Solution:

$\displaystyle \int_0^{\infty} \ln (1+e^{-ax} ) dx$ for $\displaystyle a>0$.

We use the formula: $\displaystyle \ln (1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - .... \mbox{ for }-1<x\leq 1$.

Since $\displaystyle 0<e^{-ax}\leq 1 \mbox{ on }[0,\infty)$:

$\displaystyle \int_0^{\infty} e^{-ax} - \frac{e^{-2ax}}{2} + \frac{e^{-3ax}}{3} - \frac{e^{-4ax}}{4}+ ... dx$

Integrate term-by-term*

$\displaystyle -\frac{e^{-ax}}{a} + \frac{e^{-2ax}}{2^2a} - \frac{e^{-3ax}}{3^2a} + ... \big|_0^{\infty}$

$\displaystyle \frac{1}{a} - \frac{1}{2^2a}+\frac{1}{3^2a}-\frac{1}{4a} + ... = \frac{1}{a} \left( 1 - \frac{1}{2^2}+\frac{1}{3^2} - \frac{1}{4^2}+... \right) = \frac{\pi^2}{12a}$

*)This is legal because of something called

uniform convergence.
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Note, this integral was taken from

here.