TITLE: How find this integral $I=\int_{0}^{+\infty}\frac{x}{1+e^x}dx$
QUESTION [6 upvotes]: Question:

$$I=\int_{0}^{+\infty}\dfrac{x}{1+e^x}dx$$

I know use
$$I=\int_{0}^{\infty}\dfrac{xe^{-x}}{1+e^{-x}}dx=\sum_{k=0}^{\infty}\int_{0}^{\infty}xe^{-(k+1)x}dx=\sum_{k=1}^{\infty}(-1)^{k-1}\cdot\dfrac{1}{k^2}=\dfrac{\pi^2}{12}$$
someone have other methods? Thank you,Because I want collect more methods,Thank you

REPLY [3 votes]: Another variation of the method based on using series expansion. First, we make the substitution $e^{-x}=t$, and then integrate by parts:
$$I=\int_0^{+\infty}\frac{x}{1+e^x}dx=-\int_0^1\frac{\ln t}{1+t}dt=-\int_0^1\ln t\,d\ln(1+t)=\int_0^1\frac{\ln (1+t)}{t}dt.$$
Now we use Taylor series for $\ln(1+t)$:
$$I=\int_0^1\frac{1}{t}\sum_{n=1}^\infty(-1)^{n-1}\frac{t^n}{n} dt=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n}\int_0^1 t^{n-1} dt=\sum_{n=1}^\infty(-1)^{n-1}\frac{1}{n^2}=\frac{\pi^2}{12}.$$