Convergence of $\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$ Does this integral converge to any particular value?
$$\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$$
If the answer is yes, how should I calculate its value? 
I tried to use convergence tests but I failed due to the complexity of the integral itself.

It is possible to generalize the result.
Claim: $$I=\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx=-\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right)
 $$ where $\textrm{Li}_{n}\left(x\right)
 $ is the polylogarithm and $a>0$. 
Consider $$\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\frac{\left(-1\right)^{n}e^{-nb}}{n^{a+1}}\int_{0}^{\infty}y^{a}e^{-y}dy=\frac{\Gamma\left(a+1\right)\left(-1\right)^{n}e^{-bn}}{n^{a+1}}
 $$ and recalling that $$\textrm{Li}_{k}\left(x\right)=\sum_{n\geq1}\frac{x^{n}}{n^{k}}
 $$ we have $$\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right)=\sum_{n\geq1}\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\int_{0}^{\infty}x^{a}\sum_{n\geq1}\left(-1\right)^{n}e^{-n\left(x-b\right)}dx=
 $$ $$=-\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx.$$
Maybe it's interesting to note that the function ${x^{a}}/{e^{x-b}+1}$ is the Fermi-Dirac distribution.
