How to find $ \int_0^\infty \dfrac x{1+e^x}\ dx$ How to find
$$
\int_0^\infty \dfrac x{1+e^x}\ dx=\ ...?
$$
I don't know where should I start with. The correct answer from my textbook is $\frac{\pi^2}{12}$.
This is my homework with 10 questions but I can only answer 9 questions, this one I'm stuck. Can anyone help me? Thanks.
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\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{x \over \expo{x} + 1}\,\dd x}
=\int_{0}^{\infty}x\pars{{1 \over \expo{x} + 1} - {1 \over \expo{x} - 1}}
\,\dd  x + \int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
\\[3mm]&=-2\int_{0}^{\infty}{x \over \expo{2x} - 1}\,\dd x
 + \int_{0}^{\infty}{x \over \expo{x} - 1}\,\dd x
=\half\int_{0}^{\infty}{x\expo{-x} \over 1 - \expo{-x}}\,\dd x
\\[3mm]&=-\,\half\int_{0}^{\infty}\ln\pars{1 - \expo{-x}}\,\dd x
=\half\int_{0}^{\infty}\sum_{n = 1}^{\infty}{\expo{-nx} \over n}\,\dd x
=\half\sum_{n = 1}^{\infty}{1 \over n}\
\overbrace{\int_{0}^{\infty}\expo{-nx}\,\dd x}^{\ds{=\ {1 \over n}}}
\\[3mm]&=\half\
\overbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
^{\ds{\zeta\pars{2} = {\pi^{2} \over 6}}}
=\color{#00f}{\large{\pi^{2} \over 12}}
\end{align}

$\ds{\zeta\pars{z}}$ is the Riemann Zeta Function.

A: One way to work this one out is to rewrite the integral as
$$\int_0^{\infty} dx \, \frac{x \, e^{-x}}{1+e^{-x}} $$
then expand the denominator in a Taylor series.  The representation of this integral becomes
$$\int_0^{\infty} dx \, x \sum_{k=0}^{\infty} (-1)^k e^{-(k+1) x} $$
Because the sum and integral converge, we can change the order of summation and integration:
$$\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dx \, x \, e^{-(k+1) x} = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}$$

Another way to work this, which you may or may not be aware of, is to use Cauchy's theorem.  Consider the contour integral
$$\oint_C dz \frac{z^2}{1+e^z} $$
where $C$ is the rectangle with vertices $0$, $R$, $R+i 2 \pi$, and $i 2 \pi$, with a semicircular segment of radius $\epsilon$ centered at $i \pi$ into the rectangle.  The contour integral then becomes
$$\int_0^R dx \frac{x^2}{1+e^x} + i \int_0^{2 \pi} dy \frac{(R+i y)^2}{1+e^R e^{i y}} +\int_R^0 dx \frac{(x+i 2 \pi)^2}{1+e^x} \\ +i \int_{2 \pi}^{\pi+\epsilon} dy \frac{(i y)^2}{1+e^{i y}}+ i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{(i \pi+\epsilon e^{i \phi})^2}{1-e^{\epsilon e^{i \phi}}} + i \int_{\pi-\epsilon}^0 dy \frac{(i y)^2}{1+e^{i y}}$$
As $R\to\infty$, the second integral vanishes.  As $\epsilon \to 0$, the fifth integral becomes
$$i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{-\pi^2}{-\epsilon e^{i \phi}} = -i \pi^3$$
The contour integral is then
$$-i 4 \pi \int_0^{\infty} dx \frac{x}{1+e^x} + 4 \pi^2 \int_0^{\infty} \frac{dx}{1+e^x} + i PV \int_0^{2 \pi} dy \frac{y^2}{1+e^{i y}}-i \pi^3$$
Note that $PV$ denotes the Cauchy principal value.  Also note that
$$\begin{align}i PV \int_0^{2 \pi} dy \frac{y^2}{1+e^{i y}} &= i \frac12 \int_0^{2 \pi} dy \, y^2 + \frac12 PV \int_0^{2 \pi} dy \, y^2 \tan{\left ( \frac{y}{2}\right )}\\ &= i \frac{4 \pi^3}{3} + \frac12 PV \int_0^{2 \pi} dy \, y^2 \tan{\left ( \frac{y}{2}\right )} \end{align}$$
By Cauchy's theorem, the contour integral is zero, which means that both the real and imaginary parts are zero.  From the imaginary part of the contour integral being zero, we have
$$\int_0^{\infty} dx \frac{x}{1+e^x} = \frac{\pi^2}{12} $$
A: Another way to evaluate the integral. Rewrite
$$
\int_0^{\infty} \frac{x}{1+e^{x}}\ dx=\int_0^{\infty} \frac{x \, e^{-x}}{1+e^{-x}}\ dx.
$$
Using IBP by taking $u=x$ and $dv=-\dfrac{d(e^{-x})}{1+e^{-x}}$, you will obtain
$$
\int_0^\infty\ln(1+e^{-x})\ dx.
$$
Now, expand the integrand using Mclaurin's series for natural logarithm. After integrating the series, compare the result with Dirichlet eta function and you will obtain the integral is equal to $\eta(2)=\dfrac12\zeta(2)=\dfrac{\pi^2}{12}$.
A: $$\int^\infty_0 \frac{x^{s-1}}{e^x+1}\, dx = \left(1-2^{1-s}\right)\zeta(s)\Gamma(s)$$
knowing that 
$$\zeta(2)=\sum_{n\geq 1} \frac{1}{n^2}=\frac{\pi^2}{6}$$
$$\int^\infty_0 \frac{x}{e^x+1}\, dx = \frac{1}{2}\zeta(2)\Gamma(1)=\frac{\pi^2}{12}$$
A: Hint: Let $~x=\ln t,~$ and $~u=\dfrac1t.~$ Then the integral becomes $~-\displaystyle\int_0^1\frac{\ln u}{1+u}du.~$ Now expand
$\dfrac1{1+u}~$ in to its binomial series, and switch the order of summation and integration. You will 
then recognize the expression of the Dirichlet $\eta$ function of argument $2.~$ See also Basel problem.
