Evaluate $\int_{0}^{\frac{\pi}{2}} x\ln(\tan(x))dx $ How do I evaluate:
$$\int_{0}^{\frac{\pi}{2}} x\ln(\tan(x))dx $$
I know it equals $\frac{7\zeta(3)}{8}$, but I'm not sure how to get there. I've been trying to convert it into a sum somehow, but I'm not getting anywhere. Equivalent integrals are:
$$\frac{1}{2}\int_{-\infty}^\infty x\operatorname{sech}(x)\arctan(e^x)dx$$
$$\int_0^\infty \frac{\arctan(x)\ln(x)}{1+x^2} dx$$
I've tried differentiating under the integral but to no avail.
 A: 
$$\mathfrak{I}=\int_0^{\pi/2} x\log(\tan x )dx=\frac78\zeta(3)\tag1$$

The idea of converting it into a sum is good! Appropriate in this case are the Fourier Expansions of $\log(\sin x)$ and $\log(\cos x)$ since we can rewrite $\log(\tan x)$ as $\log(\sin x)-\log(\cos x)$. These Series are given by
$$\begin{align}
\log(\sin x)&=-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\\
\log(\cos x)&=-\log(2)-\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}
\end{align}$$
Plugging these series in $\mathfrak{I}$ yields to
$$\small\begin{align}
\mathfrak{I}=\int_0^{\pi/2} x\log(\tan x )dx&=\int_0^{\pi/2} x(\log(\sin x )-\log(\cos x))dx\\
&=\int_0^{\pi/2}x\left(-\log(2)-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}+\log(2)+\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}\right)dx\\
&=\int_0^{\pi/2}x\left(\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}-\sum_{n=1}^{\infty}\frac{\cos(2nx)}{n}\right)dx
\end{align}$$
Hence the given Fourier Series converge within the interval $[0,\pi]$ we may interchange the order of summation and integration which leaves with a rather simple integral which can be done using only Integration by Parts. So we further get
$$\begin{align}
\mathfrak{I}&=\int_0^{\pi/2}x\left(\sum_{n=1}^{\infty}(-1)^n\frac{\cos(2nx)}{n}-\sum_{n=1}^{\infty}\frac{\cos(2nx)}n\right)dx\\
&=\sum_{n=1}^{\infty}\frac{(-1)^n}n\underbrace{\int_0^{\pi/2}x\cos(2nx)dx}_{=I}-\sum_{n=1}^{\infty}\frac1n\underbrace{\int_0^{\pi/2}x\cos(2nx)dx}_{=I}
\end{align}$$
The inner integral can be evaluated fairly easy as already mentioned applying Integration by Parts. It turns out that this integral equals
$$I=\int_0^{\pi/2}x\cos(2nx)dx=\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}$$
The sine term vanishs directly since $\sin(n\pi)=0~\forall n\in\mathbb N$. The $\cos(n\pi)$ on the other hand oscillates between $1$ and $-1$ depending on whether $n$ is even or odd. Putting this together leads to
$$\begin{align}
\mathfrak{I}&=\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right]-\sum_{n=1}^{\infty}\frac1n\left[\frac{\pi\sin(n\pi)}{4n}+\frac{\cos(n\pi)}{4n^2}-\frac1{4n^2}\right]\\
&=\frac14\sum_{n=1}^{\infty}\frac{(-1)^n}n\left[\frac{(-1)^n}{n^2}-\frac1{n^2}\right]-\frac14\sum_{n=1}^{\infty}\frac1n\left[\frac{(-1)^n}{n^2}-\frac1{n^2}\right]\\
&=\frac14\left(\sum_{n=1}^{\infty}\frac1{n^3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^3}+\sum_{n=1}^{\infty}\frac1{n^3}\right)\\
&=\frac12(\zeta(3)+\eta(3))\\
&=\frac78\zeta(3)
\end{align}$$
That we can split the sum here can be justified due the absolute convergence of the given series. Within the last few steps we utilized the Dirichlet Eta Function $\eta(s)$ and its quite simple relation to the Riemann Zeta Function $\zeta(s)$, namely $\eta(s)=(1-2^{1-s})\zeta(s)$. And so it follows that

$$\therefore\mathfrak{I}=\int_0^{\pi/2} x\log(\tan x )dx=\frac78\zeta(3)$$

A: Hint:
Rewrite your original integral into
$$\int_{0}^{\frac{\pi}{2}}x\ln\left(\frac{\sin(x)}{\cos(x)}\right)dx=\int_{0}^{\frac{\pi}{2}}x(\ln(\sin(x))-\ln({\cos(x)}))dx$$
And use Fourier Series
$$\ln(\sin(x))=-\ln(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}$$
$$\ln(\cos(x))=-\ln(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}$$
