I would like to learn more about this integral:
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx=\frac{1}{128}\Big(16\pi \operatorname{C}-21\zeta(3)-4\pi^2\ln(2)\Big)$$
where $\operatorname{C}$ is the Catalan's constant.
I have tried integration by parts , substitution $y=\cos x$, etc. But, it reveals nothing. WA gives the anti-derivative in polylogarithmic functions that I don't know about.
I think it is a hard integral. If you have an elementary method to guide me, it would be nice.
Thanks a lot!
Best Answer
Here is to integrate with elementary methods. \begin{align} & \int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx\\ =&\frac12 \int_0^{\frac\pi4 }x\ln\frac{\sin 2x}2 \overset{2x\to x}{dx }- \frac12\int_0^{\frac\pi4 }x\ln(\tan x) \overset{x\to \frac\pi2-x}{dx }\\ =& - \frac{\pi^2}{64}\ln2 +\frac18\int_0^{\frac\pi2}x\ln(\sin x)dx-\frac14\int_0^{\frac\pi2}x\ln(\tan x) dx -\frac\pi8 \int_0^{\frac\pi4 }\ln(\tan x)dx\tag1 \end{align} where $\int_0^{\frac\pi4 }\ln(\tan x)dx=-G$, \begin{align} & \int_0^{\frac\pi2}x\ln(\tan x) dx\\ =&\int_0^\infty \frac{\ln t \tan^{-1}t}{1+t^2}dt =\int_0^\infty \frac{\ln t}{1+t^2} \left(\int_0^1 \frac{t}{1+t^2y^2}dy\right)\overset{t^2\to t}{dt}\\ =& \>\frac14 \int_0^1 \int_0^\infty \frac{\ln t}{(1+t)(1+y^2t)}{dt}\>dy\>\>\>\>\>({t\to \frac1{y^2t}})\\ =& \>\frac14 \int_0^1 \int_0^\infty \frac{-\ln y}{(1+t)(1+y^2t)}dt\>dy =\>\frac12 \int_0^1 \frac{\ln^2 y}{1-y^2}dy=\frac78\zeta(3)\\ \\ & \int_{0}^{\frac{\pi}{2}}x\ln(\sin x)dx\\ =&\>\frac12 \int_0^{\frac\pi2 }x\ln\frac{\sin 2x}2 \overset{2x\to x}{dx }+\frac12\int_0^{\frac\pi2 }x\ln(\tan x) dx\\ =&\>\frac18\int_0^{\pi} x\ln(\sin x)\overset{x\to \pi-x}{dx}-\frac{\pi^2}{16}\ln2 +\frac12\cdot\frac7{8}\zeta(3)\\ =&\>\frac\pi{16}\int_0^{\pi} \ln(\sin x){dx}-\frac{\pi^2}{16}\ln2 +\frac7{16}\zeta(3) = -\frac{\pi^2}{8} \ln2+\frac7{16}\zeta(3) \end{align}
Substitute into (1) to obtain $$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx = -\frac{\pi^2}{32} \ln2-\frac{21}{128}\zeta(3)+\frac\pi8G $$