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I can solve this integral in a certain way but I'd like to know of other, simpler, techniques to attack it:

\begin{align*} \int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\tan \left(x\right)}\:\mathrm{d}x&\overset{ t=\sin\left(x\right)}=\int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\sin \left(x\right)}\cos \left(x\right)\:\mathrm{d}x\\[2mm] &=\int _0^1\frac{\ln \left(t\right)\ln \left(\cos \left(\arcsin \left(t\right)\right)\right)}{t}\cos \left(\arcsin \left(t\right)\right)\:\frac{1}{\sqrt{1-t^2}}\:\mathrm{d}t\\[2mm] &=\int _0^1\frac{\ln \left(t\right)\ln \left(\sqrt{1-t^2}\right)}{t}\sqrt{1-t^2}\:\frac{1}{\sqrt{1-t^2}}\:\mathrm{d}t\\[2mm] &=\frac{1}{2}\int _0^1\frac{\ln \left(t\right)\ln \left(1-t^2\right)}{t}\:\mathrm{d}t=-\frac{1}{2}\sum _{n=1}^{\infty }\frac{1}{n}\int _0^1t^{2n-1}\ln \left(t\right)\:\mathrm{d}t\\[2mm] &=\frac{1}{8}\sum _{n=1}^{\infty }\frac{1}{n^3}\\[2mm] \int _0^{\frac{\pi }{2}}\frac{\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)}{\tan \left(x\right)}\:\mathrm{d}x&=\frac{1}{8}\zeta (3) \end{align*}

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    Papa Flammy already answered this, here's the link: https://www.youtube.com/watch?v=Y6yYSS3YbD8&t=914s – euler_med Aug 17 '20 at 11:41

2 Answers2

7

Let $$I=\int _0^{\frac{\pi }{2}}\frac{\ln (\sin \left(x\right))\ln (\cos \left(x\right))}{\tan \left(x\right)}\:\mathrm{d}x$$ Let $\ln \cos x=-t$, then we'll have: $$I=-\frac{1}{2}\int_{0}^{\infty} t \ln(1-e^{-2t}) dt$$ Use $\ln(1-z)=-\sum_{k=1}^{\infty} \frac{z^k}{k} $ $$\implies I=\frac{1}{2}\sum_{k=1}^{\infty} \int_{0}^{\infty}\frac {t e^{-2kt}}{k} dt$$ $$I=\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{4k^3}=\frac{\zeta(3)}{8}$$

José Carlos Santos
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Z Ahmed
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3

Substitute $t= \sin^2x$

\begin{align} &\int _0^{\frac{\pi }{2}}\frac{\ln (\sin x)\ln (\cos x)}{\tan x}\>{d}x \\= &\frac18\int _0^{1}\frac{\ln t\ln (1-t)}{t}\>dt \overset{IBP}= \frac1{16}\int_0^1 \frac{\ln^2 t}{1-t}dt= \frac1{16}\cdot2\zeta(3)= \frac18\zeta(3) \end{align}

Quanto
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