I have the problem below:
$$\int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx$$
I have tried $u=\ln(\sin(x))$ so $dx=\tan(x)du$
so the integral becomes:
$$\int_{-\infty}^0u\ln(\cos(x))du$$
but I cannot find a simple way of getting rid of this $\ln(\cos(x))$
I also tried using the substitution $v=x-\frac{\pi}{2}$
so the integral becomes:
$$\int_{-\frac{\pi}{2}}^0\frac{\ln(\cos(v))\ln(\cos(v+\frac{\pi}{2}))}{\tan(v+\frac{\pi}{2})}dv$$
but this does not seems to lead anywhere useful
EDIT to follow up
$$B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)dx$$
$$\frac{\partial^2}{\partial_\alpha\partial_\beta}B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\ln(\sin(x))\ln(\cos(x))dx$$
so I see that when $\alpha\to1$ and $\beta\to-1$ that this is the form that we want, so do we now have to take partial derivates of the non-integral form of the beta function then take the double integral for our chosen values?
Best Answer
You may consider that $$ f(\alpha,\beta) = \int_{0}^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\,dx = \frac{\Gamma\left(\frac{a}{2}\right)\Gamma\left(1+\frac{b}{2}\right)}{2\,\Gamma\left(1+\frac{a+b}{2}\right)}$$ by Euler's Beta function. By applying $\frac{\partial^2}{\partial\alpha\,\partial\beta}$ to both sides, then considering the limit as $\beta\to 0$ and $\alpha\to 0^+$, we have
$$ \int_{0}^{\pi/2}\frac{\log\sin(x)\log\cos(x)}{\tan(x)}\,dx = -\frac{1}{12}\psi''(1) = \color{red}{\frac{\zeta(3)}{8}}.$$ By enforcing the substitutions $x\mapsto\arctan(x)$ or $x\mapsto 2\arctan(x)$ in the original integral we get interesting identities for nasty (poly)logarithmic integrals.