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I am trying to solve this integral:

$$\int \log\left(1 + \frac{1}{\pi}\exp\left(\frac{-x^2}{2a^2}\right)\right) dx$$

where $a$ is some fixed constant. The bounds of this integral are $-a$ and $a$, however, I just want to compute this integral, then I know how to substitute this values. Any suggestions, even how to simplify this integral will be very helpful.

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    $\begingroup$ You can't even find an integral for the exponential part inside the logarithm. However, similar integrals a have been solved so there's no reason to think a definite doesn't exist. (One day people will change their mindset and start assuming integrals to arbitrary functions don't exist) $\endgroup$ – Zach466920 Jun 24 '15 at 20:49
  • $\begingroup$ Have you made any attempts at solving this? If so, why don't you post your attempt. $\endgroup$ – user228288 Jun 24 '15 at 23:09
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You'll have to accept some things as fact. The Taylor expansion of $$\ln(u+1)=u-u^2/2+u^3/3...$$ and as alluded to in my comment the Gaussian integral, $$\int_{-\infty}^{\infty} {{(b \cdot exp(-x^2/(2 \cdot a^2)))^n} \over n} \ dx ={{b^n \sqrt{2} \cdot a} \over {n^{3/2}}} \cdot \sqrt{\pi}$$ Here's Wikipedia link for that.

So the key is to integrate term by term. What have to do is Taylor expand and then substitute in the exponent. Then you use the various results for the Gaussian integral mentioned above. Rather than write out the sun explicitly, we'll express the the integral in question $J$, as an infinite series... $$J=\sum_{n=1}^{\infty} (-b)^{n-1} \cdot {{\sqrt{2 \pi} \cdot a} \over {n^{3/2}}}$$

That sum is a fusion between the geometric sum and the zeta function so it makes sense the integral is equal to the poly logarithm. Just set $b={1 \over {\pi}}$

$$J=-\sqrt{2a^2 \pi} \cdot Li_{3/2}(-b)$$

(Posted from mobile)

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    $\begingroup$ Posted from mobile? Seriously? A little bit of me dies anytime I try to write latex on my phone. Kudos. $\endgroup$ – breeden Jun 24 '15 at 21:42
  • $\begingroup$ @breeden I was feeling brave today :) although yah, it's not really the best way to answer questions, hence the 3+ edits... $\endgroup$ – Zach466920 Jun 24 '15 at 21:49
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By scaling, then by numerical integration

$$\int_{x=-a}^a\log\left(1 + \frac{1}{\pi}\exp\left(\frac{-x^2}{2a^2}\right)\right) dx= a\int_{x=-1}^1\log\left(1 + \frac{1}{\pi}\exp\left(\frac{-x^2}2\right)\right) dx\\ \approx0.480805\cdots a$$

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