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Does any one know how to calculate the following integration?

$$ \int_{\mathbb{R}} \left(\exp(z \: e^{-y^2})-1\right)^2 dy=?,\quad z>0. $$

This post is related to my previous question here , based on which one can have a upper bound. Can any one give an exact solution for this integration?

Thanks a lot! Anand

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2 Answers 2

up vote 7 down vote accepted

You can expand the integrand in powers of z. The coefficient of $z^k$ is $$e^{-ky^2}\sum_{j=1}^{k-1} {1\over j!(k-j)!}=e^{-ky^2}(2^k-2)/k!.$$ This yields the following series expression for the integral: $$\sum_{k=1}^\infty \sqrt{\pi\over k}(-2+2^k)z^k/k!.$$

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Thanks Michael Renardy for your solutions! –  Anand Mar 28 '14 at 13:41

A complementary approach is to consider the limit $z \to +\infty$. Based on the shape of the integrand, the overwhelming contribution should come from the vicinity of $y=0$, where only the highest order term $\exp(2z e^{-y^2}) \sim e^{2z} \exp(-2zy^2)$ will contribute significantly. Doing the Gaussian integral gives the leading term of the large $z$ asymptotic expansion: $e^{2z}/\sqrt{2z/\pi}$. A more systematic approach would go through Laplace's method, which I believe would yield the same leading term.

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So the answer is in terms of the function $$\sum_{k=1}^\infty \frac{x^k}{k!\sqrt{k}}.$$ Has it explicit form? –  Sergei Apr 14 '14 at 19:10
Sergei, I don't think that it has an explicit form. See my previous post: mathoverflow.net/questions/161068/has-anyone-seen-this-series –  Anand Apr 17 '14 at 16:18

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