Logarithms, Trigonometry, Integrals (Part 4)

$\large{\mathbb I= \frac{1}{2} \displaystyle \int ^{\frac{\pi}{2}}_{0} \frac{\ln(\sin x) \ln (\cos x)}{\sin x \cos x}dx}$

Multiplying $\mathbb I$ by $\frac{2 \sin x \cos x }{2 \sin x \cos x}$ and put $t=\sin ^2 x$ we get

$\large{\Rightarrow \frac{1}{16}\displaystyle \int^{1}_{0} \frac{\ln(1-t)\ln t}{(1-t)t}dt}$

Let $J=\displaystyle \int^{1}_{0} \frac{\ln(1-t)\ln t}{(1-t)t}dt$

$\large{\Rightarrow J=\displaystyle \int^{1}_{0} \frac{\ln(1-t)\ln t}{t}dt+\displaystyle \int^{1}_{0} \frac{\ln(1-t)\ln t}{(1-t)}dt}$

$\large{\Rightarrow J=2 \displaystyle \int^{1}_{0} \frac{\ln(1-t)\ln t}{t}dt}$

Using Integration by parts. Taking $\ln (1-t)$ as first function.

$\large{ \Rightarrow J=2 \left(\frac{\ln(1-t) \ln^{2}(t)}{2}|^{1}_{0}+ \displaystyle \int^{1}_{0}\frac{\ln^{2}(t)}{2(1-t)}dt \right)}$

$\large{ \Rightarrow J= \displaystyle \int^{1}_{0} \displaystyle \sum^{\infty}_{r=1} t^{r-1} \ln^{2} (t) dt}$

We know that $\large{\displaystyle \int^{1}_{0} x^{a-1}=\frac{1}{a}}$

now $\large{\displaystyle \int^{1}_{0} \frac{\partial^{s}}{\partial x^{s}}x^{a-1} =\displaystyle \int^{1}_{0}x^{a-1} \ln^{s}(x)=\frac{(-1)^{s} s!}{a^{s+1}} }$

$\large{ \Rightarrow J= \displaystyle \sum^{\infty}_{r=1} \displaystyle \int^{1}_{0} t^{r-1} \ln^{2} (t) dt}$

$\large{ \Rightarrow J=\displaystyle \sum^{\infty}_{r=1} \frac{2}{r^3}}$

$\Large{ \Rightarrow J=2 \zeta(3)}$

$\Large{ \Rightarrow \mathbb I=\frac{\zeta(3)}{8}}$

$a=3$ and $b=8$

Hence $\Large{a+b=11}$