Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$ Nowadays I encounter  an integral which is difficult for me to evaluate it. Please help me to evaluate it. Thank you.

$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$$

where $\gamma$ is The Euler–Mascheroni constant.
 A: Hint: make use of the Binet's second formula http://mathworld.wolfram.com/BinetsLogGammaFormulas.html. 
$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$
$$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$
$$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-\frac{1}{2}\int_{0}^{\infty}\frac{x}{((x/2)^2+1)(e^{2\pi x}-1)} dx$$
Now, we see that 
$$\frac{\partial}{\partial z}\left(\int_0^{\infty} \frac{\arctan(x/z)}{e^{2 \pi x}-1} \ dx\right)=-\int_0^{\infty}\frac{x}{\displaystyle \left(e^{2 \pi  x}-1\right) z^2 \left(\frac{x^2}{z^2}+1\right)} \ dx$$
Can you take it from here?
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} + 1}}\,\dd x}
\\[5mm]&=\int_{0}^{\infty}{x \over \pars{x^{2} + 1}}
\pars{{1 \over \expo{2\pi x} + 1} - {1 \over \expo{2\pi x} - 1}}\,\dd x
+\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}}
\\[5mm]&=-2\int_{0}^{\infty}
{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{4\pi x} - 1}}
+\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}}
\\[5mm]&=-2\int_{0}^{\infty}
{x\,\dd x \over \pars{x^{2} + 4}\pars{\expo{2\pi x} - 1}}
+\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}}
\end{align}

With identity ${\bf 6.3.21}$:
  $$
\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + z^{2}}\pars{\expo{2\pi x} - 1}}
=\half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}
$$
  we'll get

\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} + 1}}\,\dd x}
\\[5mm]&=-2\braces{%
\half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}_{z\ =\ 2}}
+\half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}_{z\ =\ 1}
\\[5mm]&=-\ln\pars{2} + {1 \over 4} +\
\overbrace{\Psi\pars{2}}^{\dsc{\Psi\pars{1} + 1}}
-{1 \over 4} - \half\ \overbrace{\Psi\pars{1}}^{\dsc{-\gamma}}
=\color{#66f}{\large 1 - {\gamma \over 2} - \ln\pars{2}}
\end{align}
A: A Generalisation:
\begin{align}
\int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1
=&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2
=&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3
=&\int^\infty_0xe^{-x}\left(\sum^\infty_{n=1}\frac{(-1)^{n-1}}{x^2+4n^2\pi^2w^2}\right){\rm d}x\\
=&\int^\infty_0\frac{e^{-x}}{2x}-\frac{e^{-x}}{4w}\mathrm{csch}\left(\frac{x}{2w}\right){\rm d}x\tag4\\
=&-\frac{1}{2}\int^1_0\frac{1}{\ln{x}}+\frac{x^{1/2w}}{w(1-x^{1/w})}\ {\rm d}x\tag5\\
=&-\frac{1}{2}\int^1_0\frac{x^{w-1}}{\ln{x}}+\frac{x^{w-1/2}}{1-x}{\rm d}x\tag6\\
=&-\frac{1}{2}\int^0_\infty\int^1_0x^{t+w-1}+\frac{x^{t+w-1/2}\ln{x}}{1-x}\ {\rm d}x\ {\rm d}t\tag7\\
=&-\frac{1}{2}\int^0_\infty\frac{1}{t+w}-\psi_1\left(t+w+\frac{1}{2}\right)\ {\rm d}t\tag8\\
=&\boxed{\large{\color{blue}{\displaystyle\frac{1}{2}\psi_0\left(w+\frac{1}{2}\right)-\frac{1}{2}\ln{w}}}}\\
\end{align}
Explanation:
$(1)$: Substituted $x\mapsto\dfrac{x}{2\pi}$. 
$(2)$: Expanded $\dfrac{1}{1+e^{-x}}$ as a geometric series. 
$(3)$: Substituted $x\mapsto\dfrac{x}{n}$. 
$(4)$: One can show that $\displaystyle\sum^\infty_{n=-\infty}\frac{(-1)^n}{x^2+4n^2\pi^2w^2}=\frac{1}{2wx}\mathrm{csch}\left(\frac{x}{2w}\right)$ using the residue theorem. 
$(5)$: Substituted $x\mapsto-\ln{x}$. 
$(6)$: Substituted $x\mapsto x^w$. 
$(7)$: Used the fact that $\displaystyle\frac{1}{\ln{x}}=\int^0_\infty x^{t}\ {\rm d}t$. 
$(8)$: Used the integral representation of the polygamma function.

The Integral:
Let $w=1$ to get
$$\int^\infty_0\frac{x}{(x^2+1)(e^{2\pi x}+1)}{\rm d}x=\frac{1}{2}\psi_0\left(\frac32\right)=\boxed{\large{\color{red}{\displaystyle 1-\frac{\gamma}{2}-\ln{2}}}}$$
A: Here is an elementary way to evaluate the integral without involving any special functions or advance formulas. Notice that
$$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$
Hence we have
\begin{align}
\int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\int_0^\infty\int_0^\infty \frac{e^{-y}\sin(xy)}{e^{2\pi x}+1}\mathrm dy\;\mathrm dx\\[7pt]
&=\int_0^\infty e^{-y}\int_0^\infty \frac{\sin(xy)}{e^{2\pi x}+1}\mathrm dx\;\mathrm dy
\end{align}
One may use the following technique to evaluate the inner integral
\begin{align}
\int_0^\infty \frac{\sin(xy)}{e^{2\pi x}+1}\mathrm dx&=\int_0^\infty \frac{e^{-2\pi x}\sin(xy)}{1+e^{-2\pi x}}\mathrm dx\\[7pt]
&=\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty} e^{-2\pi kx}\sin (xy)\;\mathrm dx\\[7pt]
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{y}{4\pi^2k^2+y^2}\\[7pt]
&=\frac{1}{4}\left[\frac{2}{y}-\operatorname{csch}\left(\frac{y}{2}\right)\right]
\end{align}
then we obtain
\begin{align}
\int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\frac{1}{4}\int_{0}^{\infty} e^{- y}\left[\frac{2}{y}-\operatorname{csch}\left(\frac{y}{2}\right)\right]\mathrm dy\\[7pt]
&=\frac{1}{2}\int_{0}^{\infty} e^{- y}\left[\frac{1}{y}-\frac{e^{-y/2}}{1-e^{-y}}\right]\mathrm dy\\[7pt]
&=\frac{1}{2}\int_{0}^{\infty} e^{- y}\left[\frac{1}{y}-\frac{1}{1-e^{-y}}+\frac{e^{y/2}}{1+e^{y/2}}\right]\mathrm dy\\[7pt]
&=\frac{1}{2}\underbrace{\int_{0}^{\infty} \left[\frac{e^{- y}}{y}-\frac{1}{e^{y}-1}\right]\mathrm dy}_{\large\color{blue}{(*)}}+\frac{1}{2}\int_{0}^{\infty} \frac{e^{-y/2}}{1+e^{y/2}}\mathrm dy\\[7pt]
&=-\frac{\gamma}{2}+\frac{1}{2}\int_{0}^{\infty} \left[e^{-y/2}-\frac{e^{-y/2}}{1+e^{-y/2}}\right]\mathrm dy\\[7pt]
&=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large1-\frac{\gamma}{2}-\ln2}}
\end{align}
where $\color{blue}{(*)}$ is the integral representation of the Euler–Mascheroni constant.
A: I give a proof of
$$
\int^{\infty}_{0}\frac{x}{x^2+a^2}\frac{1}{e^{2\pi x}-1}dx=-\frac{1}{2}\psi(a)-\frac{1}{4a}+\frac{\log(a)}{2}
$$
We use the sine Fourier transforms:
$$
\frac{x}{x^2+a^2}\leftrightarrow \sqrt{\frac{\pi}{2}}e^{-aw}
$$
and
$$
\frac{1}{e^{2\pi x}-1}\leftrightarrow \frac{1}{\sqrt{2 \pi}}\frac{-1+w\coth(w/2)}{2w}
$$
Then we can write
$$
I=\int^{\infty}_{0}\frac{x}{x^2+a^2}\frac{1}{e^{2\pi x}-1}dx=
$$
$$
=\int^{\infty}_{0}\frac{w\coth(w/2)-2}{4w}e^{-a w}dw=
$$
$$
=-\frac{1}{4a}-\frac{1}{2}\frac{\Gamma'(a)}{\Gamma(a)}+\frac{\log(a)}{2}.
$$
Since 
$$
\int^{\infty}_{h}\frac{e^{-a w}}{w}dw=\int^{\infty}_{ah}\frac{e^{-t}}{t}dt=-e^{-ah}\log(ah)+\int^{\infty}_{ah}e^{-t}\log(t)dt\textrm{ : }\textrm{(1)}
$$
and
$$
\frac{1}{4}\int^{\infty}_{h}\coth(w/2)e^{-aw}dw=\frac{1}{4}\int^{e^{-h}}_0\frac{t^{a-1}(1+t)}{1-t}dt=
$$
$$
=\frac{1}{4}\int^{e^{-h}}_{0}\left(\frac{t^{a-1}}{1-t}-\frac{1}{1-t}\right)dt+\frac{1}{4}\int^{e^{-h}}_{0}\left(\frac{t^a}{1-t}-\frac{1}{1-t}\right)dt-\frac{1}{2}\log\left(1-e^{-h}\right)=
$$ 
$$
=\frac{1}{4}\sum^{\infty}_{k=0}\left(\frac{e^{-(k+a)h}}{k+a}-\frac{e^{-(k+1)h}}{k+1}\right)
+\frac{1}{4}\sum^{\infty}_{k=0}\left(\frac{e^{-(k+a+1)h}}{k+a+1}-\frac{e^{-(k+1)h}}{k+1}\right)-
$$
$$
-\frac{1}{2}\log\left(1-e^{-h}\right)\textrm{ : }\textrm{(2)}
$$
From (1) and (2) we have (taking $h\rightarrow 0$):
$$
I=\frac{-1}{4}\{\gamma+\psi(a)+\gamma+\psi(a+1)\}-\frac{1}{2}\lim_{h\rightarrow 0}\left(\log(1-e^{-h})-e^{-ah}\log(ah)\right)+\frac{\gamma}{2}=
$$
$$
=-\frac{1}{2}\psi(a)-\frac{1}{4a}+\frac{\log(a)}{2}.
$$
Where we have used $\int^{\infty}_{0}e^{-t}\log(t)dt=-\gamma$.
