text:
\section{Incomplete Bessel functions, their properties and representations}

Following [1],  we define the incomplete modified Bessel function $J(z,\nu,w)$, cutting  integral (1.2)

$$J(z,\nu,w) = \int_{0}^w  e^{-z\cosh (u) } \cosh(\nu u) du.\eqno(2.1)$$
On the other hand, Entry 2.4.18.13 in [3], Vol. I suggests the representation of  Macdonald's  function  in the form

$$K_\nu(x)= {1\over 2}\   e^{\nu\pi i/2} \int_{-\infty}^\infty e^{ix \sinh(u) -\nu u} du,\quad x >0,\  |{\rm Re}  \nu | < 1.\eqno(2.2)$$
Letting $\nu= i\tau,\ \tau \in \mathbb{R}$ in (2.2), we easily find the formula

$$K_{i\tau}(x)=  e^{-\pi \tau/2} \int_{0}^\infty \cos\left(x \sinh(u) - \tau u\right) du,\eqno(2.3)$$
which can be also verified via (1.3), (1.4).  In the sequel we will employ the incomplete versions of  formulas (1.3), (1.4), introducing the following functions

$$K_c\left(x,\nu,w\right)= {1\over \cos(\pi \nu/2)} \int_0^w  \cos(x \sinh (u))  \cosh (\nu u) du,\eqno(2.4)$$

$$K_s\left(x,\nu,w\right)= {1\over \sin(\pi \nu/2)} \int_0^w  \sin (x \sinh (u))  \sinh (\nu u) du.\eqno(2.5)$$
Further, the incomplete modified Bessel function (2.1) is a solution of the inhomogeneous differential equation (cf. [1])

$$  z^2{d^2u\over dz^2}  + z{du\over dz} - (z^2+\nu^2)u = - \left(\nu \sinh(\nu w) + z \cosh(\nu w)\sinh(w) \right) e^{-z\cosh(w)}.\eqno(2.6)$$

In particular,  letting  $w=\pi, \ \nu = in,\ n \in \mathbb{N}$ in (2.1), (2.4) and integrating  by parts, we obtain, respectively,

$$ J(x, in, \pi) =  {x\over n} \int_0^\pi e^{-x\cosh(u)} \sinh(u) \sin (n u) du,\quad x  >0,\eqno(2.7)$$

$$K_c (x, in, \pi)= {x\over n \cosh(\pi n/2)} \int_0^\pi \sin (x \sinh (u))  \cosh(u) \sin (n u) du,\quad   x   >0.\eqno(2.8)$$
Moreover, the theory of Fourier series suggests  the following equalities for all $x >0$ and $u \in [0,\pi]$

$${\pi\over 2} e^{-x\cosh(u)} =  {1\over 2}\ J (x,0, \pi)   + \sum_{n=1}^\infty  J(x, in, \pi) \cos(nu),\eqno(2.9)$$

$${\pi x\over 2 } e^{-x\cosh(u)} \sinh(u) =  \sum_{n=1}^\infty  n  J (x, in, \pi) \sin(nu),\eqno(2.10)$$

$${\pi\over 2} \cos(x\sinh(u))  =  {1\over 2}\  K_c(x, 0, \pi)  + \sum_{n=1}^\infty  \cosh\left({\pi n\over 2}\right)  K_c(x, in, \pi) \cos(nu),\eqno(2.11)$$

$$ {\pi x\over 2 } \sin (x \sinh (u))  \cosh(u) = \sum_{n=1}^\infty  n  \cosh\left({\pi n\over 2}\right) K_c(x, in, \pi)  \sin(nu),\eqno(2.12)$$

$${\pi\over 2} \sin(x\sinh(u))  =   \sum_{n=1}^\infty   \sinh\left({\pi n\over 2}\right)  K_s(x, in, \pi)  \sin(nu),\eqno(2.13)$$

$${\pi\over 2}\int_0^\pi  e^{-2 x\cosh(u)} du  =   {\pi\over 2}  \ J(2x, 0, \pi) =  {1\over 2}\  J^2 (x,0,\pi)  + \sum_{n=1}^\infty  J^2 (x, in, \pi),\eqno(2.14)$$

$${\pi x^2\over 2 }\int_0^\pi  e^{-2 x\cosh(u)} \sinh^2 (u) du  =   {\pi x^2 \over 4} \left[ J (2x, 2, \pi) -  J (2x,0, \pi) \right] $$

$$= \sum_{n=1}^\infty   n^2 J^2(x, in, \pi),\eqno(2.15)$$

$${\pi\over 2}  \int_0^\pi \cos^2 (x\sinh(u)) du   =  {\pi\over 4}  \left[ \pi +  K_c (2x, 0, \pi) \right] $$

$$= {\left[ K_c (x, 0, \pi)\right]^2  \over 2}  + \sum_{n=1}^\infty  \cosh^2\left({\pi n\over 2}\right)  \left[ K_c (x,in, \pi) \right]^2 ,\eqno(2.16)$$

$$ {\pi x^2\over 2 }  \int_0^\pi \sin^2 (x \sinh (u))  \cosh^2(u) du  = \sum_{n=1}^\infty  n^2  \cosh^2\left({\pi n\over 2}\right) \left[ K_c(x, in, \pi) \right]^2,\eqno(2.17)$$

$$ {\pi \over 2 }  \int_0^\pi \sin^2 (x \sinh (u)) du  = \sum_{n=1}^\infty    \sinh^2\left({\pi n\over 2}\right) \left[ K_s(x, in, \pi) \right]^2.\eqno(2.18)$$

The following lemma establishes biorthogonality of sequences of functions

$$ \{ K_{in}(x) \}_{ n\in \mathbb{N}},  \{ J (x, in, \pi) \}_{ n\in \mathbb{N}},\eqno(2.19)$$

$$\{ K_{in}(x) \}_{ n\in \mathbb{N}}, \  \{ K_c (x, in, \pi) \}_{ n\in \mathbb{N}},\eqno(2.20) $$

$$ \{ K_{i\tau}(n) \}_{ n\in \mathbb{N}},\  \{ K_s \left(n, i\tau, \sinh^{-1} (\pi)\right) \}_{ n\in \mathbb{N}},\eqno(2.21)$$
where we use the notation $ \sinh^{-1}(u) = \log\left( u+ \sqrt{1+ u^2}\right).$

Indeed, we have

{\bf Lemma 1}. {\it  Sequences of functions $(2.19), (2.20)$  are biorthogonal over $\mathbb{R}_+$ with respect to the measure ${dx\over x}$, i.e.

$$I_1= \int_0^\infty  K_{in}(x) J(x,im, \pi) {dx\over x} = { \pi^2\over 2} \ {  \delta_{n,m}\over n \sinh(\pi n)},\quad  n, m \in \mathbb{N},\eqno(2.22)$$

$$I_2= \int_0^\infty  K_{in}(x) K_c(x, im, \pi) {dx\over x} = { \pi^2\over 2} \ {  \delta_{n,m}\over n \sinh(\pi n) },\quad  n, m \in \mathbb{N},\eqno(2.23)$$
where $\delta_{n,m}$ is the Kronecker symbol.  Besides, sequences $(2.21)$ are biorthogonal over $\mathbb{R}_+$ with respect to the measure $\tau \sinh(\pi\tau) d\tau$, namely,}

$$I_3= \int_0^\infty   K_{i\tau}(n)  K_s\left(m, i\tau, \sinh^{-1} (\pi)\right) \tau \sinh(\pi\tau) d\tau =  { \pi^2 n \over 2} \  \delta_{n,m},\quad  n, m \in \mathbb{N}.\eqno(2.24)$$

\begin{proof}   In fact, using (2.7), we substitute the corresponding integral into (2.22), having the estimate

$$ |I_1| \le {1\over m} \int_0^\infty  \left| K_{in}(x) \right| \int_0^\pi e^{-x\cosh(u)} \sinh(u)  \left| \sin (m u) \right| du dx $$

$$\le  {1\over m} \int_0^\infty  K_{0}(x)  \int_0^\pi e^{-x\cosh(u)} \sinh(u)  du dx =  {1\over m} \int_0^\infty  K_{0}(x) \left( 1- e^{-x\cosh(\pi)}\right) {dx \over x}  < \infty.$$
Therefore Fubini's theorem permits the interchange of the order of integration to write $I_1$  as follows

$$I_1 =   {1\over m} \int_0^\pi  \sinh(u)  \sin (m u)   \int_0^\infty  K_{in}(x) e^{-x\cosh(u)} dx du.\eqno(2.25)$$
But the   integral with respect to $x$ on the right-hand side of (2.25) is calculated in [3], Vol. II, Entry 2.16.6.1,  and we have

$$ \int_0^\infty  e^{-x\cosh(u)} K_{in}(x)  dx = {\pi \sin( nu) \over \sinh (u) \sinh(\pi n)}.\eqno(2.26)$$
Hence (2.25) becomes

$$I_1 =   {\pi \over m \sinh(\pi n)} \int_0^\pi  \sin (n u)   \sin( mu) du =  { \pi^2\over 2} \ {  \delta_{n,m}\over n \sinh(\pi n)},$$
which proves (2.22).

Analogously, employing (2.8) and the estimate

$$ |I_2| \le  {1\over m \cosh(\pi m/2)} \int_0^\infty  \left| K_{in}(x) \right|   \int_0^\pi \left| \sin (x \sinh (u))  \cosh(u) \sin (m u) \right| du  dx $$

$$\le   {\pi \cosh(\pi) \over m \cosh(\pi m/2)} \int_0^\infty   K_{0}(x) dx < \infty,$$
we change the order of integration by Fubini's theorem to obtain $I_2$  in the form

$$I_2  =  {1\over m \cosh(\pi m/2)}  \int_0^\pi   \cosh(u) \sin (m u)   \int_0^\infty  K_{in}(x) \sin (x \sinh (u)) dx du.$$
The latter integral with respect to $x$ is calculated in   [3], Vol. II, Entry   2.16.14.1, and we get

$$ \int_0^\infty  K_{in}(x) \sin (x \sinh (u)) dx =  {\pi\over 2}  { \sin (n u)\over \sinh(\pi n/2) \cosh(u) }.\eqno(2.27)$$
Therefore

$$I_2=  {\pi \over 2 m \cosh(\pi m/2)  \sinh(\pi n/2)}  \int_0^\pi  \sin (n u)  \sin (m u) du =    {\pi^2 \delta_{n,m}  \over 2 n \sinh(\pi n)} $$
and we establish equality (2.23).

Finally, in order to prove (2.24) we appeal to (2.5) to write

$$K_s \left(m, i\tau, \sinh^{-1} (\pi) \right)= {1\over \sinh(\pi \tau/2)} \int_0^{\sinh^{-1} (\pi) } \sin(m \sinh (u))  \sin (\tau u) du.$$
Integration by parts and simple substitutions on the right-hand side of the latter equality then give

$${1\over \sinh(\pi \tau/2)} \int_0^{\sinh^{-1} (\pi) } \sin(m \sinh (u))  \sin (\tau u) du =  {m \over \tau \sinh(\pi \tau/2)} \int_0^{\sinh^{-1} (\pi) } \cos(m \sinh (u)) $$

$$\times  \cos (\tau u) \cosh(u) du =  {m \over \tau \sinh(\pi \tau/2)} \int_0^{\pi} \cos(m u)  \cos \left(\tau \sinh^{-1} (u) \right) du.$$
Hence

$$I_3 = 2m \int_0^\infty   K_{i\tau}(n)  \cosh \left({\pi\tau\over 2}\right)  \int_0^{\pi} \cos(m u)  \cos \left(\tau \sinh^{-1} (u)\right) du d\tau$$

$$= 2m \lim_{N\to \infty } \int_0^N   K_{i\tau}(n)  \cosh \left({\pi\tau\over 2}\right)  \int_0^{\pi} \cos(m u)  \cos \left(\tau \sinh^{-1} (u)\right) du d\tau$$

$$= 2m \lim_{N\to \infty }   \int_0^{\pi} \cos(m u)  \int_0^N   K_{i\tau}(n)  \cosh \left({\pi\tau\over 2}\right)  \cos \left(\tau \sinh^{-1} (u)\right) d\tau du,\eqno(2.28)$$
where the latter interchange of the order of integration is due to the absolute and uniform convergence.  The problem now is to motivate the passage to the limit under the integral sign on the right-hand side of the latter equality. Then we use the value of the improper integral with respect to $\tau$ (cf. [4])

 $$ \int_0^\infty   K_{i\tau}(n)  \cosh \left({\pi\tau\over 2}\right)  \cos\left(\tau  \sinh^{-1} (u) \right) d\tau = {\pi\over 2} \cos(n u)\eqno(2.29)$$
to complete the proof of Lemma 1.   So, choosing a big enough $N >0$ and fixing $n \in \mathbb{N}$,  we use asymptotic behavior of the Macdonald function with respect to $\tau$ [4]

$$ K_{i\tau}(n) = \left({2\pi\over \tau}\right)^{1/2} e^{-\pi\tau/2} \sin\left( {\pi\over 4} + \tau \log\left({2\tau\over n} \right) - \tau \right) \left( 1+ O\left({1\over \tau}\right) \right),\ \tau \to +\infty$$
to get the remainder of the integral (2.28) in the form

$$ \int_N^\infty   K_{i\tau}(n)  \cosh \left({\pi\tau\over 2}\right)  \cos\left(\tau  \sinh^{-1} (u) \right) d\tau $$

$$= \sqrt{{\pi\over 2}}  \int_N^\infty   \sin\left( {\pi\over 4} + \tau \log\left({2\tau\over n} \right) - \tau \right)  \cos\left(\tau \sinh^{-1} (u)\right) {d\tau\over \sqrt \tau }
+  O\left({1\over \sqrt N}\right). $$
As we see via trigonometric identities the latter integral is associated with the following ones

$$G_1(N)= \int_N^\infty   \sin\left(  \tau \left( \log\left({2\tau\over n} \right) - 1 \mp \sinh^{-1} (u) \right)\right) {d\tau\over \sqrt \tau }, $$

$$G_2(N)= \int_N^\infty   \cos\left(  \tau \left( \log\left({2\tau\over n} \right) - 1 \mp \sinh^{-1} (u) \right) \right) {d\tau\over \sqrt \tau }.$$
Taking, for instance, the sine integral, we integrate by parts to derive

$$ G_1(N)=  {1\over \sqrt N} \ \frac{ \cos\left(  N \left( \log\left({2N/ n} \right) - 1 \mp \sinh^{-1} (u) \right)\right)}{ \log( 2N /n) \mp \sinh^{-1} (u) }$$

$$- {1\over 2} \int_N^\infty  { \cos \left(  \tau \left( \log\left({2\tau/ n} \right) - 1 \mp \sinh^{-1} (u) \right)\right) \over \left(\log(2\tau/n) \mp \sinh^{-1} (u) \right) \tau^{3/2}  } d\tau$$

$$-  \int_N^\infty  { \cos \left(  \tau \left( \log\left({2\tau/ n} \right) - 1 \mp \sinh^{-1} (u) \right)\right) \over \left(\log(2\tau/n) \mp \sinh^{-1} (u)  \right)^2 \tau^{3/2}  } d\tau.$$
Hence,

$$| G_1(N)| \le   { 1 \over  \sqrt N}\  \left( \log( 2N /n) - \sinh^{-1} (\pi)  \right)^{-1}  $$

$$ +   \int_N^\infty {\left(\log(2\tau/n) -  \sinh^{-1} (\pi) \right) /2 +1\over  \left(\log(2\tau/n) - \sinh^{-1} (\pi)  \right)^2 \tau^{3/2}  } d\tau \to 0,\ N \to \infty.$$
Analogously the integral $G_2(N)$ can be treated.  Thus the passage to the limit under the integral sign in (2.28) is allowed. Then in view of formula (2.29), we readily end up with the biorthogonality (2.24).

\end{proof}

To complete this section, we will exhibit below our main results, which will be proved in the sequel. It involves  the following expansions  of   sequences and functions under certain conditions,  giving  rise to  discrete analogs of the Kontorovich - Lebedev transform (1.1), namely
$( n\in  \mathbb{N},\ x >0 )$

$$a_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty J (x, in, \pi) \sum_{m=1}^\infty a_m K_{im}(x)  {dx\over x} ,\eqno(2.30)$$

$$a_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty K_{in} (x)  \sum_{m=1}^\infty a_m J (x, im, \pi)  {dx\over x},\eqno(2.31)$$

$$a_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty K_c(x, in,  \pi)  \sum_{m=1}^\infty a_m K_{im}(x)  {dx\over x},\eqno(2.32)$$

$$a_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty K_{in} (x)  \sum_{m=1}^\infty a_m K_c (x, im, \pi)  {dx\over x},\eqno(2.33)$$

$$a_n =  {4\over \pi^2\ n} \lim_{\alpha\to  {\pi\over 2} -} \int_0^\infty   \tau  \sinh\left({\pi \tau\over 2}\right) \cosh\left(\alpha \tau \right)  K_s\left(n, i\tau, \sinh^{-1} (\pi) \right) \sum_{m=1}^\infty a_m K_{i\tau}(m)  d\tau,\eqno(2.34)$$

$$a_n =  {2\over \pi^2\ n}  \int_0^\infty   \tau  \sinh(\pi \tau )    K_{i\tau}(n) \sum_{m=1}^\infty a_m K_s \left(m, i\tau, \sinh^{-1} (\pi)  \right)  d\tau,\eqno(2.35)$$

$$  f(x)  =  {2\over x \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  J (x, in, \pi)  \int_0^\infty  K_{in} (y) f(y) dy,\eqno(2.36)$$

$$f(x)  =  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_{in} (x)   \int_0^\infty  J (y, in, \pi) f(y) {dy\over y},\eqno(2.37)$$

$$  f(x)  =  {2\over x \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_c(x, in,  \pi)  \int_0^\infty  K_{in} (y)  f(y) dy,\eqno(2.38)$$

$$f(x)  =  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_{in} (x)  \int_0^\infty K_c (y, in, \pi) f(y) {dy\over y},\eqno(2.39)$$

$$f(x)  =  {4\over \pi^2\ } \  x \sinh\left({\pi x\over 2}\right)    \sum_{n=1}^\infty  {1\over n} \ K_s \left(n, ix,  \sinh^{-1}(\pi) \right) $$

$$\times \int_0^\infty   \cosh\left({\pi \tau\over 2}\right )  K_{i\tau}(n)  f(\tau) d\tau,\eqno(2.40)$$

$$f(x)  =  {2\over \pi^2\ }   \sum_{n=1}^\infty  { K_{ix}(n) \over n} \int_0^\infty   \tau \sinh\left(\pi \tau\right )  K_s\left(n, i\tau,  \sinh^{-1}(\pi)\right)   f(\tau) d\tau.\eqno(2.41)$$

\section {Expansion theorems}

We begin with

{\bf Theorem 1.} {\it Let the sequence $\{a_n\}_{n\in \mathbb{N}}$ be such that the following series converges

$$\sum_{n=1}^\infty  |a_n|  e^{-\pi n/2} < \infty.\eqno(3.1)$$
Then  expansion $(2.30)$ holds, where the iterated series and integral converge absolutely.}

\begin{proof}  In fact, recalling representation (2.7) of the kernel $J(x, in, \pi)$, substituting this expression on  the right-hand side of (2.30) and employing Lebedev's inequality (1.9),  we derive the estimates

$$ {2\over \pi^2}  \sinh(\pi n) \int_0^\infty    \int_0^\pi \left| e^{-x\cosh(u)} \sinh(u) \sin (n u) \right| du  \sum_{m=1}^\infty \left| a_m K_{im}(x) \right| dx $$

$$\le  {2 A\over \pi^2}  \sinh(\pi n) \int_0^\infty   x^{-1/4}  \int_0^\pi e^{-x\cosh(u)} \sinh(u) du dx  \sum_{m=1}^\infty  {| a_m | \over \sqrt{ \sinh (\pi m)} } $$

$$  \le {2 \sqrt 2\over \pi^2} \  \Gamma(3/4) A \sinh(\pi n) \int_0^\pi { \sinh(u) \over \cosh^{3/4} (u)} du \sum_{m=1}^\infty  |a_m|  e^{-\pi m/2} $$

$$ =   {8 \sqrt 2\over \pi^2} \    \Gamma(3/4) A \sinh(\pi n) \left[ \cosh^{1/4} (\pi) -1\right]  \sum_{m=1}^\infty  |a_m|  e^{-\pi m/2} < \infty $$
owing to the condition (3.1) of the theorem. Hence the interchange of the order of integration and summation in (2.30) is guaranteed by Fubini's theorem and we find

$$ {2\over \pi^2}  \sinh(\pi n) \int_0^\infty   \int_0^\pi e^{-x\cosh(u)} \sinh(u) \sin (n u) du \sum_{m=1}^\infty  a_m K_{im}(x)  dx $$

$$ = {2\over \pi^2}  \sinh(\pi n) \sum_{m=1}^\infty  a_m   \int_0^\pi  \sinh(u) \sin (n u)  \int_0^\infty  e^{-x\cosh(u)} K_{im}(x)  dx du.$$
Then appealing to formula (2.26),  we have finally

$${2\over \pi^2}  \sinh(\pi n) \sum_{m=1}^\infty  a_m   \int_0^\pi  \sinh(u) \sin (n u)  \int_0^\infty  e^{-x\cosh(u)} K_{im}(x)  dx du$$

$$=  {2\over \pi}  \sinh(\pi n) \sum_{m=1}^\infty  {a_m \over \sinh(\pi m)}  \int_0^\pi  \sin (n u) \sin (m u) du = a_n.$$
Theorem 1 is proved.

\end{proof}

Concerning expansion (2.31), we have

{\bf Theorem 2.} {\it Let the sequence $\{a_n\}_{n\in \mathbb{N}}$ be such that the following series converges

$$\sum_{n=1}^\infty  {|a_n| \over n}  < \infty.\eqno(3.2)$$
Then  expansion $ (2.31)$ holds valid, where the iterated series and integral converge absolutely.}

\begin{proof}  Doing in the same manner, we appeal to (1.9) and (2.7)  to get the estimate

$$ {2\over \pi^2}  n \sinh(\pi n) \int_0^\infty  \left| K_{in}(x) \right|  \sum_{m=1}^\infty  { |a_m| \over m}  \int_0^\pi  \left| e^{-x\cosh(u)} \sinh(u) \sin (m u)  \right| du   dx $$

$$\le  {2 A\over \pi^2} n \sqrt{\sinh(\pi n)} \int_0^\infty   x^{-1/4}  \int_0^\pi e^{-x\cosh(u)} \sinh(u) du dx  \sum_{m=1}^\infty  {| a_m | \over m } $$

$$  =  {2  \Gamma(3/4) A\over \pi^2} n  \sinh(\pi n) \int_0^\pi { \sinh(u) \over \cosh^{3/4} (u)} du \sum_{m=1}^\infty { |a_m| \over m}  $$

$$ =   {8  \Gamma(3/4) A\over \pi^2}  n \sinh(\pi n) \left[ \cosh^{1/4} (\pi) -1\right]  \sum_{m=1}^\infty  {|a_m| \over m} < \infty $$
due to the condition (3.2) of the theorem. Therefore, as above,  we change  the order of integration and summation by Fubini's theorem and recall  equality (2.26) to  arrive at (2.31), completing the proof of Theorem 2.

\end{proof}

{\bf Remark 1}.  Expansions (2.30), (2.31) generate two pairs of discrete Kontorovich-Lebedev transforms

$$f(x)= \sum_{m=1}^\infty  a_m K_{im} (x),\quad  x >0,\eqno(3.3)$$

 $$g(x)= \sum_{m=1}^\infty  b_m J (x,im, \pi),\quad  x >0.\eqno(3.4)$$
Coefficients $a_n,\ b_n,\ n \in \mathbb{N}$ are calculated accordingly

$$a_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty J (x,in, \pi)  f(x)   {dx \over x},\eqno(3.5)$$

$$b_n =  {2\over \pi^2} \ n  \sinh(\pi n) \int_0^\infty K_{in} (x)   g(x)   {dx \over x}.\eqno(3.6)$$
Integral  (3.6) is the discrete  Kontorovich-Lebedev transform  whose existence conditions are well known in different spaces of functions (see [4]).  A simple conditions is based on the inequality $ | K_{in} (x)  | \le K_0 (x),\ x > 0$. Thus if $g$ is Lebesgue integrable with respect to the measure $K_0(x) dx/x$, i.e. $ g \in L_1\left(\mathbb{R}_+;    K_0(x) dx/x \right)$, the Kontorovich-Lebedev transform (3.6) exists.    The corresponding existence conditions for discrete transform (3.5) follow from the behavior of the kernel $J (x,in, \pi) $.  For instance, representation (2.7) yields the equality $J (x,in, \pi) = O(x)$ as $ x \to 0$ and  simple inequality

$$\left| J (x,in, \pi) \right| \le {x\over n} \int_0^\pi e^{-x\cosh(u)} \sinh(u) du =   {1\over n} \left( e^{-x}- e^{- x\cosh (\pi)}\right) \le  {2\over n}\   e^{-x},\ x > 0.$$
Hence,  if $f \in L_1\left(\mathbb{R}_+;    e^{-x} dx \right)$, the discrete transform (3.5) exists.

{\bf Remark 2.}  Another way to prove the validity of expansions (2.30), (2.31) is to use the biorthogonality (2.22), verifying first   the interchange of the order of integration and summation under conditions (3.1), (3.2), respectively.

{\bf Theorem 3}. {\it Let the sequence $\{a_n\}_{n\in \mathbb{N}}$ satisfy the condition

$$ \sum_{m=1}^\infty \left| a_m \right| e^{-\delta m} < \infty,\ \delta \in \left[0, \  {\pi\over 2} \right).\eqno(3.7)$$
Then expansions  $(2.32), (2.33)$ hold, where the iterated   series and integral converge absolutely.   Moreover,  expansion $(2.33)$  holds for a wider class of sequences under the condition }

$$ \sum_{m=1}^\infty {\left| a_m \right|\over m}  e^{-\pi m/2} < \infty.\eqno(3.8)$$

\begin{proof}  Indeed, appealing to the inequality for the modified Bessel function (see [4], p. 15)

$$ \left| K_{i\tau} (x) \right| \le e^{-\delta \tau} K_0\left( x \cos(\delta) \right), \quad x, \tau >0,\ \delta \in \  \left[0, \  {\pi\over 2} \right),$$
we use integral representation (2.8) for the kernel $K_c(x,in, \pi)$ and the following estimates to justify the interchange of the order of integration and summation in (2.32), ( 2.33) under conditions of the theorem.  Thus

$$\int_0^\infty {1\over x} \  | K_c (x,in, \pi) | \sum_{m=1}^\infty \left| a_m K_{im}(x) \right| dx \le  {1\over n \cosh(\pi n/2)}  \int_0^\infty  K_0\left( x \cos(\delta) \right) dx $$

$$\times  \int_0^\pi \left| \sin (x \sinh (u))  \cosh(u) \sin (n u)\right| du  \sum_{m=1}^\infty \left| a_m  \right|   e^{-\delta m}$$

$$\le  {\sinh(\pi) \over n \cosh(\pi n/2)}  \int_0^\infty  K_0\left( x \cos(\delta) \right) dx  \sum_{m=1}^\infty \left| a_m  \right|   e^{-\delta m} $$

$$= { \pi \sinh(\pi) \over 2 n \cosh(\pi n/2) \cos(\delta)}   \sum_{m=1}^\infty \left| a_m  \right|   e^{-\delta m}< \infty, \quad \delta \in \  \left[0, \  {\pi\over 2} \right),$$

$$\int_0^\infty { | K_{in}(x) |\over x} \sum_{m=1}^\infty \left| a_m K_c (x, im, \pi)  \right| dx $$

$$\le 2  \sinh(\pi) \int_0^\infty  K_0\left( x  \right) dx  \sum_{m=1}^\infty {\left| a_m \right| \over m }  e^{-\pi m/2}$$

$$\le \pi  \sinh(\pi) \sum_{m=1}^\infty \left| a_m \right|   e^{- \delta m}< \infty,\quad \delta \in \  \left[0, \  {\pi\over 2} \right).$$
Therefore, changing the order of integration and summation on the right-hand side of (2.32), (2.33) and employing biorthogonality  (2.23), we establish the desired equalities.

\end{proof}

Concerning expansions (2.34), (2.35), we have

{\bf Theorem 4}. {\it  Let the sequence $\{a_n\}_{n \in \mathbb{N} } \in l_1$, i.e. $\sum_{m\ge 1} |a_m| < \infty$. Then  its general term  can be expanded with respect to  $(2.33)$ with  the convergence  in the sense of Abel.  Moreover, under  stronger  condition
$$\sum_{m=1}^\infty  m \left| a_m \right|  < \infty\eqno(3.9)$$
expansion $(2.35)$  holds, where the convergence is in the improper sense.}

\begin{proof} In fact, from the proof of Lemma 1 we have the equality

$$ K_s\left(n, i\tau, \sinh^{-1}(\pi) \right) =   {n \over \tau \sinh(\pi \tau/2)}  \int_0^{\pi} \cos(n u)  \cos \left(\tau  \sinh^{-1}(u)\right) du.\eqno(3.10)$$
Substituting this expression on the right-hand side of (2.34), we get

$${4\over \pi^2\ n} \lim_{\alpha \to {\pi\over 2} -}  \int_0^\infty   \tau  \sinh\left({\pi \tau\over 2}\right )\cosh \left(\alpha \tau\right)   K_s\left(n, i\tau, \sinh^{-1}(\pi) \right)$$

$$\times  \sum_{m=1}^\infty a_m K_{i\tau}(m)  d\tau = {4\over \pi^2} \lim_{\alpha \to {\pi\over 2} -}  \int_0^\infty   \cosh \left(\alpha \tau\right)  \int_0^{\pi} \cos(n u)  \cos \left(\tau \sinh^{-1}(u)  \right) du $$

$$\times  \sum_{m=1}^\infty a_m K_{i\tau}(m)  d\tau.\eqno(3.11)$$
From  inequality (1.9) we have the estimate

$$\int_0^\infty   \cosh \left(\alpha \tau\right)  \int_0^{\pi} \left| \cos(n u)  \cos \left(\tau \sinh^{-1}(u) \right)\right| du   \sum_{m=1}^\infty \left| a_m K_{i\tau}(m) \right| d\tau $$

$$\le \pi A  \int_0^\infty   {\cosh \left(\alpha \tau\right)  \over \sqrt{\sinh(\pi\tau)}} \ d\tau \sum_{m=1}^\infty   \left| a_m \right| < \infty,\quad  0 < \alpha < {\pi\over 2},$$
which validates  the interchange of the order of integration and summation on the right-hand side of (3.11). Thus we obtain

$${ 4\over \pi^2} \lim_{\alpha \to {\pi\over 2} -}  \int_0^\infty   \cosh \left(\alpha \tau\right)  \int_0^{\pi} \cos(n u)  \cos \left(\tau \sinh^{-1}(u) \right) du $$

$$\times  \sum_{m=1}^\infty a_m K_{i\tau}(m)  d\tau = { 4\over \pi^2} \lim_{\alpha \to {\pi\over 2} -}    \sum_{m=1}^\infty a_m  \int_0^{\pi} \cos(n u)  $$

$$\times \int_0^\infty   \cosh \left(\alpha \tau\right) K_{i\tau}(m)  \cos \left(\tau \sinh^{-1}(u) \right) d\tau du.$$
But the integral with respect to $\tau$ is calculated in [3], Vol. II, Entry 2.16.48.20 and we find the equality

$$ \int_0^\infty   \cosh \left(\alpha \tau\right) K_{i\tau}(m)  \cos \left(\tau \sinh^{-1}(u) \right) d\tau = {\pi\over 2}\  e^{-m\cos(\alpha) (1+u^2)^{1/2}} \cos( m u\sin(\alpha)).\eqno(3.12)$$
Therefore, combining with (3.11), it gives

$$ {4\over \pi^2\ n} \lim_{\alpha \to {\pi\over 2} -}  \int_0^\infty   \tau  \sinh\left({\pi \tau\over 2}\right )\cosh \left(\alpha \tau\right)   K_s\left(n, i\tau, \sinh^{-1}(\pi) \right)$$

$$\times  \sum_{m=1}^\infty a_m K_{i\tau}(m)  d\tau =  {2\over \pi} \lim_{\alpha \to {\pi\over 2} -} \sum_{m=1}^\infty a_m$$

$$\times   \int_0^{\pi}  e^{-m\cos(\alpha) (1+u^2)^{1/2}}  \cos(n u) \cos( m u\sin(\alpha)) du. \eqno(3.13)$$
Hence equality (2.34) will be proved,  if we motivate the passage to the limit under the series sign on the right-hand side of (3.13).  But this fact is an immediate consequence of the estimate

$$\sum_{m=1}^\infty \left| a_m \int_0^{\pi}  e^{-m\cos(\alpha) (1+u^2)^{1/2}}  \cos(n u) \cos( m u\sin(\alpha)) du \right| \le \pi \sum_{m=1}^\infty  \left| a_m \right|.$$
In order to establish  (2.35),  we appeal to the biorthogonality result (2.24) together with  (2.28) to justify changing   order of integration and summation.   The completion of the proof is left to the reader.

\end{proof}

Let us argue  the validity of expansions  (2.36), (2.37).   We have

{\bf Theorem 5}. {\it Let $f$ be a complex-valued function on $\mathbb{R}_+$ which is represented by the integral

$$f(x)= \int_{-\pi}^\pi e^{-x\cosh(u)} \varphi(u) du,\ x >0,\eqno(3.14)$$
where $ \varphi(u) = \psi(u)\sinh(u)$ and $\psi$ is a  $2\pi$-periodic function, satisfying the Lipschitz condition on $[-\pi, \pi]$, i.e.

$$\left| \psi(u) - \psi(v)\right| \le C |u-v|, \quad  \forall \  u, v \in  [-\pi, \pi],\eqno(3.15)$$
where $C >0$ is an absolute constant.  Then expansion $(2.36)$ holds for all $x >0$.  Further, if $f$ can be expanded in terms of the series $(3.3)$
of a suitable   sequence $\{a_m\}_{m \in \mathbb{N}}$, satisfying condition $(3.1)$, then expansion $(2.37)$ holds valid. }

\begin{proof}  Let $S_N(x) $ denote  a partial sum of the series (2.36), substitute the value of $f(x)$ by integral (3.14) and  interchange the order of integration with the use of the equality (2.26) to obtain

$$S_N(x)  = {2\over \pi  x} \   \sum_{n=1}^N   n \  J (x, in, \pi)  \int_{-\pi}^\pi   {\varphi(u) \over \sinh(u)} \sin(nu) du.\eqno(3.16)$$
But the right-hand side of (3.16) can be rewritten, employing representation (2.7) to get

$$S_N(x)  =   {1\over \pi} \   \sum_{n=1}^N  \int_{-\pi}^\pi e^{-x\cosh(t)} \sinh(t) \sin (n t) dt  \int_{-\pi}^\pi   {\varphi(u) \over \sinh(u)} \sin(nu) du.\eqno(3.17)$$
Now, calculating the sum via the known formula, equality (3.17) becomes

$$  S_N(x)  =  {1\over 4 \pi} \   \int_{-\pi}^\pi e^{-x\cosh(t)} \sinh(t)   \int_{-\pi}^{\pi}   {\varphi(u)+ \varphi(-u)  \over \sinh(u)} \  {\sin \left((2N+1) (u-t)/2 \right)\over \sin( (u-t) /2)}  du dt $$

$$=   {1\over 4 \pi} \   \int_{-\pi}^\pi e^{-x\cosh(t)} \sinh(t)   \int_{-\pi}^{\pi}  \left[ \psi(u)- \psi(-u) \right]  \  {\sin \left((2N+1) (u-t)/2 \right)\over \sin( (u-t) /2)}  du dt.\eqno(3.18)$$
Since $\psi$ is $2\pi$-periodic, we treat  the latter integral with respect to $u$ as follows

$$  \int_{-\pi}^{\pi}  \left[ \psi(u)- \psi(-u) \right]  \  {\sin \left((2N+1) (u-t)/2 \right)\over \sin( (u-t) /2)}  du $$

$$=  \int_{ t-\pi}^{t+ \pi}  \left[ \psi(u)- \psi(-u) \right]  \  {\sin \left((2N+1) (u-t)/2 \right)\over \sin( (u-t) /2)}  du $$

$$=  \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(-u-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du. $$
Moreover,

$$ {1\over 2\pi} \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(-u-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du - \left[ \psi(t)- \psi(-t) \right] $$

$$=  {1\over 2\pi} \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(t) + \psi (-t) - \psi(-u-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du.$$
When  $u+t > \pi$ or  $u+t < -\pi$ then we interpret  the value  $\psi(u+t)- \psi(t)$ by  formulas

$$\psi(u+t)- \psi(t) = \psi(u+t-2\pi)- \psi(t - 2\pi),$$

$$\psi(u+t)- \psi(t) = \psi(u+t+ 2\pi)- \psi(t +2\pi),$$
respectively.  Analogously, the value  $\psi(-u-t)- \psi(-t)$  can be treated.   Then   due to the Lipschitz condition (3.15) we have the uniform estimate
for any $t \in [-\pi,\pi]$

$${\left|  \psi(u+t)- \psi(t) + \psi (-t) - \psi(-u-t) \right| \over | \sin( u /2) |}  \le 2C \left| {u\over \sin( u /2)} \right|.$$
Therefore,  owing to the Riemann-Lebesgue lemma

$$\lim_{N\to \infty } {1\over 2\pi} \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(-u-t)  - \psi(t) + \psi (-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du =  0\eqno(3.19)$$
for all $ t\in [-\pi,\pi].$    Besides, returning to (3.18), we estimate the iterated integral

$$ \int_{-\pi}^\pi e^{-x\cosh(t)} |\sinh(t)  | \int_{ -\pi}^{\pi} \left| \left[ \psi(u+t)- \psi(-u-t)  - \psi(t) + \psi (-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  \right| du dt$$

$$\le  4 C \int_{0}^\pi e^{-x\cosh(t)}  \sinh(t) dt   \int_{ -\pi}^{\pi}   \left| {u\over \sin( u /2)} \right| du < \infty,\ x >0.$$
Consequently, via  the dominated convergence theorem it is possible to pass to the limit when $N \to \infty$ under the  integral sign, and recalling (3.19), we derive

$$  \lim_{N \to \infty}   {1\over 4 \pi}  \int_{-\pi}^\pi e^{-x\cosh(t)} \sinh(t)  \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(-u-t)  - \psi(t) + \psi (-t) \right] $$

$$\times  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du dt =  {1\over 4 \pi}  \int_{-\pi}^\pi e^{-x\cosh(t)} \sinh(t)  $$

$$ \times \lim_{N \to \infty}  \int_{ -\pi}^{\pi}  \left[ \psi(u+t)- \psi(-u-t)  - \psi(t) + \psi (-t) \right]  \  {\sin \left((2N+1) u/2 \right)\over \sin( u /2)}  du dt = 0.$$
Hence, combining with (3.18),  we obtain  by virtue of  the definition of $\varphi$ and $f$

$$ \lim_{N \to \infty}  S_N(x) =   {1\over 2} \   \int_{-\pi}^\pi e^{-x\cosh(t)}    \left[ \varphi (t)+ \varphi(-t) \right] dt = f(x),$$
where the integral (3.14) converges since $\varphi \in C[-\pi,\pi]$.  Thus we established  (2.36). In order to prove expansion (2.37), we plug the series (3.3) into (2.37)  and recall the representation (2.7) of the kernel $J(y, in, \pi)$ to deduce

$$ {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_{in} (x)   \int_0^\infty J (y, in, \pi) f(y) {dy\over y} $$

$$= {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_{in} (x)   \int_0^\infty \int_0^\pi e^{-y\cosh(u)} \sinh(u) \sin (n u)   \sum_{m=1}^\infty {a_m \over m}\ K_{im} (y)  du dy. \eqno(3.20)$$
Observing that the interchange of the order of integration and summation is permitted due to the estimates in the proof of Theorem 1, we find the equality

$$ {2\over \pi^2} \int_0^\infty \int_0^\pi e^{-y\cosh(u)} \sinh(u) \sin (n u)   \sum_{m=1}^\infty {a_m\over m}  \ K_{im} (y)  du dy $$

$$= {2\over \pi} \sum_{m=1}^\infty {a_m \over  m \sinh(\pi m)} \int_0^\pi  \sin (n u) \sin (m u) du= {a_n\over n \sinh (\pi n)}.$$
Hence, substituting this  result on the right-hand side of (3.20), we end up with (2.37).

\end{proof}

{\bf Example 1}.  Let $\psi(u)= \sin(u).$ Then integral (3.14) can be easily calculated by parts to obtain via (2.1) $f(x)= 2 J(x,i,\pi)/x.$  Hence we find the expansion

$$ J(x,i,\pi)  =  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  J (x, in, \pi)  \int_0^\infty  K_{in} (y) J(y,i,\pi) {dy\over y},\quad x >0.$$

{\bf Example 2}.  Let $\psi $ be the $2\pi$-periodic extension of the function $\psi(u)= u, u \in (-\pi,\pi).$  Then, analogously,  $f(x)= 2 \left(J(x,0,\pi) - \pi e^{-x\cosh(\pi)}\right) /x$  and

$$ J(x,0,\pi) - \pi e^{-x\cosh(\pi)}  =  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  J (x, in, \pi) $$

$$\times  \int_0^\infty  K_{in} (y) \left( J(y,0,\pi) - \pi e^{-y\cosh(\pi)} \right) {dy\over y},\quad x >0.$$

Next we give sufficient conditions of the validity of expansions (2.38), (2.39).

{\bf Theorem 6}. {\it Let $f$ be a complex-valued function on $\mathbb{R}_+$ which is represented by the integral

$$f(x)= \int_{-\pi}^\pi \sin (x \sinh (u)) \varphi(u) du,\ x >0,\eqno(3.21)$$
where $ \varphi(u) = \psi(u)\cosh(u)$ and $\psi$ is a  $2\pi$-periodic function, satisfying the Lipschitz condition $(3.15)$.
Then expansion $(2.38)$ holds for all $x >0$.  Further, if $f$ can be expanded in terms of the series $(3.3)$
of an arbitrary  sequence $\{a_m\}_{m \in \mathbb{N}}$, satisfying condition $(3.1)$, then the expansion $(2.39)$ holds valid. }

\begin{proof} In fact,  proceeding similar to (3.18) and taking into account formulas (2.8), (2.27),  we have for a partial sum $S_N(x)$ of the series in (2.38)

$$S_{N}(x) =  {1\over \pi} \   \sum_{n=1}^N  \int_{-\pi}^\pi \sin (x \sinh (t)) \cosh(t) \sin (n t) \int_{-\pi}^\pi {\varphi(u)\over \cosh(u)} \sin (n u) dt du$$

$$=   {1\over  4\pi} \   \int_{-\pi}^\pi \sin (x \sinh (t))  \cosh(t)   \int_{-\pi}^{\pi}  \left[ \psi(u)- \psi(-u) \right]  \  {\sin \left((2N+1) (u-t)/2 \right)\over \sin( (u-t) /2)}  du dt.$$
Hence the same ideas and justifications as in the proof of Theorem 5 drive us to the equalities

$$  \lim_{N\to \infty}   S_{N}(x) = {1\over  2} \   \int_{-\pi}^\pi \sin (x \sinh (t))  \left[ \varphi(t) - \varphi(-t) \right] dt  =  \int_{-\pi}^\pi \sin (x \sinh (t)) \varphi(t) dt = f(x),$$
which establish  (2.38).   Next, doing in the manner as in the proof of Theorem 5, we substitute series (3.3) in the integral on the right-hand side of (2.39) and  change the order of integration owing to condition (3.1). Then, appealing to equalities (2.8), (2.27), we find

$$  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_{in} (x)  \int_0^\infty K_c(y, in, \pi) f(y) {dy\over y} $$

$$= {2\over \pi}  \lim_{N \to \infty}  \sum_{n=1}^N   \sinh \left({\pi n\over 2}\right)  K_{in} (x)   \sum_{m=1}^\infty   {a_m \over   \sinh \left(\pi m/2\right)}\int_0^\pi  \sin (m u)   \sin (n u) du $$

$$=  \sum_{n=1}^\infty a_n  K_{in} (x)  = f(x).$$
Theorem 6 is proved.

\end{proof}

{\bf Example 3}.  Let $\psi(u)= \sin(u).$ Then integral (3.21) can be easily calculated by parts to obtain via (2.4) $f(x)= 2 \cosh(\pi/2) K_c(x,i,\pi)/x.$  Hence we find the expansion

$$ K_c(x,i,\pi)  =  {2\over \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\pi n)  K_c(x, in,  \pi)  \int_0^\infty  K_{in} (y)  K_c(y,i,\pi) {dy\over y},\quad x >0.$$

Finally we will prove an expansion theorem which concerns equalities (2.40), (2.41).

{\bf Theorem 7.} {\it Let  $f$ have the following series representation

$$f(\tau)= \tau \sinh\left({\pi \tau\over 2}\right)  \sum_{m=1}^\infty   K_s\left(m, i\tau, \sinh^{-1}(\pi)\right) a_m,\quad \tau >0,\eqno(3.22)$$
where the sequence $\{a_m\}_{m \ge 1}$ satisfies condition $(3.9)$. Then the expansion $(2.40)$ holds for all $x >0$. If, in turn, $f$ is given by series $(3.3)$ under condition $\{a_m\}_{m\ge 1} \in l_1$ then expansion $(2.41)$ takes place for all $x > 0$, where the corresponding integral is understood in the Abel   sense $(2.34)$.}

\begin{proof}  Indeed, recalling  (2.28), (2.29) and (3.10), we plug the series (3.22) into  (2.40) to obtain under condition (3.9) the chain of equalities

$$ {4\over \pi^2\ } \  x \sinh\left({\pi x\over 2}\right)   \lim_{N\to \infty}  \sum_{n=1}^N  {1\over n} \ K_s\left(n, ix, \sinh^{-1} (\pi) \right) $$

$$\times \int_0^\infty   \cosh\left({\pi \tau\over 2}\right )  K_{i\tau}(n)  f(\tau) d\tau$$

$$= {2\over \pi^2\ } \  x \sinh\left({\pi x\over 2}\right)   \lim_{N\to \infty}  \sum_{n=1}^N  {1\over n} \ K_s\left(n, ix, \sinh^{-1}(\pi) \right) $$

$$\times \int_0^\infty   \tau \sinh\left(\pi \tau\right )  K_{i\tau}(n) \sum_{m=1}^\infty   K_s\left(m, i\tau,  \sinh^{-1}(\pi) \right) a_m d\tau$$

$$= {2\over \pi } \  x \sinh\left({\pi x\over 2}\right) \lim_{N\to \infty}  \sum_{n=1}^N  {1\over n} \ K_s\left(n, ix, \sinh^{-1}(\pi) \right) $$

$$\times \sum_{m=1}^\infty   m  a_m  \int_0^\pi \cos(mu) \cos(nu) du =  x \sinh\left({\pi x\over 2}\right)  \sum_{n=1}^\infty  K_s\left(n, ix, \sinh^{-1}(\pi) \right) a_n = f(x). $$
In order to prove (2.41) we write under conditions of the theorem and equalities (3.10), (3.12)

$${2\over \pi^2 }     \sum_{n=1}^\infty  { K_{ix}(n) \over n}   \int_0^\infty   \tau \sinh\left(\pi \tau\right )  K_s\left(n, i\tau, \sinh^{-1}(\pi)\right)   f(\tau) d\tau$$

$$= {4\over \pi^2 }     \sum_{n=1}^\infty  { K_{ix}(n) \over n}   \lim_{\alpha \to {\pi\over 2} -}   \int_0^\infty   \tau \sinh\left({\pi \tau\over 2}\right ) \cosh\left(\alpha \tau \right ) K_s\left(n, i\tau, \sinh^{-1}(\pi) \right) $$

$$\times  \sum_{n=1}^\infty  K_{i\tau}(m) a_m   d\tau = {4\over \pi^2 }     \sum_{n=1}^\infty   K_{ix}(n)  \lim_{\alpha \to {\pi\over 2} -}   \int_0^\infty   \cosh\left(\alpha \tau \right ) \int_0^{\pi} \cos(n u)  \cos \left(\tau \sinh^{-1}(u) \right) du $$

$$\times  \sum_{n=1}^\infty  K_{i\tau}(m) a_m   d\tau = {2\over \pi }     \sum_{n=1}^\infty   K_{ix}(n)  \lim_{\alpha \to {\pi\over 2} -}    \sum_{n=1}^\infty  a_m   \int_0^{\pi}  e^{-m\cos(\alpha) (1+u^2)^{1/2}}  \cos(n u) \cos( m u\sin(\alpha))  du $$

$$=   {2\over \pi }     \sum_{n=1}^\infty   K_{ix}(n)   \sum_{n=1}^\infty  a_m   \int_0^{\pi}    \cos(n u) \cos( m u)  du =  \sum_{n=1}^\infty   K_{ix}(n)  a_n = f(x).$$
where all interchanges of the summation,  integration and the passage to the limit are allowed via the absolute and uniform convergence.   Theorem 7 is proved.

\end{proof}

Let us demonstrate a few examples of the absolutely convergent series and their values which are based on expansions (2.38), (2.40).

{\bf Example 4}.  Let $f(x)= e^{-x} x^\alpha,\  \alpha > -1.$  Then the integral in (2.38) is calculated via relation (8.4.23.3) in [3], Vol. III, and we find the formula

$$   \sum_{n=1}^\infty   n  \sinh(\pi n) \left| \Gamma( 1+\alpha+ in)\right|^2  K_c(x, in, \pi) =  \pi^{3/2} 2^\alpha  \Gamma\left({3\over 2} +\alpha\right)  x^{\alpha+1} e^{-x},\quad x >0,$$
where $\Gamma(z)$ is the Euler gamma function [3], Vol. III.

{\bf Example 5}.  Let $f(x)= K_\nu(x) x^{\alpha-1},\  \alpha > |\nu|,\ \nu \in \mathbb{R}.$  Then the integral in (2.38) is calculated by formula  (2.16.33.2) in [3], Vol. II,  and we get the equality

$$   \sum_{n=1}^\infty   n  \sinh(\pi n) \left| \Gamma\left( {\alpha+ \nu+ in\over 2}\right) \Gamma\left( {\alpha- \nu+ in\over 2}\right)\right|^2  K_c(x, in, \pi) =   \pi^{2}  2^{2-\alpha}  \Gamma\left(\alpha\right)  x^{\alpha} K_\nu(x),\quad x >0.$$

{\bf Example 6}.  Let $f(x)= \hbox{sech} (\pi x/ 2).$  Then the integral in (2.40) is calculated by formula  (2.16.48.1) in [3], Vol. II,  and we obtain

$$   \sum_{n=1}^\infty   {e^{-n} \over n} \ K_s \left(n, ix, \sinh^{-1} (\pi) \right)=   {\pi\over x \sinh(\pi x) },\quad x >0.$$

{\bf Example 7}.  Let $f(x)=  x \tanh \left(\pi x/ 2\right) \hbox{sech}\left(\pi x/ 2\right).$  Then the integral in (2.40) is calculated by formula  (2.16.48.14) in [3], Vol. II,  and we derive

$$   \sum_{n=1}^\infty   K_0(n) K_s\left(n,  ix, \sinh^{-1} (\pi) \right)=   {\pi^2\over 4} \hbox{sech}^2 \left(\pi x/ 2\right),\quad  x \in \mathbb{R}.$$

\section{Boundary value problem}

In this section we will employ expansion (2.36) to solve explicitly a Dirichlet boundary value problem in the upper half-plane for the inhomogeneous Helmholtz equation

$$\Delta u - u = h(x,y),\eqno(4.1)$$
where $\Delta$ is the Laplacian,  $(x,y) \in \mathbb{R}_+\times \mathbb{R}_+$, $r= \sqrt{x^2+ y^2}$,

$$h(x,y) =  {2\sinh(\pi)   \over \pi^2 r } e^{- \cosh(\pi) r} \sum_{n=1}^\infty  (-1)^{n+1}  n  \sinh\left( n \cos^{-1} \left({x\over r}\right)  \right)  a_n\eqno(4.2)$$
and $\{a_n\}_{n \ge 1}$ is a suitable sequence.  Writing (4.1) in polar coordinates $(r,\theta),\ r > 0,\ \theta \in [0, \pi]$, it has

$$  {\partial^2u\over \partial r^2}  + {1\over r}  {\partial u\over \partial r} + {1\over r^2}  {\partial^2 u\over \partial \theta^2}-  u$$

$$ = {2 \sinh(\pi)   \over \pi^2 r } e^{- \cosh(\pi) r} \sum_{n=1}^\infty  (-1)^{n+1}  n  \sinh\left( n \theta \right)  a_n.\eqno(4.3)$$

{\bf Theorem 8}. {\it Let  $r >0, \ 0 \le \theta \le \theta_0 < \pi$ and $\{a_n\}_{n \ge 1}$ satisfy the condition

$$\sum_{n=1}^\infty |a_n|\   n^2 e^{\theta_0 n}  < \infty.\eqno(4.4)$$

Then the function

$$u(r,\theta) =  {2\over  \pi^2} \   \sum_{n=1}^\infty   n  \sinh(\theta n)  J (r, in, \pi) a_n\eqno(4.5)$$

is a  solution of equation $(4.3)$, vanishing at infinity.}

\begin{proof}  Indeed, it follows immediately, substituting (4.5) into (4.3) and  fulfilling the differentiation under the series sign due to the absolute and uniform convergence, which is allowed   by virtue of assumption (4.4) and the  estimate  $|J (r, in, \pi) | \le 2 e^{-r}/ n$ in Remark 1.  Then, appealing to the differential equation (2.6) for the incomplete modified Bessel function, we see that $u(r,\theta)$ is a solution of (4.3), which tends to zero  when $r \to \infty$.

\end{proof}

 {\bf Theorem 9}. {\it Let the sequence $\{a_n\}_{n \ge 1}$ under assumption $(4.4)$ be the Kontorovich-Lebedev transform of discrete variable

 $$a_n= \int_0^\infty K_{in} (x) f(x) {dx\over x},\quad n \in \mathbb{N},\eqno(4.6)$$
where $f(x)/x,\ x >0$ satisfies conditions of Theorem $5$. Then  $u(r,\theta)$ given by formula $(4.5)$ solves the inhomogeneous Helmholtz equation $(4.3)$ in the interior of the upper half-plane $(r,\theta),\ r >0, \ 0 < \theta < \pi$, vanishes at infinity  and on the real axis  takes the following prescribed values}

$$  u(r, 0) = 0,  \quad\quad    u(r,\pi) = f(r).\eqno(4.7)$$

\begin{proof} The proof follows from  the previous theorem.  To verify boundary conditions (4.7), we appeal to (4.6), Theorem 5 and expansion (2.36). Moreover, $u(r,\theta)$ is continuous on the boundary. In fact, it follows from the uniform convergence with respect to $\theta \in [0,\pi]$ of the series (4.5) by virtue of the Abel test.

\end{proof}

\bigskip
\centerline{{\bf Acknowledgments}}
\bigskip

\noindent The work was partially supported by CMUP, which is financed by national funds through FCT (Portugal)  under the project with reference UIDB/00144/2020.   The author is sincerely indebted to  referees for them  careful reading of the manuscript and constructive comments and suggestions   that greatly improved its form and content.

\bigskip
\centerline{{\bf References}}
\bigskip
\baselineskip=12pt
\medskip
\begin{enumerate}

\item[{\bf 1.}\ ]  D.S. Jones, Incomplete Bessel functions. I,  {\it Proc. Edinb. Math. Soc.} {\bf 50} (2007), N 1,  173-183.

\item[{\bf 2.}\ ] N.N. Lebedev,   Sur un formule d'inversion,  {\it C.R. (Doklady) Acad. Sci. URSS (N.S.)} {\bf 52} (1946),  655-658 (in French).

\item[{\bf 3.}\ ] A.P. Prudnikov, Yu.A. Brychkov and O.I. Marichev, {\it Integrals and Series}. Vol. I: {\it Elementary
Functions}, Vol. II: {\it Special Functions}, Gordon and Breach, New York and London, 1986, Vol. III : {\it More special functions},  Gordon and Breach, New York and London,  1990.

\item[{\bf 4.}\ ] S. Yakubovich, {\it Index Transforms}, World Scientific Publishing Company, Singapore, New Jersey, London and
Hong Kong, 1996.

\end{enumerate}

\vspace{5mm}

\noindent S.Yakubovich\\
Department of  Mathematics,\\
Faculty of Sciences,\\
University of Porto,\\
Campo Alegre st., 687\\
4169-007 Porto\\
Portugal\\
E-Mail: syakubov@fc.up.pt\\

\end{document}
\end

redpajama_set_name: RedPajamaArXiv


text:
IHG: Introduction to Tools for Life and Web Accessibility - February 2020
Revision as of 14:20, 18 February 2020 by Vholder3 (talk | contribs)
Tools for Life: Your Assistive Technology Act Program! and Digital Accessibility and User Experience
InterContinental Hotels Group, web designers and architects
Tools for Life (TFL) is funded by the Federal Tech Act, which is legislation to promote direct AT services to individuals of all ages and disabilities in each state. Tools for Life increases independence for Georgians with disabilities by providing access to and acquisition of assistive technology devices and services. The TFL Network is fulfilling the intent of the Tech Act through AT assessments, AT lending libraries, device demonstrations, trainings for individuals and groups, AT reuse, and funding education and assistance. This presentation explores information about the Tools for Life program, network and community partners including funding resources that assist with purchasing assistive technology devices.
The Center for Inclusive Design and Innovation (CIDI) provides services in the area of digital accessibility evaluations, trainings and technical assistance to organizations across the street and around the globe. Accessibility in this case means "easily used or accessed by people with disabilities." A review of digital accessibility standards, including the Web Content Accessibility Guidelines (WCAG), was also provided to IHG, along with a demonstration of the IHG website from the perspective of a screen reader user.
Tori Holder, Outreach Specialist, Tools for Life
John Rempel, QA Accessibility Analyst, Center for Inclusive Design and Innovation
Resources for You!
PDF Presentation
Visit the Tools for Life website to learn more about Georgia's assistive technology services.
Holder, T. and Rempel, J. (2020) Tools for Life: Your Assistive Technology Act Program! and Digital Accessibility and User Experience. Presented at the Center for Inclusive Design and Innovation, Atlanta, GA, 2020.
Retrieved from "http://gatfl.gatech.edu/tflwiki/index.php?title=IHG:_Introduction_to_Tools_for_Life_and_Web_Accessibility_-_February_2020&oldid=5015"

redpajama_set_name: RedPajamaCommonCrawl


text:
AIIB and BRICS Bank: Complements or Competition?
Leaders of the 5 BRICS founding member countries meet in Brazil.
The Asian Infrastructure Investment Bank (AIIB) is emerging at a time when China's economic and diplomatic ties are strengthening throughout the western world. Not only has it recently launched its own infrastructural institution in AIIB, but in 2011, China co-founded the New Development (NDB), formerly known as the "BRICS Development" Bank. These advancements force us to consider whether new, non-western institutions like AIIB and BRICS are meant to complement, or compete, with one another.
The birth of new regional lending institutions, like AIIB and NDB, was provoked partly by the US Congress' refusal to reallocate IMF voting shares in 2012. Conceived out of a desire for a diplomatic status equivalent to their growing economic clout, the emerging economies of Brazil, Russia, India, China, and South Africa coalesced in 2012 and founded the NDB. Aimed at mobilizing "resources for infrastructure and sustainable development projects in emerging and developing economies," BRICS offers alternatives to American and European-led lending institutions like the International Monetary Fund (IMF) or World Bank. While seemingly diverse, we must now speculate whether these distinctly non-western projects will support one another or increase development competition.
NDB's launch took considerably more time than AIIB's. First proposed in 2012, the bank took just over three years to launch, celebrating its official establishment in July 2015. AIIB, however, was first conceived in 2014 and will launch in early 2016. The five founding members of BRICS endured three years of negotiations in order to consolidate their structure. Contrastingly, AIIB's unilateral founding has enabled its speedier, year and a half formation. Perhaps indicative of superior organization and efficiency, critics worry that AIIB may outpace the development of BRICS.
This concern is furthered by the comparison of voting shares between the two banks. Within BRICS, all five founding members have contributed an equal share of $10 billion out of its $100 billion in assets. In AIIB, China easily dominates, having contributed nearly $30 billion of its $100 billion capital base. This in turn, positions China with the greatest number of voting shares held by AIIB member countries. Critics caution that China will possess considerably more voting power than its fellow member states, and will utilize that power in determining the organizational structure of AIIB.
Though analysts hypothesize that AIIB and NDB may compete, others suggest that the two institutions are meant to complement one another. Aimed at alleviating the gaps in Asia's infrastructural needs, the two banks could benefit from cooperation. Aligned with the NDB, AIIB has potential to make a greater contribution within international debate. Although there is skepticism that China will not be able to comply with human rights or environmental regulations, supporters of NDB emphasize that the influence of fellow emerging economies, like Brazil or Russia, will motivate China to recognize western standards.
Alongside emerging markets that are similar in economic and diplomatic statures, BRICS and AIIB may act independently of western lending institutions. Analysts stress that cooperation between AIIB and NDB could even break the 'monopoly' held by the IMF and World Bank. Proponents hope that potential partnerships between AIIB and NDB will compel western-led institutions to act more democratically. While these banks formed in response to the unequal allocation of voting shares in western-led institutions, they may still desire to reform voting powers, and thereby further integrate emerging markets into the international order.
Gabrielle Torres is an undergraduate studying Government and Spanish at the University of Texas at Austin.


Product prices and availability are accurate as of 2019-04-18 18:28:42 BST and are subject to change. Any price and availability information displayed on http://www.amazon.com/ at the time of purchase will apply to the purchase of this product.
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I would like to render subset of OpenStreetMap features over big area into one image, no sub-tiles.
Does Envelope() match the .osm coordinates? Yes.
Any ideas what I made wrong? How are you rendering one-big-image maps?
If my way is completely wrong, please suggest better way. The ultimate goal is to render for example all roads and all rivers into one big image for A1 poster.
I strongly recommend you to look at QGIS for this type of work (https://qgis.org). A GIS (Geographic Information System) like QGIS is actually designed for this type of work: to style and layout large maps and output them to an image format or PDF for printing purposes.
Most tools in the OpenStreetMap world, are not geared to this kind of work, but solely create tiled output for webservices, which means that putting them to work for printing purposes, always results in a more or less hackish solution.
QGIS does not suffer from these drawbacks if you use it to access the PostGIS data directly using QGIS's PostGIS database connection options, as you will be accessing and styling the vector data in your OpenStreetMap database directly in that case. It should allow you to relatively easily style and output your desired map to the custom format you desire.
Not the answer you're looking for? Browse other questions tagged postgresql openstreetmap mapnik linux or ask your own question.


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It's rare for us to have a grappling/BJJ tournament right in our own backyard. More often than not we end up driving a team either to NJ or down to places like Virginia Beach and Florida.
We got lucky this time because The Good Fight Tournament held it's event right in Severn Maryland, a stone's throw away from us here in Arnold.
The Good Fight is a smaller, local tournament circuit that holds events most on the upper east coast. We've been sending students to compete for a while and we know the guys who run it pretty well.
This particular tournament featured matches in the "No Time Limit" format. That means matches go until someone gets the tap out. Some of these matches can last for hours, I've seen it many times.
Competing under these rules can be extra challenging and good cardio is a must for success. Good cardio and for many it also means making adjustments to your game plan and match tactics.
Thankfully our students have always thrived in this type of submission only tournament. The style we teach here at IL seems to translate well and because of this, our winning percentage for submission only events is really high.
Mr. Garlington snagged another pair of gold medals to add to his growing collection.
Everyone had fun and did well, which is always a plus. For students like Hannah and CJ, just being able to get that experience competing in front of a crowd makes it totally worth the time.
For others, like Micheal "Mouth" Garlington, it's just another day at the office and one more tournament leading up to Pan-Ams in March. No matter what, seasoned competitor or new to the tournament scene, great times were had.
The standout performance for the day was without a doubt 14 year old Magnus "Magneto" Benz.
Magnus looked like his old self out on the mats, winning match after match using an in-your-face, aggressive style that when combined with his solid technical skills, was too much for his opponents to handle.
I'd like to point out that he won the adult white belt division, competing against grown men in their 20's and 30's, at just 14 years old. That's pretty impressive for a kid his age to pull off.
Next up for Magnus will be either Pan-Kids in California or NAGA NJ. I'd really like to see the Magnus I saw at Good Fight make a return in the very near future..stay tuned.
Next up for Ivey League is the Atlanta Open on February 4th and on the same day, the PA NAGA. We follow that up with the most high level kids tournament in the world, Pan-Kids.
Be sure to check back to this blog for the latest news and events related to IL.
Hannah Goldman Came Back With a Bronze Medal in Her First Grappling Tournament.


The year-end brings no greater pleasure than the opportunity to express to you season's greetings and good wishes. On behalf of all of us at GlobalVision International, Inc., may the peace and joy of the holiday season be with you throughout the coming year. Here are a few Localization Lessons to recap over the past year.
As we put the year behind us, it is important that we reflect upon the localization lessons acquired throughout the year before we forge ahead into the future.
In this article, we discuss ten New Year's localization resolutions that you can make – and keep – to ease your localization efforts throughout 2008 and beyond. Click here to read them.
Most content publishers, software or other, perform updates on their product, manuals, website, and marketing literature yearly or on a more regular basis. They update their source language files and engage their localization vendor or staff to update all the supported target languages.
Changes to the source files take the shape of additions (new text), removal of obsolete text (deletions), or edits (modifications to existing text). The new text requires new translations for each target language. The deleted text is disregarded. The edited or modified text will require updating in all target languages.
How can you objectively account for the additional work that is needed to budget and plan for it? Click here to get your answer.


text:
Teacher who gave sedative pills to students fired
Menatalla Ibrahim
An investigation was launched last week following a parent's complaint to authorities after it emerged that students at a private school in Qatar were given drugs by a teacher without…
Schools open up as Qatar drops blended learning system
The decision comes as Qatar sees a significant drop in Covid-19 cases that has prompted the country to ease restrictions and measures. The Ministry of Education and Higher Education has…
Arabic, Islamic studies now compulsory subjects for private schools in Qatar
Qatar's Ministry of Education and Higher Education announced that Arabic and Islamic education is now compulsory in private schools. Arabic, and Islamic studies are now mandatory subjects at all private…
Schools limited to 2% price hike, ministry official says
Applications to increase prices at several schools were rejected, officials announced. Qatar's ministry of education confirmed fee hikes for private schools are limited to 2%, restricting several institutions from substantially…
Schools prepare for exam season with stricter Covid-19 measures
The Ministry of Education and Higher Education will ensure that schools adhere to precautionary measures as exam season begins. With midterm exams for students from grades 1 to 11 starting…
Ministry of Education releases guidelines for exams
Sana Hussain
Public school exams to last 16 days Authorities have issued their guidelines on how exams should be conducted in public schools across Qatar as the end of the first semester…
Junk food banned from Qatar's independent school canteens
Victoria Scott
Banned foods include chocolate, chips, fried foods, processed meats and sugary drinks, said Hamad al Rumaihi, Chief of Canteens at the Ministry of Education and Higher Education.
Qatar school removes textbook calling Palestinians terrorists
Lesley Walker and Riham Sheble
The International School of Choueifat (ICS) in Onaiza was ordered to remove all copies of the book and to reimburse students for its cost, the Ministry of Education and Higher Education said yesterday.
Qatar's Education Ministry denies banning abayas in schools
An online furore erupted earlier this week following the publication on Twitter of a notice issued by state-funded primary school which said it had been ordered by the ministry to ban the abaya on campus.


Testimony before the US Senate Judiciary Committee on Immigration, Border Security, and Citizenship.
"Mr. Chairman, Senator Kennedy, Members of the Subcommittee.
President Bush has challenged us to fix our broken immigration system. In presenting his vision, he has demonstrated remarkable political courage and a willingness to think outside of the box that has been unmatched by any other president in recent memory—and has created an opening for the US Congress and all of us to be equally thoughtful and courageous.
The task is daunting. We are asked to re-engineer an immigration system riddled with bad or no longer relevant ideas and weighed down with the cumulative inconsistencies of more than a century of political expedience and "delicate" compromises. Furthermore, we all understand that we must do so in ways that are both politically feasible and can be administered with relatively few resources and little taste for regulation and enforcement. Finally, we must do so in a climate that is ambivalent at best—both about immigration and about immigrants. A tall order indeed, but one to which we must nonetheless respond. [...]"


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Gregory of Tours
b. November 30, AD 538 – d. November 17, AD 594 • Bishop • Historian
The Oxford Dictionary of the Christian Church
Gregory of Tours, St
Gregory of Tours, St (538/9–94), Bp. of Tours and historian of the Franks. He came of a Gallo-Roman senatorial family which had already produced several bishops, and was elected Bp. of Tours in 573. Though he had at first to oppose King Chilperic, he readily supported the 'good king', Guntram, against
Ecclesiastical Dictionary: Containing, in Concise Form, Information upon Ecclesiastical, Biblical, Archæological, and Historical Subjects
Gregory of Tours (St.)—Bishop and historian. Was born of a noble family at Clermont, in the Province of Auvergne, in 539. Members of his father's and mother's families had held high offices in both Church and State. His education was directed by his uncle, St. Gall, Bishop of Clermont, and by Avitus,
Encyclopedia of Ancient Christianity
GREGORY of Tours (ca. 538–594). George Florentius, later called Gregory after one of his forefathers, was born around 538 in Clermont-Ferrand into a family of Roman origin. He entered an ecclesiastical career and was bishop of *Tours from 573 until his death (15 November 594). He was deeply involved
Who's Who in Christian History
Gregory of Tours (c. 538–594)Bishop of Tours; historian of medieval Gaul (France)Born Georgius Florentius, he took the name Gregorius Turonesis ("of Tours"). From his birth in Clermont-Ferrand, he was associated with ecclesiastical life. After his father died, Gregory's mother moved to Cavaillon, where


Redwood Shores, Calif. -- Closely following the recent alliance with Wingcast, and commitment to integrate wireless and voice access across all product lines, Oracle Corp. (NASDAQ: ORCL) today shipped the mobile-enabled Oracle(r) E-Business Suite. Effective immediately, customers can access key e-business applications -- such as field sales, field service, customer intelligence, financials, supply chain and warehouse management -- from the desktop, the Web and a variety of mobile and wireless devices, such as Palms, Web-enabled phones, smartphones and other PDAs. In addition, Oracle announced that it is working with strategic partners, including Compaq, Motorola, Palm and others, to deliver solutions for the mobile enterprise that encompass everything customers need to get up and running, including software, hardware, devices and consulting services. Additionally, such solutions can be delivered to customers using the Mobile Blueprint framework through Oracle's strategic partnership with Motorola.
The mobile-enablement of the Oracle E-Business Suite is the final component of Oracle's strategy to offer mobile access throughout a complete, integrated set of technology and applications products and services. By delivering on this strategy, Oracle enables companies to provide customers, employees and partners with mobile and wireless access to critical business applications from any interface -- Web, wireless or voice -- thereby helping improve productivity and ultimately reducing the cost of doing business.
Mobile-enabled Enterprise Applications: An Underserved Market With the mobile CRM market alone expected to grow to $1.7 billion by 2005 according to Datamonitor, the mobile enterprise applications market represents a significant growth opportunity for Oracle. One of the first customers to embrace mobile information access across the enterprise is Kyocera Wireless Corp. Kyocera plans to provide employees with access to Oracle E-Business Suite applications such as CRM, Financials and Manufacturing on the Kyocera QCP 6035 smartphone.
"At Kyocera Wireless Corp., we firmly believe that companies can realize significant benefits by accessing enterprise applications from devices such as the Kyocera QCP 6035 smartphone," said Masahiro Inoue, president and CEO of Kyocera Wireless Corp., a leading manufacturer of wireless handsets. "We plan to be one of the first companies to combine the QCP 6035 with the mobile-enabled Oracle E-Business Suite across our company to make our employees significantly more efficient and improve our business processes."
"The enterprise market for mobile business applications is an untapped growth opportunity," said Stephen Drake, research manager of IDC's Wireless and Mobile Enterprise Access service. "The winners in this market will include those suppliers who deliver comprehensive mobile solutions addressing critical technology and application needs, leverage businesses' existing technology resources and provide improved efficiency to the mobile workforce."
Oracle E-Business Suite Goes Mobile The Oracle E-Business Suite leverages the integrated wireless functionality of Oracle9i Application Server to provide mobile access to key business applications. As a result, companies using the mobile-enabled Oracle E-Business Suite can benefit from the enhanced capabilities and increased productivity that come from mobile functionality such as location, personalization, messaging and notification-key features of Oracle9iAS that can be leveraged across the entire suite of applications. Mobile access is immediately available to key applications such as field sales, field service, customer intelligence, property manager, Internet expenses, supply chain and warehouse management.
In addition to the immediate availability of mobile access to the applications listed above, additional applications of the Oracle E-Business Suite -- such as asset management, treasury management, receivables, time management -- and vertical applications for healthcare and pharmaceuticals, are expected to be mobile-enabled by early 2002.


Once you've completed the college's general admissions process, you can register. The Registrar's office assists you with registration, graduation and requests for transcripts and other student-related records. The Veterans Services office can assist with these actions on a case by case basis.
If you have a complete admissions file, you may register online or through your advisor on campus.
A copy of your course schedule must be provided to the Veterans Services office.
If you have fewer than 24 hours of credit, you must see an advisor before registering for classes.
Chapter 30 – Student payment by payment deadline is required.
Chapter 31 - Veterans Service Office will notify Cashier that VA payment is anticipated.
Chapter 33 - Veterans Service Office will notify Cashier that VA payment is anticipated.
Chapter 35 – Student payment by payment deadline is required.
Chapter 1606 – Student payment by payment deadline is required.
Hazlewood – Veteran Services office will notify Cashier that tuition exemption is authorized. Students will be required to pay the student services fee for each semester enrolled while using the Hazlewood Act Exemption. The current rate is $7 per semester.
Penalty Fees – Late registration, returned checks, repeated courses and any other fees that are not required of every student.
Each semester that you have registered for classes you must come to the Veteran Services office to request certification of your enrollment. Your course schedule can be printed at the Veteran Services office. The request for certification form or Hazelwood application will be completed and submitted at this time. If you make any changes to your course schedule after the Veteran Services office has processed your benefits, you must notify the Veteran Services office immediately to avoid any overpayments or debts that may be generated as a result of these changes. It is recommended that you contact the Veteran Services office prior to making any changes to your schedule.
Any time your course schedule changes you must immediately notify the Veteran Services Office. Title 38, Code of Federal Regulations, Section 21.4203 requires that all changes in enrollment credit must be reported in a timely manner. Reducing the number of semester hours you are attempting within a given semester may cause a decrease in your VA benefit amount and could generate an overpayment from the VA. If overpayment occurs, you will become indebted to the VA and your future benefit payments could be reduced by the amount of the overpayment.
If you are a first-time student who entered college in Fall 2007 or after, you cannot drop more than six courses, including any course you have dropped at another college in Texas, according to Senate Bill 1231. There are exceptions to this policy if you can show good cause for dropping a course(s). For further information, contact the Registrar's Office.
You should be aware that dropping courses may affect your eligibility for financial aid. You should contact Financial Aid Services prior to dropping a course or completely withdrawing from school. If you stop attending class without officially withdrawing from the College, then the grade is an automatic "F." Students receiving Veterans Benefits for education should contact the Veteran Services Office for specific policies concerning drops and withdrawals. These changes may have a direct effect on your VA benefits.
If you are receiving veteran's educational benefits, you must make satisfactory academic progress by maintaining a 2.0 grade point average (GPA) each enrollment period.
The first term you fall below a 2.0 grade point average will result in your being placed on probation. If you fail to meet the minimum standard the next semester of enrollment, you will be placed on suspension and unable to utilize your VA education benefits. Failure to achieve the required GPA will cause the benefits to be terminated until the GPA is satisfactory.
To regain eligibility, you must register for at least half-time enrollment in the next semester that you attend at your own expense and earn at least a 2.0 grade point average. By earning a 2.0 GPA, you will be removed from suspension and reinstated eligibility to your VA education benefits.
In accordance with Texas Administrative Code Title 40 Part 15 Chapter 461 Subchapter A, students using the Hazlewood Act Exemption must meet the GPA requirement of the institution's satisfactory academic progress (SAP) policy in a degree or certificate program as determined by the institution's financial aid policy, except that they are not required to enroll in a minimum course load.
Students who wish to utilize the Hazlewood Act Exemption but do not meet the SAP requirements as indicated above, are advised to contact the Financial Aid Department to obtain a SAP appeal package.
The above provisions do not apply to a spouse or child of a veteran who has died from a service-related injury or illness or who was classified as missing in action (MIA) or killed in action (KIA).
Important note: VA Education benefits do not pay out of state/non-resident tuition. In accordance with Texas Education Code Section 54.241 (k) A person is entitled to pay tuition and fees at an institution of higher education at the rates provided for Texas residents without regard to the length of time the person has resided in this state if the person files with the institution at which he/she intends to register a letter of intent to establish residence in this state and resides in this state while enrolled in the institution and the person is eligible for Post 9/11 VA education benefits or any other federal law authorizing educational benefits for veterans.
In order to qualify for in-state and in-district tuition, if you are a military service member or dependent who is a resident of another state but assigned in the State of Texas or if you are a Texas Resident assigned outside of the state of Texas, you must submit a letter from the commanding officer or from the individual assigned to handle such duty, verifying Texas as the state of duty station. This verification must be submitted once per year to Del Mar College on or before registration. Forms are available from the Veteran Services Office, the Student Enrollment Center and the Registrar's Office.
Page last updated February 2, 2018.


text:
Firefly: The Train Job
Mal: "They tell you never hit a man with a closed fist. But it is, on occasion, hilarious."
Very exciting, frenetic episode, full of action as well as exceptionally clever lines. The opening Unification Day barfight in particular was so much fun that I wanted to quote three-quarters of the dialogue. But this episode was also very much about moral choices, an important concept to establish when your heroes are outlaws.
Mal was so much lighter here than in the pilot, much more quippy -- in the barfight, on the train job, even the Indiana Jones-like scene of kicking Crow into the manifold. Mal and Zoe bounced wonderful lines off each other during the train job scenes. And he and Inara did some interesting semi-hurtful bantering, while still managing to convey that they cared about each other.
But it was Jayne who stole the show. He was utterly hilarious, menacing, and endearing at the same time. Jayne really does think that Mal will eventually turn River in for the money because he can't imagine anyone not doing that. And yet, half-conscious on the stair landing, Jayne still joined in the fight and saved the day. You might say he's a bit of a contradiction.
He's also not that bright, as evidenced by his vocabulary issues. In the bar fight, BadAss said "ass-picious"; Jayne thought he meant "suspicious" instead of "auspicious." He referred to "kosherized" rules, and called Simon "fancible." I particularly loved the scenes where Jayne was drugged, both in the control room and on the stair landing. When he was trying to catch the little angels he saw in the air, I immediately thought of fireflies.
Inara rescued Mal and Zoe from the botched train job, but Simon showed unexpected boldness and intelligence by rescuing everyone else from Jayne. The fact that Simon gave up a great deal to save his sister was emphasized, as well as how hard it was to take care of her because she was "whimsical in the brainpan." We also saw some very nasty Clockwork Orangy flashbacks to River being tortured. She kept saying, "Two by two, hands of blue," and son of a gun, two guys wearing blue gloves showed up at the end. Stay tuned.
There was way too much exposition for a regular episode, repetition of what we learned about the characters in the pilot. Yes, I know the reason for that: this episode originally aired as the quicker and funnier pilot instead of the longer and more serious "Serenity." Doesn't matter about the repetitious exposition, though, because "The Train Job" is still terrific.
-- Serenity is such a friendly ship. Easy to maneuver, like a big honking helicopter. And it doesn't have big guns. River said, "Midbulk transport. Standard radion accelerator core, class-code 03-K64. Firefly." Just so we would know.
-- Loved the holographic bar window.
-- Again, we got the multicultural mix, with the belly dancer and the bar waitresses in kimonos. And the train was a mix, too: wildly modern on the outside, but like a spaghetti western on the inside.
-- It was established that Inara can choose her clients, and that appearance isn't the most important thing to her.
-- Dan noticed that the Alliance soldiers were wearing the same costumes as in Starship Troopers. In the commentary, Joss Whedon mentioned this and confirmed that they were the costumes from Starship Troopers; they had no money to design new costumes.
-- River pointed out that "Mal" means "bad" in Latin. I knew that.
-- Space monkeys?
-- There was more mention of God. Mal told Shepherd: "You're welcome on my boat. God ain't." But Inara told Shepherd to pray for Mal, and implied that she does it herself. "Don't tell him. I never do."
-- One of the blue hand guys was another Buffyverse alumnus.
Quotable quotes, and there were a lot of them in this episode:
Bad Ass: "Six years today, the Alliance sent the browncoats running, pissing their pants. (pause) You know, your coat is kind of a brownish color..."
Mal: "It was on sale."
Mal: "This is why we lost, you know. Superior numbers."
Zoe: "Thanks for the re-enactment, sir."
Mal: "Drunks are so cute."
Inara: "What did I say to you about barging into my shuttle?"
Mal: "That it was manly and impulsive?"
Simon: "So what are we doing?"
Kaylee: "Crime."
Kaylee: "No call to be snappy, Jayne."
Jayne: "Are you about to jump onto a moving train?"
Jayne: "You know what the chain of command is? It's a chain I go get and beat you with 'til you understand who's in ruttin' command here."
Inara: "Zoe, what would your husband say if he knew you were here?"
Zoe: "I... I was weak." It was Gina Torres' expression that made this one work.
Four out of four stars,
Billie Doux loves good television and spends way too much time writing about it.
Firefly reviews
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Tags: Billie Doux, Firefly
ChrisB Friday, March 2, 2012 at 9:24:00 AM EST
This is fun! And it never hurts when your heroes have such a strong moral compass -- it even makes it all right for them to throw someone out into space.
Love all the characters, but I agree with Billie that Jayne is the most interesting to watch. Yes, he's a bit dim and a bit rough, but I get the strong sense that he is the hero in his own mind.
What Whedon does well is write strong women characters. All the women on this ship are smart and capable, but still women. I can think of more than one show currently on the air that could learn from this.
And, how do you not love Mal? One minute kicking ass, the next whining about getting stitched up. Smile!!!
Josie Kafka Wednesday, July 30, 2014 at 2:13:00 AM EDT
Yes, I know the reason for that: this episode originally aired as the quicker and funnier pilot instead of the longer and more serious "Serenity."
Netflix streaming includes the very weird voiceover saga sell at the beginning of this episode; I'd never heard it before, and I'm not sure it would have lured me in if I were watching this live.
While I like this episode--I like all the episodes--this might be my least favorite. In places it sells itself too hard, especially compared to the more subtle real pilot. But it's still fun, especially little moments, like no one caring that Simon drugged Jayne.
sunbunny Tuesday, January 26, 2021 at 4:49:00 AM EST
This episode was mandatory. When trying to get us to root for criminals, you have to establish that they aren't bad people. At least most of them aren't (Hi Jayne!).
What a terrible pilot though. I'm fortunate enough to have watched the show in its intended order not its idiotic run order so now the info dumps just seem a bit strange and out of place instead of vital information as to what the hell was going on. I'm watching on Hulu this time around (too lazy to find the DVDs) and am experiencing the saga sell for the first time. Totally insufficient introduction to the Fireflyverse. I get that this was before streaming and even probably DVR and networks wanted nice, episodic television but jeez. Well, it's not like Fox is new to being evil.


text:
Smoking in Colombia was banned in May 2008 in public transport vehicles, hospitals, kindergartens and schools, bars, clubs, restaurants, business centers and airports. Smoking areas in enclosed public places were also banned.

Prevalence
There are over 5 million smokers in Colombia. Teenage smokers account for nearly a third of smokers, and their numbers continue to rise. The ministry of social security and health reported that diseases related to smoking killed an average of 68 Colombians per day before the implementation of the ban.

References

Health in Colombia

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text:
HomeTV ShowsAmerican IdolMonday Morning Mediabase 09/25/2017
Monday Morning Mediabase 09/25/2017
09/25/2017 Kirsten American Idol, Charts, X Factor, X Factor UK 3
Eight milestones this Monday Morning to report.
Niall Horan becomes the second One Directioner to top the Pop charts. He does this with his second single "Slow Hands" from his debut album "Flicker" (out October 20th). He's likely to be followed by Liam Payne whose "Strip That Down" (featuring Quavo) is firmly ensconced at second on that chart. "Slow Hands" is also currently top 4 on HAC and top 20 on AC. The song also added a new format this week, Rhythmic.
Who was the first One Directioner to hit the top of the Pop chart? Zayn with his debut single "Pillowtalk". Zayn has his own milestone this week as his lead single for his second album climbs onto the HAC chart. "Dusk Till Dawn"'s appearance on the chart marks the first time all 5 One Directioners have been on the same airplay chart. Also there are Niall Horan with "Slow Hands", Liam Payne with "Strip That Down", Harry Styles with "Two Ghosts" and Louis Tomlinson with "Back To You".
Speaking of X-Factor Bands, Fifth Harmony is back on the Pop charts. "He Like That" is their second single from their "Fifth Harmony" album released in August.
Kelly Clarkson is our second artist to have two milestones this week. "Love So Soft", debut single from her upcoming album, just keeps knocking down these milestones each week since its release. This week it goes top 30 on Pop and top 20 on AC.
Also from American Idol is Mandisa. "Bleed the Same" is the third single released from her fifth studio album. This week, it goes top 40 on CAC.
Finally, we have Survivor Chase Rice. His "Three Chords & The Truth" make it onto the Country charts just ahead of his album release on September 29th.
Billboard reports that the Foo Fighters have their second number one album and BTS becomes the first K-Pop group to score a US top 10 album.
1 (NEW) Foo Fighters "Concrete and Gold" 127K, 120K sales
2 (4) Lil Uzi Vert "Luv is Rage 2" 50K (-11%)
3 (1) Tomas Rhett "Life Changes" 39K (-68%)
4 (8) Kendrick Lamar "DAMN" 34K (-2%)
5 (6) XXXTentacion "17" 33K (-17%)
6 (9) Khalid "American Dream" 31K (+1%)
7 (NEW) BTS "Love Yourself and Her" 31K, 18K in sales, 4K in TEA, 10K in SEA
8 (13) Imagine Dragon "Evolve" 28 (+8%)
9 (NEW) Big and Rich "Did it for the Party" 27K , 26K in sales
10 (17) Post Malone "Stoney" 26K (+24%)
Camila Cabello:
"Havana" featuring Young Thug: ^31 Pop (39)
featured with Travis Scott and Quavo on Major Lazar's "Know No Better": ^26 Pop (27); ^32 RHY (33)
featured on Keith Urban's "The Fighter": 45 HAC (42); ^12 AC (12)
Chase Rice:
"Three Chords & The Truth": ^48 Country (51)
Danny Gokey:
"The Comeback": 10 CAC (10)
Fifth Harmony:
"He Like That": ^42 Pop (53)
Harry Styles:
"Two Ghosts": ^32 Pop (31), ^32 HAC (36)
James Arthur:
"Can I Be Him": 35 HAC (26)
"Say You Won't Let Go": 5 AC (3)
Kelly Clarkson:
"Love So Soft": ^27 Pop (32); ^13 HAC (13); ^20 AC (24)
Lauren Alaina:
"Doin' Fine": 43 Country (47)
featured on "What Ifs" by Kane Brown: ^5 Country (5)
Liam Payne:
"Strip that Down" featuring Quavo: ^2 Pop (4); ^15 HAC (14); 13 RHY (11)
Louis Tomlinson:
"Back to You" featuring Bebe Rexha & Digital Farm Animals: 41 Pop (29); 34 HAC (31)
Lindsey Stirling:
"Love's Just a Feeling" f. Rooty: 31 AC (25); ^42 HAC (47)
Mandisa:
"Bleed the Same": ^34 CAC (46)
Niall Horan:
"Slow Hands": ^1 Pop (2); ^4 HAC (4); ^19 AC (19); ^49 RHY (52)
Phillip Phillips:
"Miles": ^21 HAC (21); 47 AC (47); ^25 AAA (27)
Scotty McCreery:
"Five More Minutes": ^35 Country (39)
Zayn:
"Dusk Till Dawn" featuring Sia: ^41 HAC (65)
Upcoming Adds:
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Stats collected on Monday Morning.
About Kirsten 3052 Articles
Kirsten has had a long love affair with numbers. Marry that with her love of cheese and the Numbers Threads at MJs were born.
The Voice's Meghan Linsey Kneels After Singing National Anthem (VIDEO)
Headlines: The Voice 13 Premieres, Kelly Clarkson Behind the Scenes


text:
/*************************************************************************/
/*  json.h                                                               */
/*************************************************************************/
/*                       This file is part of:                           */
/*                           GODOT ENGINE                                */
/*                      https://godotengine.org                          */
/*************************************************************************/
/* Copyright (c) 2007-2021 Juan Linietsky, Ariel Manzur.                 */
/* Copyright (c) 2014-2021 Godot Engine contributors (cf. AUTHORS.md).   */
/*                                                                       */
/* Permission is hereby granted, free of charge, to any person obtaining */
/* a copy of this software and associated documentation files (the       */
/* "Software"), to deal in the Software without restriction, including   */
/* without limitation the rights to use, copy, modify, merge, publish,   */
/* distribute, sublicense, and/or sell copies of the Software, and to    */
/* permit persons to whom the Software is furnished to do so, subject to */
/* the following conditions:                                             */
/*                                                                       */
/* The above copyright notice and this permission notice shall be        */
/* included in all copies or substantial portions of the Software.       */
/*                                                                       */
/* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,       */
/* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF    */
/* MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.*/
/* IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY  */
/* CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT,  */
/* TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE     */
/* SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.                */
/*************************************************************************/

#ifndef JSON_H
#define JSON_H

#include "core/object/ref_counted.h"
#include "core/variant/variant.h"

class JSON : public RefCounted {
	GDCLASS(JSON, RefCounted);

	enum TokenType {
		TK_CURLY_BRACKET_OPEN,
		TK_CURLY_BRACKET_CLOSE,
		TK_BRACKET_OPEN,
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		TK_IDENTIFIER,
		TK_STRING,
		TK_NUMBER,
		TK_COLON,
		TK_COMMA,
		TK_EOF,
		TK_MAX
	};

	enum Expecting {
		EXPECT_OBJECT,
		EXPECT_OBJECT_KEY,
		EXPECT_COLON,
		EXPECT_OBJECT_VALUE,
	};

	struct Token {
		TokenType type;
		Variant value;
	};

	Variant data;
	String err_str;
	int err_line = 0;

	static const char *tk_name[];

	static String _make_indent(const String &p_indent, int p_size);
	static String _stringify(const Variant &p_var, const String &p_indent, int p_cur_indent, bool p_sort_keys, Set<const void *> &p_markers, bool p_full_precision = false);
	static Error _get_token(const char32_t *p_str, int &index, int p_len, Token &r_token, int &line, String &r_err_str);
	static Error _parse_value(Variant &value, Token &token, const char32_t *p_str, int &index, int p_len, int &line, String &r_err_str);
	static Error _parse_array(Array &array, const char32_t *p_str, int &index, int p_len, int &line, String &r_err_str);
	static Error _parse_object(Dictionary &object, const char32_t *p_str, int &index, int p_len, int &line, String &r_err_str);
	static Error _parse_string(const String &p_json, Variant &r_ret, String &r_err_str, int &r_err_line);

protected:
	static void _bind_methods();

public:
	String stringify(const Variant &p_var, const String &p_indent = "", bool p_sort_keys = true, bool p_full_precision = false);
	Error parse(const String &p_json_string);

	inline Variant get_data() const { return data; }
	inline int get_error_line() const { return err_line; }
	inline String get_error_message() const { return err_str; }
};

#endif // JSON_H

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