# Integration of Exponential of Gaussian [duplicate]

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?

## marked as duplicate by Chill2Macht, tired, tilper, Zain Patel, ervxJul 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.

• Man. I read it a bit fast. What's with the exponential within the exponential? – Patrick Da Silva Jul 29 '16 at 12:33
• 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. – titanium Jul 29 '16 at 12:35
• well, where would you need to know the result of this integral? Does it show up anywhere? – Patrick Da Silva Jul 29 '16 at 12:36
• 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 $0.87$. – titanium Jul 29 '16 at 12:36
• no, sir!!!!!!!! – tired Jul 29 '16 at 12:37

## 1 Answer

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...