<s>[INST] Integral $\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx$ 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!
 [/INST] Not an answer but some transformations which might be useful. Didn't go beyond double series.
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos(x))dx= \color{blue}{(\cos x=t)} $$
$$ =\int_{1/\sqrt{2}}^1 \arccos t \ln t \frac{dt}{\sqrt{1-t^2}}=\color{blue}{(1-t^2=u^2)} $$
$$ = \frac12 \int_0^{1/\sqrt{2}} \frac{du}{\sqrt{1-u^2}} \arcsin u \ln (1-u^2)=$$
$$=\frac12 \int_0^{1/\sqrt{2}} \int_0^1 \frac{udv du}{\sqrt{1-u^2}\sqrt{1-u^2 v^2}} \ln (1-u^2)= \color{blue}{(u=\sqrt{w})}$$
$$=\frac14 \int_0^{1/2} \int_0^1 \frac{dv dw}{\sqrt{1-w}\sqrt{1-w v^2}} \ln (1-w)= $$
$$ =-\frac14 \sum_{k=1}^\infty \frac{1}{k} \int_0^1\int_0^{1/2}  \frac{w^k dw dv}{\sqrt{1-w}\sqrt{1-w v^2}}=\color{blue}{(w=\frac{s}{2})} $$
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \int_0^1\int_0^1  \frac{s^k ds dv}{\sqrt{1-s/2}\sqrt{1-s v^2/2}}=$$ 
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \sum_{n=0}^\infty \binom{2n}{n} \frac{1}{8^n} 
 \int_0^1\int_0^1  \frac{s^{k+n} v^{2n}  ds dv}{\sqrt{1-s/2}}=$$
$$=-\frac18 \sum_{k=1}^\infty \frac{1}{k2^k} \sum_{n=0}^\infty \binom{2n}{n} \frac{1}{(2n+1)8^n} \int_0^1  \frac{s^{k+n}  ds }{\sqrt{1-s/2}}=$$
$$=-\frac14 \sum_{k=1}^\infty \frac{1}{k} \sum_{n=0}^\infty \binom{2n}{n} \frac{B_{1/2} \left(\frac12,k+n+1 \right)}{(2n+1)4^n}=$$
$$=- \sum_{k=1}^\infty \frac{4^k}{k} \sum_{n=0}^\infty \binom{2n}{n} \frac{B_{1/2-1/\sqrt{8}} \left(k+n+1,k+n+1 \right)}{2n+1}$$
Where we have incomplete Beta function. Note that in this case incomplete Beta is a polynomial with order $n+k+2$.
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