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.
$$\mathfrak{I}=\int_0^{\pi/2} x\log(\tan x )dx=\frac78\zeta(3)\tag1$$
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
$$\therefore\mathfrak{I}=\int_0^{\pi/2} x\log(\tan x )dx=\frac78\zeta(3)$$
^^
– mrtaurho
Dec 24 '18 at 01:42
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 $$\ln(\sin(x))=-\ln(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}$$ $$\ln(\cos(x))=-\ln(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}$$
