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I am looking for a solution to the following integral and finding it quite hard to find one ($|x| < a$):

$$ \int e^{\frac{1}{x^2 - a^2}} dx $$

I've tried to solve it with several substitutions, such as $u := x^2 - a^2$, but it yielded no other integral that seemed easier to solve. I've also tried to separate the exponent into something like $\frac{A}{x-a} + \frac{B}{x+a}$ and then integrating by parts, but also this approach didn't work out.

The reason for this question is the following: I need to modify this function, such that the integral from $-a$ to $a$ equals 1. That's why I came up with this integral, but if you see a way to find the required modification without having to integrate, that'll be fine as well. Thanks!

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  • $\begingroup$ Can this be one of those functions that you cannot possibly integrate like $\int \frac{1}{e^{x^2}}$ (which is somewhat similar to the error function) $\endgroup$ Commented Apr 14, 2013 at 21:32
  • $\begingroup$ @StringerBell How did you come up with this? I don't think an antiderivative for the integrand function is an elementary function. $\endgroup$
    – Git Gud
    Commented Apr 14, 2013 at 21:33
  • $\begingroup$ I don't think this has a primitive that can be expressed in terms of elementary functions. $\endgroup$
    – Pedro
    Commented Apr 14, 2013 at 21:33
  • $\begingroup$ @GitGud I need to modify this function, such that the Integral from $-a$ to $a$ equals 1, so actually it is not an undefined integral, but I don't think this makes it easier :) $\endgroup$
    – kafman
    Commented Apr 14, 2013 at 21:37
  • $\begingroup$ @StringerBell I suggest you add that to that question, whilst clarifying what you mean with modifying the function. $\endgroup$
    – Git Gud
    Commented Apr 14, 2013 at 21:40

1 Answer 1

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Well, if you just want $-a$ to $a$, use change of variable $u=-1/(x^2-a^2 )$ as follows: $$ \int_{-a}^a \exp\frac{1}{x^2-a^2}\;dx = 2 \int_{0}^a \exp\frac{1}{x^2-a^2}\;dx = \int_{1/a^2}^\infty \frac{e^{-u}}{u^{3/2}\sqrt{a^2u-1}}\;du = \frac{e^{-1/(2a^2)}}{a}\left(K_1\left(\frac{1}{2a^2}\right)-K_0\left(\frac{1}{2a^2}\right)\right) $$

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    $\begingroup$ What are $K_0$ and $K_1$? Or rather, how are they called? $\endgroup$
    – Julien
    Commented Apr 14, 2013 at 22:08
  • $\begingroup$ Did you subsitute u with $\exp(-1/(x^2 - a^2))$ or only $u = -1/(x^2 - a^2)$? And why the the minus? Shouldn't you then be integrating from $-1/a^2$ to $\infty$? $\endgroup$
    – kafman
    Commented Apr 14, 2013 at 22:10
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    $\begingroup$ Bessel functions ... en.wikipedia.org/wiki/… $\endgroup$
    – GEdgar
    Commented Apr 14, 2013 at 22:11
  • $\begingroup$ @GEdgar Thank you. It's never too late to learn...+1. $\endgroup$
    – Julien
    Commented Apr 14, 2013 at 22:12
  • $\begingroup$ The minus is so $u$ is positive when $0<x<a$. $\endgroup$
    – GEdgar
    Commented Apr 14, 2013 at 22:19

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