calculus - Show that $int_{0}^{pi/2}frac {log^2sin xlog^2cos x}{cos xsin x}mathrm{d}x=frac14left( 2zeta (5)-zeta(2)zeta (3)right)$

Show that :
$$\int_{0}^{\Large\frac\pi2} {\ln^{2}\left(\vphantom{\large A}\cos\left(x\right)\right) \ln^{2}\left(\vphantom{\large A}\sin\left(x\right)\right) \over \cos\left(x\right)\sin\left(x\right)}\,{\rm d}x ={1 \over 4}\, \bigg[2\,\zeta\left(5\right) - \zeta\left(2\right)\zeta\left(3\right) \bigg]$$

I can only do non squared one. Anyone has a clue?

Answer

Related problems: (I), (II), (III), (IV), (V), (6). Use the change of variables $\ln(\cos(x))=t$ to transform the integral to

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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