Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$. Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan  xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$.
I think it's natural to use series, but if I take $u=\tan x$ the integral becomes: $\int_{0}^{+\infty}\frac{\arctan u\ln u}{u^2+1}du$.
If I expand $f(x)=\frac{\arctan u}{u^2+1}$ to series, the result is not convergent.  However, the answer is so simple! Is there really a direct method to solve it?
 A: \begin{align}
&\int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx\\
=&\int_0^1  \int_0^\infty  \frac{x\ln x}{(1+x^2)(1+t^2x^2)} \overset{x\to \frac1{tx}}{dx}dt = \frac12 \int_0^1 \frac{\ln^2 t}{1-t^2}dt\\
=& \ \frac12 \sum_{n=0}^\infty \int_0^1  t^{2n}\ln^2 t \ dt
= \sum_{n=0}^\infty \frac1{(2n+1)^3}
=\frac78\zeta(3)
\end{align}
A: We can make quick work of this by using the Fourier series for $\ln(\sin(x))$ and $\ln(\cos(x))$.
\begin{align*}
 \int_0^{\pi/2}x\ln(\tan(x))\,\mathrm dx &= \int_0^{\pi/2}x\ln(\sin(x))\,\mathrm dx-\int_0^{\pi/2}x\ln(\cos(x))\,\mathrm dx \\
 &= \int_0^{\pi/2}x\left(-\ln(2)-\sum_{n=1}^\infty\frac{\cos(2nx)}{n}\right)\,\mathrm dx + \int_0^{\pi/2}x\left(\ln(2)+\sum_{n=1}^\infty(-1)^n\frac{\cos(2nx)}{n}\right)\,\mathrm dx \\
 &= -2\sum_{k=0}^\infty\frac{1}{2k+1}\int_0^{\pi/2}x\cos(2(2k+1)x)\,\mathrm dx,\,\text{terms of even index cancel} \\
 &= -2\sum_{k=0}^\infty\frac{1}{2k+1}\left[\frac{(2(2k+1))x\sin((2(2k+1))x)+\cos((2(2k+1))x)}{(2(2k+1))^2}\right]\Bigg\vert_0^{\pi/2},\\
&\qquad \text{integrate by parts with }u=x\text{ and }\mathrm dv=\cos(2(2k+1)x)\,\mathrm dx \\
&= -2\sum_{k=0}^\infty\frac{1}{2k+1}\cdot\frac{-2}{4(2k+1)^2} \\
&= \sum_{k=0}^\infty\frac{1}{(2k+1)^3} \\
&= \sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3} \\
&= \zeta(3)-\frac{1}{8}\zeta(3) \\
&= \frac{7}{8}\zeta(3).
\end{align*}
A: Let's consider a more general case and denote
$\displaystyle I(\alpha)=\int_0^\frac{\pi}{2}\tan^\alpha (x)\,x\,dx=\int_0^\infty t^\alpha\arctan t\,\frac{dt}{1+t^2};\,\alpha\in(-2;1)\tag*{}$
Then
$$I_k=\int_0^{\frac{\pi}{2}}x\ln^k(\tan  x)dx=\int_0^\infty \ln^kt\,\arctan t\,\frac{dt}{1+t^2}=\frac{d^k}{d\alpha^k}I(\alpha)\,\bigg|_{\alpha=0}, \,k=0,1, 2...\tag{0}$$
Using $\, \arctan t=\int_0^1\frac{t\,dx}{1+(xt)^2}$, we can write $I(\alpha)$ in the form
$\displaystyle I(\alpha)=\int_0^1\frac{dx}{x^2}\int_0^\infty\frac{t^{\alpha+1}}{(t^2+1)\big(t^2+\frac{1}{x^2}\big)}dt=\int_0^1\frac{dx}{x^2}J(\alpha, x)\tag*{(1)}$
Using complex integration (along the keyhole contour)
$\displaystyle \Big(1-e^{2\pi i\alpha}\Big)J(\alpha, x)=2\pi i\underset{z=\pm i; \,\pm\frac{i}{x}}{\operatorname{Res}}\frac{z^{\alpha+1}}{(z^2+1)\big(z^2+\frac{1}{x^2}\big)}\tag*{}$
The residues evaluation is straightforward. We get
$\displaystyle J(\alpha, x)=\frac{i\,x^2}{1-x^2}\Big(1-\frac{1}{x^\alpha}\Big)\frac{e^\frac{\pi i\alpha}{2}+e^\frac{3\pi i\alpha}{2}}{1-e^{2\pi i\alpha}}=\frac{\pi}{2\sin\frac{\pi\alpha}{2}}\frac{x^2}{1-x^2}\frac{1-x^\alpha}{x^\alpha}\tag*{(2)}$
Putting (2) into (1)
$\displaystyle I(\alpha)=\frac{\pi}{2\sin\frac{\pi\alpha}{2}}\int_0^1\frac{1-x^\alpha}{x^\alpha}\frac{dx}{1-x^2}=\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\int_0^1\Big(t^{-\frac{1+\alpha}{2}}-t^{-\frac{1}{2}}\Big)\frac{dt}{1-t}\tag*{}$
To perform integration, we use regularization and the main presentation of beta-function
$\displaystyle I(\alpha)=\lim_{\epsilon\to 0}\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\int_0^1\Big(t^{-\frac{1+\alpha}{2}}-t^{-\frac{1}{2}}\Big)(1-t)^{\epsilon-1}dt\tag*{}$
$\displaystyle =\lim_{\epsilon\to 0}\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\Gamma(\epsilon)\bigg(\frac{\Gamma\big(\frac{1-\alpha}{2}\big)}{\Gamma\big(\frac{1-\alpha}{2}+\epsilon\big)}-\frac{\Gamma\big(\frac{1}{2}\big)}{\Gamma\big(\frac{1}{2}+\epsilon\big)}\bigg)\tag*{}$
Denoting $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ - digamma-function, we get the answer:
$\displaystyle I(\alpha)=\frac{\pi}{4}\,\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}\tag*{(3)}$
$\displaystyle \boxed{\,I_k=\frac{d^k}{d\alpha^k}I(\alpha)\,\bigg|_{\alpha=0}=\frac{\pi}{4}\,\frac{d^k}{d\alpha^k}\,\bigg|_{\alpha=0}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}\,\,}\tag*{(4)}$
Now, disclosing the uncertainty,
$\displaystyle I_0=I(0)=\lim_{\alpha\to 0}\frac{\pi}{4}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1}{2}\big)+\psi^{(1)}\big(\frac{1}{2}\big)\frac{\alpha}{2}+O(\alpha^2)}{\sin\frac{\pi\alpha}{2}}=\frac{\psi^{(1)}\big(\frac{1}{2}\big)}{4}=\frac{\pi^2}{8}\tag*{}$
as it should be.
The desired integral $I_1$
$$I_1=\frac{\pi}{4}\,\frac{d}{d\alpha}\,\bigg|_{\alpha=0}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}=-\,\frac{\psi^{(2)}\big(\frac{1}{2}\big)}{8}=\sum_{k=1}^\infty\frac{1}{(2k-1)^3}=\frac{7}{8}\zeta(3)$$
