# Show that $\int_{0}^{\pi/2}\frac {\log^2\sin x\log^2\cos x}{\cos x\sin x}\mathrm{d}x=\frac14\left( 2\zeta (5)-\zeta(2)\zeta (3)\right)$

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

I can only do non squared one :( Anyone got a clue? :)

-
What is so nice about it? –  Kaster Jan 30 '13 at 2:48
It is interesting, in that neither of $\zeta(3),\ \zeta(5)$ are "known", in the sense that $\zeta(2)=\pi^2/6$ is known. –  coffeemath Jan 31 '13 at 9:03

Related problems: (I), (II), (III), (IV), (V). 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

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

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

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

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.

-
Thx for the answering ! I think since there are a lots of zeta involves. Maybe we use the series at the start? :) –  Ryan Feb 1 '13 at 23:40
@Ryan:You are welcome. Note that, you can get different forms of solutions for this integral. –  Mhenni Benghorbal Feb 2 '13 at 3:11
:You may hav a look at this question too :) math.stackexchange.com/questions/291850/… –  Ryan Feb 2 '13 at 14:26
@MhenniBenghorbal Nice and concise! (+1) –  O.L. May 28 '13 at 19:12
@MhenniBenghorbal, how exactly did you take the beta function limit (excellent answer, by the way)? –  Alyosha Dec 9 '13 at 18:16