A nice log trig integral

Question

Show that : $$\int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\cos x{{\ln }^{2}}\sin x}{\cos x\sin x}}\text{d}x=\frac{1}{4}\left( 2\zeta \left( 5 \right)-\zeta \left( 2 \right)\zeta \left( 3 \right) \right)$$ I can only do non squared one :( Anyone got a clue? :)

Answer 1

A related problem. Use the change of variables $\ln(\cos(x))=t$ to transform the integral to

$$ \int_{0}^{\frac{\pi }{2}}{\frac{{{\ln }^{2}}\cos x{{\ln }^{2}}\sin x}{\cos x\sin x}}\text{d}x = \frac{1}{4}\,\int _{-\infty }^{0}\!{\frac {{t}^{2} \left( \ln \left( 1-{ {\rm e}^{2\,t}} \right)\right) ^{2}}{1-{{\rm e}^{2t}} }}{dt}.$$

Follow it by another change of variables $ 1-e^{2t}=z $ gives

$$\frac{1}{4}\,\int _{-\infty }^{0}\!{\frac {{t}^{2} \left( \ln \left( 1-{ {\rm e}^{2\,t}} \right) \right) ^{2}}{1- {{\rm e}^{2t}} }}{dt}= \frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln \left( 1-z \right) \right) ^{2} \left( \ln \left( z \right) \right) ^{2}}{z \left( 1- z\right) }}{dz}$$

$$= \frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln \left( 1-z \right) \right) ^{2} \left( \ln \left( z \right) \right) ^{2}}{z }}{dz}+\frac{1}{32}\,\int _{0}^{1}\!{\frac { \left( \ln\left( 1-z \right) \right) ^{2} \left( \ln \left( z \right) \right) ^{2}}{ \left( 1- z\right) }}{dz} $$

$$ = \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln \left( 1-z \right) \right) ^{2} \left( \ln \left( z \right) \right) ^{2}}{z }}{dz}\longrightarrow (1). $$

Getting the exact result: Integral (1) can be evaluated as

$$ \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln \left( 1-z \right) \right)^{2} \left( \ln \left( z \right) \right)^{2}}{z }}{dz}=\frac{1}{16} \lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\int_{0}^{1} (1-z)^{w}z^{s-1}dz $$

$$ = \frac{1}{16}\lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\beta(s,w+1)=\frac{1}{16}\lim_{w\to 0}\lim_{s\to 0^+}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}$$

$$ =\frac{1}{4}\left( 2\zeta \left( 5 \right)-\zeta \left( 2 \right)\zeta \left( 3 \right) \right) \longrightarrow (*), $$

where $\beta(u,v)$ is the beta function.

Other forms for the solution 1: Using integration by parts with $u=\ln^2(1-z)$, integral $(1)$ can be written as

$$ \frac{1}{16}\,\int _{0}^{1}\!{\frac { \left( \ln \left( 1-z \right) \right)^{2} \left( \ln \left( z \right)\right)^{2}}{z }}{dz}=\frac{1}{24}\,\int _{0}^{1}\!{\frac{ \ln\left( 1-z \right)\left( \ln \left( z \right) \right)^{3}}{1-z}}{dz} $$

$$ = -\sum_{n=0}^{\infty}(\psi(n+1)+\gamma)\int_{0}^{1}z^n\ln^3(z)dz = \frac{1}{4}\sum_{n=0}^{\infty}\frac{\psi(n+1)+\gamma}{(n+1)^4}. $$

$$ = \frac{1}{4}\sum_{n=1}^{\infty}\frac{\psi(n)}{n^4}+\frac{\gamma}{4}\zeta(4)\sim 0.02413779000 \longrightarrow (**). $$

You can use the identity $ H_{n-1}=\psi(n)+\gamma $, where $H_n$ are the harmonic numbers, to write the result as

$$ \frac{1}{4}\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^4} \longrightarrow (***). $$

Other forms for the solution 2: We can have the following form for the solution

$$ \frac{1}{16}\sum_{n=1}^{\infty}\frac{H^2_{n}}{n^3}+\frac{1}{16}\sum_{n=1}^{\infty}\frac{\psi'(n+1)}{n^3}-\frac{1}{16}\zeta(2)\zeta(3)\longrightarrow (****). $$

Note 1: we used the power series expansion of the function $ \frac{\ln(1-z)}{1-z}, $

$$ -\sum _{n=0}^{\infty } \left( \psi \left( n+1 \right) + \gamma \right){x}^{n}=-\sum _{n=0}^{\infty } H_{n}{x}^{n}. $$

Note 2: Try to tackle integral $(1)$ using the technique used in solving your previous question.