Elegant solutions to $ \int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln (\tan x) d x $ .

Question

Letting $x\mapsto \frac{\pi}{2} -x$ converts the integral $\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x \tag*{} $ Using the identity $ \displaystyle \tan x=\frac{\sin 2 x}{1+\cos 2 x} $ , we get $$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}}\left[\frac{\sin 2 x}{\cos 2 x} \ln (\sin 2 x)-\frac{\sin 2 x}{\cos 2 x} \ln (1+\cos 2 x)\right] d x \tag*{} $$ Letting $2x\mapsto x$ yields $\displaystyle I=\frac{\pi}{8}\left[\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (\sin x)}_{J} d x-\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (1+\cos x) d x}_{K}\right]\tag*{} $ As $ \displaystyle J \stackrel{x \rightarrow \pi-x}{=}-J \Rightarrow J=0 $ , therefore $\displaystyle I=-\frac{\pi}{8} K. $

By my post , $\displaystyle \begin{aligned}K \stackrel{y=\cos x}{=}& \int_{-1}^{1} \frac{\ln (1+y)}{y} d y=\frac{\pi^{2}}{4}\end{aligned}\tag*{} $ Now we can conclude that $\displaystyle \boxed{I=-\frac{\pi^{3}}{32}}\tag*{} $

Request for elegant solutions. Your suggestion and alternative methods are warmly welcome!

Answer 1

Substitute $t=\tan^2x$, along with $\tan 2x =\frac{2\tan x}{1-\tan^2x}$\begin{align} & I = \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x = \frac\pi8 \int_0^\infty\frac{\ln t}{1-t^2}dt =\frac\pi8\left(-\frac{\pi^2}4\right)=-\frac{\pi^3}{32} \end{align} where $\int_0^\infty \frac {\ln t}{1-t^2}dt=-\frac{\pi^2}4 $

Answer 2

$$I=\int \tan (2 x) \log (\tan (x)) \,dx $$ $$x=\tan ^{-1}(t)\quad \implies \quad I=-2\int\frac{ t }{t^4-1}\log (t)\,dt$$ $$\frac{ t }{t^4-1}=\frac 14 \Bigg[\frac{1}{t+1}+\frac{1}{t-1} -\frac{1}{t-i}-\frac{1}{t+i}\Bigg]$$ So, four simple integrals since $$\int \frac {\log(t)}{t+a}\,dt=\text{Li}_2\left(-\frac{t}{a}\right)+\log (t) \log \left(1+\frac{t}{a}\right)$$ $$I=\frac{1}{2} \left(\text{Li}_2\left(-t^2\right)+2 \text{Li}_2(1-t)-2 \text{Li}_2(-t)+2 \log (t) \,\log \left(\frac{t^2+1}{t+1}\right)\right)$$ Using the bounds $$I=-\frac{\pi ^2}{24}-\frac{\pi ^2}{12}=-\frac{\pi ^2}{8}$$