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?

\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}
