}\right).
$$
\end{prop}

In the Ricci-flat case, which by \cite{as} corresponds to the case when $\alpha=-\beta$, these calculation agree with those the recent preprint \cite{al}.

redpajama_set_name: RedPajamaArXiv


text:
Q: VALGRIND INTERNAL ERROR: Valgrind received a signal 11 (SIGSEGV) - exiting I am integrating valgrind in AOSP for android marshmellow device.
+ Device Architecture- linux-x86
+ kernel Arch - 3.14.55-x86_64
+ valgrind version - valgrind-3.11.0.SVN.aosp
+ Host Architecture - linux-x86_64 (on this machine I am building image for device)

when I run command "valgrind ls -l" for testing. Its showing following error-:
    root@ET50T:/ # valgrind ls -l
    ==5197== Memcheck, a memory error detector
    ==5197== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al.
    ==5197== Using Valgrind-3.11.0.SVN.aosp and LibVEX; rerun with -h for copyright info
    ==5197== Command: ls -l
    ==5197==
    --5197-- VALGRIND INTERNAL ERROR: Valgrind received a signal 11 (SIGSEGV) - exiting
    --5197-- si_code=80;  Faulting address: 0x0;  sp: 0x82a4bb7c

    valgrind: the 'impossible' happened:
    Killed by fatal signal

    host stacktrace:
    ==5197==    at 0x3816250E: ??? (in /system/lib/valgrind/memcheck-x86-linux)
    ==5197==    by 0x3807A2F3: ??? (in /system/lib/valgrind/memcheck-x86-linux)
    ==5197==    by 0x380B2B12: ??? (in /system/lib/valgrind/memcheck-x86-linux)
    ==5197==    by 0x380C987E: ??? (in /system/lib/valgrind/memcheck-x86-linux)

    sched status:
     running_tid=1

    Thread 1: status = VgTs_Runnable (lwpid 5197)
    ==5197==    at 0x400D4AB: __dl___linker_init (in /system/bin/linker)
    ==5197==    by 0x40115FE: __dl__start (in /system/bin/linker)
    ==5197==    by 0x1: ???

    Note: see also the FAQ in the source distribution.
    It contains workarounds to several common problems.
    In particular, if Valgrind aborted or crashed after
    identifying problems in your program, there's a good chance
    that fixing those problems will prevent Valgrind aborting or
    crashing, especially if it happened in m_mallocfree.c.

    If that doesn't help, please report this bug to: www.valgrind.org

    In the bug report, send all the above text, the valgrind
    version, and what OS and version you are using.  Thanks.

same error is showing for all other tools(massif, hellgrind etc) also.

A: I assume the issue with the integration. Can you try with other version of valgrind, let's say "valgrind-3.10.1"?
Not sure, this might resolve the issue, but just to check the error whatever you had seen has anything to do with valgrind tool version.
I hope the valgrind tool is cross compiled properly, please cross verify by running 'file valgrind' command on the command prompt.

redpajama_set_name: RedPajamaStackExchange


text:
New York: A Love Story is the debut studio album by American R&B singer Mack Wilds. It was released on September 30, 2013, by Louder Than Life and RED Distribution. The album features guest appearances from Method Man, Raekwon and Doug E. Fresh. The album was nominated for Best Urban Contemporary Album at the 56th Grammy Awards, which was held on January 26, 2014.

Background
In July 2013, in an interview with Bossip, Mack Wilds spoke about why he decided to name the album New York: A Love Story, saying: "We had a bunch of different titles. We had a bunch of different names, and thoughts and ideas. But it really just turned into what it sounds like. It sounded like New York and it sounded like a love story, so it really just kind of gave itself that name. We just let everybody know." In the same interview, he spoke about whose idea it was to use classic samples on the album, saying: "Yes. It was a mix of both our ideas and thoughts. Salaam [Remi] came to the table and first it was moreso like a test, he wanted to see what I could do with certain songs. I think the first song I recorded was "Own It". And he just wanted to see what I could do with that, and we did it. Then we just started talking about different beats we wanted to try and it was a real moment, us just trying to figure out different things we could flip around like "Yo, what if we flip this beat like this?" or "Yo, what if we just pull this from this and did that? Yo, that would be crazy." I think one of the first beats that we flipped was "Hennesy" with the Mobb Deep sample. And it just really came together. He played it for Havoc, Havoc loved it, gave him the original beat for it. Havoc twisted it, fixed it up. It was dope. It really just came together from there."

In November 2013, in an interview with Complex, he spoke about if he was surprised by the positive feedback he got from the album, saying: "I was and I think it's mainly because I didn't know what to expect. You do a project, you know it feels good to yourself, you know what it is but to really have it be respected everywhere, by everyone. I really haven't heard anything bad as of yet." He also spoke about how he wanted to represent New York on the album, saying: "We really wanted to just re-represent that iconic sound of New York. I feel like it right now we are in a soundscape  where real musicality is coming back but in hip-hop, there is such a focus on trap rap. We wanted to do something where we fully established what the sound is in New York. Things have changed, years have gone by, but I feel like in certain aspects New York is still the same New York was when hip-hop was becoming hip-hop. New York still has the same heartbeat as it always has." He also spoke about how it was working with producer Salaam Remi on the album, saying: "Amazing. I think the best thing about it was that it wasn't even just a work relationship. Salaam is like my big brother. I've known since I was 18. From 18 to literally this past December, we were just kicking it. He would bring me to the studio and let me hear some stuff. I would let him hear the things I was working on and he would critique it. As I started to grow as an artist, he finally got to a point where he was like, "Come down to Miami, let's see what you can actually do." We went out there for a week and we knocked out like seven records in seven days and four of those records made the album."

He also spoke about remaking Michael Jackson's "Remember the Time", saying: "See, he tricked me with that! I say he tricked me because, I was in the room writing "Henny" and he called me into the other room, he was like, "Yo! Just record 'Remember the Time' over that." I'm like, "Are you serious?" He was like, "Yeah, yeah just record it." I'm thinking we are just playing around because we are in the studio, just having fun. I recorded it, he's listening to it and he's like, "That's good, we're going to keep it." "We are going to keep it? Do what? We're keeping a Michael Jackson's record?" I almost threw up. It was crazy. But it's amazing the way that it turned out especially after rerecording it, touching it up and hearing how the different aspects that he placed in it, like the orchestration or even the harps, it really turned it into a whole another monster."

Singles
On June 24, 2013, the album's first single "Own It" was released. On July 30, 2013, the music video was released for "Own It". On December 16, 2013, the remix to "Henny" featuring Mobb Deep and French Montana, was released. On December 23, 2013, the music video was released for "Henny".

Critical response

Upon its release, New York: A Love Story was met with generally positive reviews from music critics. Andy Kellman gave the album four out of five stars, saying "Tristan Wilds was only four years old when Enter the Wu-Tang was released, but he snared fellow Staten Island native Method Man for the first song of his debut album, in which The Wire and 90210 actor proclaims, "The Shaolin is back again." While Wilds is predominantly a singer who couldn't possibly be termed hardcore, music of the early to mid-'90s—specifically post-new jack R&B, or hip-hop soul—informs much of New York: A Love Story. Wilds even re-envisions Michael Jackson's 1992 single "Remember the Time" by softening and slowing it into something of a 1994-1995 throwback. Other high points include the breezy "My Crib," the harder "Henny" and "Own It," and "U Can Cry to Me," which would lose none of its appeal without the guest appearances from Raekwon and Doug E. Fresh. This is a promising first step from a smooth, thoughtful, and understated voice."

Track listing

Charts

References

2013 debut albums
Albums produced by Salaam Remi
Contemporary R&B albums by American artists
Albums produced by Pete Rock
Albums produced by Ne-Yo
Albums produced by Rico Love
Albums produced by Havoc (musician)
Albums produced by DJ Premier
Albums produced by Bink (record producer)
Albums produced by James Poyser
Albums produced by James Fauntleroy

redpajama_set_name: RedPajamaWikipedia


text:
Хутор Охотники — деревня в составе Большекоровинского сельского поселения Захаровского района Рязанской области. Расположена на правом берегу реки Вожа. Первое упоминание относится к 1862 году.

Своё название деревня получила из-за переселенцев, которых назвали охотниками, согласившихся переселиться в данную местность. Про переселения бытует две версии. По одной из них, после отмены крепостного права две семьи из сельца Рыбкина(по другой версии из д. Надеждино) получили в данной местности наделы земли (хутор), по другой один из владельцев, владения которого располагались севернее реки Вожи, предложил крестьянам, проживавшим на левом берегу реки, переселиться к нему, выделив соответствующие земельные наделы.

В 30-е годы XX века жители деревни создали колхоз «За темпы», просуществовавший до 60-х годов. В колхозе было несколько амбаров, пожарная часть, пасека. В результате реформ Хрущёва по укреплению колхозов, «За темпы» влились в колхоз «Прогресс», который просуществовал до начала 90-х годов XX века и был преобразован в ЗАО «Новый век», просуществовавший до 2003 года.

В настоящее время в деревне насчитывается 30 дворов.

Население

Примечания

Населённые пункты Захаровского района


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text:
Q: Add DCMTK library and compile for Xcode 12.4+ I'm trying to install and add the DCMTK library to my Xcode project.  The library can be found here https://dicom.offis.de/training.php.en
I ran the command to install via home-brew brew install dcmtk but I'm not sure how to add the library and start using it in my project.  I have no idea how to add this to my dependencies in Xcode.  I managed to successfully compile Horus which uses the framework, but it gives a BAD_ACCESS error after running and I'm not sure how they managed to get it to work? It appears they run it as a build target that then runs CMake behind the scenes? I'm not familiar with this style of doing it.  This is the link to the Horos Project that uses the library in the way I need to (https://github.com/horosproject/horos)
I see many questions regarding this with older versions of Xcode that no longer compile with the newest Xcode which is why I'm asking this question here.  Any help is greatly appreciated!


text:
cancer center/cancer a-z list/cancer prevention: what works? article
Cancer Prevention: What Works?
Soy, fish oil, and alcohol are still debated, but some things are certain: Stop smoking, lose weight, and exercise.
By Richard Trubo
Reviewed By Michael Smith
Cancer has surpassed heart disease to become the No. 1 cause of death in the U.S. in people under age 85. No wonder more people than ever are turning their attention to cancer prevention strategies that might protect them from becoming a statistic. But the question lingers, which approaches really work? And which may be little more than hype or wishful thinking?
If you're concerned about cancer prevention, one of your challenges is to sort through the research reported in the media and make sense of findings from highly publicized studies that are often contradictory.
"One week, there might be a study that suggests that high coffee intake causes cancer, and the next week, we'll hear about another study concluding that coffee consumption has no relationship to cancer," says Scott Litin, MD, editor-in-chief of the Mayo Clinic Family Health Book and professor of medicine at Mayo Medical School. "News organizations want to report the latest information, but this doesn't always mean it's the most reliable or the best done study. There is a lot of confusion out there, and in many areas, the medical profession doesn't yet know what's going to pan out and what won't."
In some areas of cancer prevention, carefully conducted and controlled studies just haven't been performed yet. "Many claims have been made, but until the research is done, we just don't know which reduce the risk of cancer, which have no effect, and which are harmful," says Cynthia Stein, MD, MPH, of the Harvard Center for Cancer Prevention at Harvard Medical School.
What's Fact? What's Not?
Take high-fat diets. "Ten years ago, cutting down on fat in the diet to lower cancer risk was one of our major recommendations," says Melanie Polk, MMSc, RD, director of nutrition education at the American Institute for Cancer Research in Washington, D.C. "At the time, the evidence indicated that excessive fat in the diet could promote the cancer process."
But, says Polk, the relative importance of fat restriction in cancer prevention is changing as more studies are done. "The research now supports a mostly plant-based diet emphasizing fruits and vegetables as protective of cancer," she says. "The evidence in this area far outweighs the research into fat. So while we still believe that eating a lower-fat diet is one approach to lowering your cancer risk, even more important is to eat a mostly plant-based diet. At least 20% of cancer cases could be avoided by eating at least five servings of fruits and vegetables a day."
In particular, says Stein, there is evidence that by increasing your consumption of fruits and vegetables, you may be able to lower your risk of cancers of the bladder, esophagus, pancreas, lung, and oral cavity.
Therese Bevers, MD, medical director of clinical cancer prevention at the University of Texas M.D. Anderson Cancer Center in Houston, concurs that a healthy diet is important and may reduce the risk of developing cancer. But, she adds, doctors are awaiting more definitive data on the optimal diet for cancer prevention. "Some studies have shown that reducing your dietary fats can lower your risk of breast cancer, but others have shown no such benefit," she says.
Weighing the Evidence
There is plenty of hype in many areas of cancer prevention, even when the scientific evidence isn't persuasive. For example:
The joy of soy? Women in Asia have lower rates of breast cancer than their counterparts in the Western world -- and they also have a much higher intake of soy. That kind of news has driven some women to take soy to extremes.
"There are people eating soy milk at breakfast, soy burgers at lunch, and tofu at dinner, and snacking on soy nuts throughout the day because they think that soy is going to save their lives," says Polk. "But we just don't have long-term studies to support the intake of these huge amounts of soy. If you enjoy partaking in soy, go ahead, but when it comes to eating it in very large amounts, we really don't know enough to say if it is definitely helpful or harmful."
Looking to the sea. Fish has been touted as a weapon in cancer prevention, but many experts aren't yet convinced that the oils in fish -- called omega-3 fatty acids -- have really turned the tide in human studies. Nevertheless, some animal research suggests that omega-3s may reduce cancer risk. If you want to give fish a try, concentrate on the fatty fish such as salmon, herring, mackerel, and sardines.
Drinking to Your Health? Although you have probably heard that a daily glass or two of wine or other alcoholic beverages may improve your heart health, don't count on that bottle of Bordeaux for cancer prevention. In fact, the opposite may be true.
"As little as one drink per day (one beer, one glass of wine, or one shot of liquor) increases the risk of breast and colon cancer," says Stein. "It also increases the risk of cancers of the mouth, larynx, esophagus, and liver."
So if you do enjoy an alcoholic drink now and then, don't overdo it. "We usually advise limiting your consumption to one alcoholic beverage or less per day on average," says Bevers.
Controversial and Unproven
Although you might read about various links between cancer and certain lifestyle choices, many of these claims haven't impressed the serious scientists. Internet emails, for example, have proclaimed that chemicals in underarm antiperspirants are absorbed through the skin and interfere with normal circulation, causing toxins to build up and trigger breast cancer. But according to the American Cancer Society (ACS), there is "no experimental or epidemiological evidence to support this rumor."
One popular urban legend has implicated underwire bras as a cause of breast cancer, claiming that they obstruct lymph circulation, although the ACS insists that there is no credible research that supports this claim. There is probably no need to give up coffee, either -- at least not now. Caffeine may increase symptoms of fibrocystic breast lumps, but there is no proof that this benign breast disease increases the likelihood of breast cancer.
What Makes Sense?
So what should you do for cancer prevention? "At least 50% of cancers in the U.S. could be prevented just based on what we already know about risk factors," says Stein. Here are some recommendations to take seriously:
Stop smoking or using tobacco products. In 2002, tobacco use caused more than 430,000 deaths from cancer and other serious diseases. Tobacco is a major cause of preventable disease and premature death. It increases the risk for cancer of the lung, esophagus, pancreas, kidney, bladder, and uterine cervix. For cancer prevention, throw out your cigarettes and avoid secondhand or environmental tobacco smoke, too.
Protect your skin in the sun. The majority of skin cancers are caused by excess ultraviolet radiation exposure, most of which comes from the sun. Wear SPF 15 (or higher) sunscreen and a wide-brimmed hat to guard against skin cancer. Avoid direct sunlight during the hours when the sun's ultraviolet rays are the strongest (between 10 a.m. and 4 p.m.). The American cancer Society also recommends avoiding tanning beds and sun lamps.
Lose any excess pounds. About 64% of men and women in the U.S. were overweight or obese in 1999, based on data from the surgeon general, and over 25% of the U.S. population is obese. According to the ACS, about a third of all cancer deaths from 2002 were expected to be related to nutrition, physical inactivity, obesity, and other lifestyle factors and could have been prevented. An ACS study in the April issue of The New England Journal of Medicine suggests that about 14% of all cancer deaths in men and 20% of those in women are associated with excess body weight.
"If there were one perfect way to lose weight, you wouldn't go into the bookstore and find a hundred different diet books," says Litin. "It boils down to this: Healthy eating, consuming less calories than you burn up, and doing more exercise."
Fiber up. Although doctors continue to debate the cancer prevention effects of a diet rich in fiber (emphasizing whole-grain products, cereals, vegetables, and fruits), two major studies published in the British medical journal The Lancet in May found that the risk of colorectal cancer was reduced with high intakes of dietary fiber. In one of those studies, conducted in 10 European countries, the individuals who consumed the most dietary fiber (averaging 35 grams a day) experienced a 40% lower risk of colon cancer than subjects eating the least fiber (averaging 15 grams a day).
Get active. Exercise may help regulate hormones and "growth factors" that have been linked to cancer. According to Stein, all adults should accumulate at least 30 minutes of moderate-intensity activity on most days of the week, including walking, bicycling, gardening, dancing, and cleaning. This activity, she says, "does not need to be done all at once -- 10 minutes here and there add up."
Consider "chemoprevention." If you're at high risk for breast cancer (because of a strong family history, for example), your doctor may recommend taking a prescription drug called tamoxifen to reduce your chances of developing the disease. However, in a report in July 2002, the U.S. Preventive Services Task Force advised that the drug not be given to women with a low or average breast cancer risk.
Get tested. Annual screenings are advisable for all adults. Early detection, says Bevers, offers the best chance of successful treatment.
Originally published May 23, 2003.
Medically updated Jan. 24, 2005.
SOURCES: Therese Bevers, MD, University of Texas M.D. Anderson Cancer Center, Houston. Scott Litin, MD, Mayo Medical School, Rochester, Minn. Melanie Polk, MMSc, RD, American Institute for Cancer Prevention, Washington, D.C. Cynthia Stein, MD, MPH, Harvard Center for Cancer Prevention, Harvard Medical School, Boston.

redpajama_set_name: RedPajamaCommonCrawl


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Graphic: Erin Krogh / The Ebbtide
Kirk Meyers, News Editor
Origins: True Tales of the Immigrant Experience
7:30-9:30 p.m., June 11 & 12; YouTube livestream
The Ray Howard Library YouTube channel will be hosting live presentations of "Origins: True Tales of the Immigrant Experience," a series of performances influenced by the real-life experiences of SCC students as immigrants.
Spindrift 2021 Release Event
1-5 p.m., June 12; Ravenna Third Place Books, 6504 20th Ave. NE, Seattle, WA 98115
The staff of Spindrift, SCC's art and literary publication, will be hosting an event at Ravenna Third Place Books to celebrate the release of the journal's 2021 issue. Copies will be available for purchase.
12:35-1:25 p.m., June 15 & 17; Zoom meeting
Students, staff and faculty are invited to participate in a 50-minute Zoom yoga session. No prior experience necessary.
Commencement: 2021 Drive-Thru Graduation Celebration
1-3 p.m., June 19; SCC campus (Register here)
SCC graduates are invited to take part in the college's first-ever drive-thru graduation celebration.
One vehicle is permitted per graduate, including family and friends. Students that do not have access to a vehicle may participate by riding in the college shuttle.
For more information, contact Sundi Musnicki at [email protected].
Unemployment Q&A Sessions
10-11 a.m., June 22; Virtual meeting (Register here)
At this Q&A event, participants will have the chance to ask a state Employment Security Department representative for advice on topics such as job searching, job training and unemployment claims and appeals.
For more details on any of these events, visit the SCC calendar.
Kirk Meyers is a former News Editor of The Ebbtide. He covered political, breaking and feature stories. Meyers is a motorcycle enthusiast, who enjoys spending...
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ASG in Review: Oct. 25, 2021
News Briefs: Oct. 15, 2021


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White House condemns Cairo car bombing that killed Egyptian prosecutor
Published: JUNE 29, 2015 21:33
WASHINGTON - The White House on Monday condemned a car bomb attack in Cairo that killed Egypt's top public prosecutor and injured nine people in his convoy.
"The United States stands by Egypt at this difficult time, as we continue to work together to fight the scourge of terrorism," the White House said in a statement. Prosecutor-General Hisham Barakat is the most senior Egyptian official to die at the hands of militants since an Islamist president was toppled two years ago.


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Home / News / WoW Classic Servers: New Fresh Realms Angerforge and Jin'do Added Ahead of WotLK
WoW Classic Servers: New Fresh Realms Angerforge and Jin'do Added Ahead of WotLK
Plagued by long server lines and log-in times for World of Warcraft Classic, Blizzard Entertainment has announced two new "Fresh Start" areas ahead of Wrath of the Lich King's release later this month. Don't hesitate to buy WOW classic gold from Goldpiles!
Two of wow's most popular regions, North America and Europe, will get another new "fresh start" area to try to offset the superserver demographic.
WoW Classic Server: Two new Fresh start areas are now active
In June 2022, Blizzard introduced the New Beginning Area in preparation for the launch of Wrath of the Lich King and the influx of new players. The purpose of these servers is to give players a fresh and unique experience without any additional benefits and some restrictions.
These areas are new and designed specifically for players who are worried about being left behind. They are the Wrath of the Lich King servers from the beginning, which means that once a new character is created, they will have all the new talent trees and skills in the new expansion.
However, during the pre-patch of WotLK Classic, both New Start areas and normal areas became increasingly full, causing very long queues and log-in times for players. In response, Blizzard has locked down a number of "super realms", including Pharina, Blessing, Grobbulus on US, and added a number of brand new realms, including Elanix, Skyrage, and Maradas, with some special transfer restrictions.
As this problem persists, Blizzard has slowed things down by adding two new starting areas, Angevogi on the US server and King & Wood on the EU server.
Like other initially released Newstart fields, Angleforge and King & Wood have special rules and restrictions. This means that players can only play newly created characters, server transfers are not allowed for the first 90 days, upgrades are completely restricted, and the Death Knight's restrictions require a level 55 character in the domain to create one.


Do you have a vague understanding of human interaction? Well Speaking Simulator is for you!
Take control of a robot with a prototype tongue and mouth struggling to form words to have conversations. Our unfortunate story begins soon after you are activated, becoming alive and sentient. You are given a purpose by your AI master: Infiltrate Humanity. To achieve this you will face the incredibly dangerous and difficult challenge of talking to people.
Master the disgusting art of correctly moving your tongue and lips to create human speech. Squash and stretch your mouth into fantastically ugly and absurd shapes to emit different sounds. Manipulate your piston powered tongue to hit the proper buttons inside your robotic mouth. Please try to keep the coughing, stuttering, and awkward breathing to a minimum.
Often you will become overwhelmed with the task of speaking and you will start to make errors. This may cause your emotional systems to shut down and you will start to overheat. Please try to avoid this as humans find it a little suspicious.
Make sure you smile when youre meant to smile, and frown when youre meant to frown. Humans are very picky and find it strange if you stare into their eyes with a wide grin while they tell you about their personal tragedies.
Learn the mannerisms, culture, and habits of humans.
Master social scenarios such as job interviews, doctor examinations, and romantic conversations.
Degrade yourself by partaking in various human activities.
Dynamically create sounds, vowels and consonants with your mouth.
Surprisingly good portrayal of social anxiety.
Additional Notes: Requires keyboard and mouse.


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Finance World Economy Business & Tech Environment Politics & Law Society & Culture
CX Live is Caixin Global's real-time news portal, featuring 24-hour breaking news, short-form analysis, and roundups from business and social media in China.
Beijing Breathes Cleanest Air on Record
China Considers Establishing Individual Bankruptcy System
Cofco Raises $2.1 Billion in China's First Sustainability Loan
China Considers Packing In Decades-Old Zip Codes
TSMC Counts on New iPhones for Revival After Trade War Hit
Chinese Homebuyers Lead Sharp Drop in Foreign Demand in U.S.
U.S. Tech Supplier Cuts Tie with Chinese Drone Makers
Huawei Chairman Urges U.S. to Remove Company From Entity List
By Tianyu M. Fang / Jul 12, 2019 02:30 PM / Business & Tech
Huawei's Liang Hua. Photo: IC Photo
The U.S. says it is easing restrictions on Huawei sales, but there is no progress in sight, the Chinese telecom company's chairman said at a press event Friday.
Easing restrictions on Huawei sales isn't enough, Liang Hua said. Washington should remove Huawei from the Commerce Department's so-called Entity List, which effectively blacklisted the Chinese manufacturer from conducting business with U.S. companies.
In a retaliatory move, the Chinese government announced that it would establish its own "unreliable entity list" in June, before the temporary trade truce during the G20 Summit in Osaka, when U.S. President Donald Trump promised to relieve pressure on Huawei.
U.S. Secretary of Commerce Wilbur Ross announced Tuesday that the U.S. government would issue licenses for certain Huawei products, but did not specify when it would take such actions.
Read more on Huawei


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package org.jboss.aerogear.crypto.password;

import javax.crypto.SecretKey;
import java.security.spec.InvalidKeySpecException;

/**
 * Provides a level of abstraction for PBKDF2
 */
public interface Pbkdf2 {

    /**
     * Generate the secret key
     * @param password
     * @param salt
     * @param iterations
     * @return secret key
     * @throws InvalidKeySpecException
     */
    byte[] encrypt(String password, byte[] salt, int iterations) throws InvalidKeySpecException;

    /**
     * Generate the secret key
     * @param password
     * @param salt
     * @return secret key
     * @throws InvalidKeySpecException
     */
    byte[] encrypt(String password, byte[] salt) throws InvalidKeySpecException;

    /**
     * Generate the secret key
     * @param password
     * @return secret key
     * @throws InvalidKeySpecException
     */
    byte[] encrypt(String password) throws InvalidKeySpecException;

    /**
     * Validate the generated secret key
     * @param password
     * @param encryptedPassword
     * @param salt
     * @return secret key
     * @throws InvalidKeySpecException
     */
    boolean validate(String password, byte[] encryptedPassword, byte[] salt) throws InvalidKeySpecException;

    public SecretKey generateSecretKey(String password, byte[] salt, int iterations) throws InvalidKeySpecException;

    public SecretKey generateSecretKey(String password) throws InvalidKeySpecException;

}

redpajama_set_name: RedPajamaGithub