Show that ∫π/20log2sinxlog2cosxcosxsinxdx=14(2ζ(5)−ζ(2)ζ(3))\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 :
π20ln2(Acos(x))ln2(Asin(x))cos(x)sin(x)dx=14[2ζ(5)ζ(2)ζ(3)]

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=π20ln2cosxln2sinxcosxsinxdx=140t2(ln(1e2t))21e2tdt.

Follow it by another change of variables 1e2t=z gives

140t2(ln(1e2t))21e2tdt=13210(ln(1z))2(ln(z))2z(1z)dz

=13210(ln(1z))2(ln(z))2zdz+13210(ln(1z))2(ln(z))2(1z)dz

I=11610(ln(1z))2(ln(z))2zdz(1).

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

11610(ln(1z))2(ln(z))2zdz=116lim

= \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.

Attribution
Source : Link , Question Author : Ryan , Answer Author :
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