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I am interested in the following integral $$\int_{-\infty }^{\infty } \left[1-\exp\left(-\frac{e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2 \pi\sigma^2 }}\right)\right] \, dx$$

Does any one know if an analytical form of this integral exists?

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marked as duplicate by Chill2Macht, tired, tilper, Zain Patel, ervx Jul 29 '16 at 15:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Man. I read it a bit fast. What's with the exponential within the exponential? $\endgroup$ – Patrick Da Silva Jul 29 '16 at 12:33
  • $\begingroup$ What do you mean by "What's with the exponential within the exponential?"? Notice that this integral definitely converges, as at large $x$, the integrand goes to zero. $\endgroup$ – titanium Jul 29 '16 at 12:35
  • $\begingroup$ well, where would you need to know the result of this integral? Does it show up anywhere? $\endgroup$ – Patrick Da Silva Jul 29 '16 at 12:36
  • $\begingroup$ Why should it diverge? The integrand clearly decays quickly to 0 at larger $x$. Try NIntegrate with mathematica for $\sigma=1$, and it returns you $0.87$. $\endgroup$ – titanium Jul 29 '16 at 12:36
  • $\begingroup$ no, sir!!!!!!!! $\endgroup$ – tired Jul 29 '16 at 12:37
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This is a long comment and could serve as a possible route:

First note that $$\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}.$$ Call your integral $I$, then \begin{align} I&=-\int_{-\infty}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^j}{j!}\left(\frac{e^{-\frac{x^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}\right)^jdx\\ &=-\sum_{j=1}^{\infty}\frac{(-1)^j}{(\sqrt{2\pi\sigma^2})^jj!}\int_{-\infty}^{\infty}e^{-\frac{jx^2}{2\sigma^2}}dx\\ &=-\sqrt{2\pi\sigma^2}\sum_{j=1}^{\infty}\frac{(-1)^j}{(\sqrt{2\pi\sigma^2})^jj!\sqrt{j}}\\ \end{align} The last is convergent...

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