Integral of logarithm of exponential function

Question

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.

Answer 1

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)

Answer 2

By scaling, then by numerical integration

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