## Patching parametric integrals to degenerate properly

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While reading the question about fitting all answers to the question of getting a single formula which, in the limit, captures all the answers to $\int x^{a}dx$, I thought of an obvious generalization (of the question):

What are sufficient conditions on a family of functions $f(x,a)$ so that $\int f(x,a) dx$ can be expressed as a single formula $F(x,a)+C(a)$ valid (in the limit) for all a?

From the point of view of FToC, $C(a)$ is completely irrelevant, but nevertheless such an answer has a certain elegance since $C(a)$ really captures the boundary cases of $f(x,a)$.

[Edit: although I like Dylan's answer below, it is not very algorithmic -- can one do better?]

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## 1 Answer

I don't know what is meant by a "single formula" but if, for instance, $f(x,a)$ is continuous, then $f(x,a) dx$ is a continuous 1-form, so has a well-defined path integral along any reasonable path. The integral will be path dependent if $f(x,a)$ depends at all on $a$, but let's take a path that moves first in the $a$ direction, then in the $x$ direction, i.e., set $F(x,a) = \int_{x_0}^x f(x,a)dx$ for some reasonable choice of basepoint $x_0$. Then if $f$ is differentiable in both variables, by Stokes' theorem $\frac{\partial F}{\partial a}(x,a) = \int_{x_0}^x \frac{\partial f}{\partial a}(x,a)dx$, which is still continuous.

Concretely, for the example $\int x^a dx$, take $x_0 = 1$. Then for $a \ne -1$, we have $\int_1^x x^a dx = \frac{x^{a+1}-1}{a+1}$, which approaches $\log x$ as $a \to -1$. (This is the compound interest limit.)

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Indeed.  So the answer is 1) choose a good base point, 2) 'indefinite integration' is a weird concept anyways, one should use definite integration instead.  I have long believed in #2, now I have an even better reason for it. – Jacques Carette Mar 30 2010 at 15:45
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