Evaluating $\int_0^{\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx$ I need to solve
$$
\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx
$$
I tried to use symmetric properties of the trigonometric functions as is commonly used to compute
$$
\int_0^{\Large\frac\pi2}\ln\sin x\ dx = -\frac{\pi}{2}\ln2
$$
but never succeeded. (see this for example)

$$\begin{align*}
I &= \int_0^{\frac\pi2} \frac{\log(\cos(x)) \log(\sin(x))}{\tan(x)} \, dx \\[1ex]
&= \int_0^\infty \frac{\log\left(\frac1{\sqrt{x^2+1}}\right) \log\left(\frac x{\sqrt{x^2+1}}\right)}{x} \, \frac{dx}{1+x^2} \tag{1} \\[1ex]
&= \frac14 \int_0^\infty \frac{\log^2(x^2+1)-\log(x^2)\log(x^2+1)}{x} \, \frac{dx}{1+x^2} \\[1ex]
&= \frac18 \int_0^\infty \frac{\log^2(x+1)-\log(x)\log(x+1)}{x(1+x)} \, dx \tag{2} \\[1ex]
&= \frac18 \int_0^1 \left(\frac{\log^2(x+1)}{x} - \frac{\log(x)\log(x+1)}x\right) \, dx \tag{3} \\[1ex]
&= \frac14 \int_0^1 \frac{\log(x)\log(x+1)}x \, dx - \frac12 \int_0^1 \frac{\log(x+1)\log(x)}{x+1} \, dx \tag{4} \\[1ex]
&= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}n \int_0^1 x^{n-1} \log(x) \, dx + \frac12 \sum_{n=1}^\infty (-1)^n H_n \int_0^1 x^n \log(x) \, dx \tag{5} \\[1ex]
&= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}{n^3} + \frac12 \sum_{n=1}^\infty \frac{(-1)^n H_n}{(n+1)^2} \tag{4} \\[1ex]
&= \boxed{\frac{\zeta(3)}8} \tag{6}
\end{align*}$$


*

*$(1)$ : substitute $x\mapsto\arctan(x)$

*$(2)$ : substitute $x\mapsto\sqrt x$

*$(3)$ : split up the integral at $x=1$; substitute $x\mapsto\frac1x$ in the integral over $[1,\infty)$; partial fractions

*$(4)$ : integrate by parts

*$(5)$ : recall the power series $-\log(1-x)=\sum\limits_{n\ge1}\frac{x^n}n$ and $-\frac{\log(1-x)}{1-x}=\sum\limits_{n\ge1} H_n x^n$, where $H_n$ is the $n^{\rm th}$ harmonic number

*$(6)$ : see e.g. section $7$ for a contour-integration method to evaluating the Euler sum; I'm sure there are more efficient means

