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

Question

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 1

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.

Answer 2

Mhenni Benghorbal gave a way to solve the problem. Unfortunately, he did not show how to get $$ \lim_{w\to0}\lim_{s\to0}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}. $$ I want to finish the missed part which is not easy to get. In order to evaluate this limit, we have to use $$ \Gamma'(x)=\Gamma(x)\psi_0(x), \psi_n'(x)=\psi_{n+1}(x). $$ It is not hard to get \begin{eqnarray*} \frac{d^2}{dw^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}&=&\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}(\psi_0^2(w+1)-2\psi_0(w+1)\psi_0(s+w+1)\\ &&+\psi_0^2(s+w+1)+\psi_1(w+1)-\psi_1(s+w+1)). \end{eqnarray*} Note $$\psi_0(1)=-\gamma, \psi_1(1)=\frac{\pi^2}{6}$$ and hence \begin{eqnarray*} \lim_{w\to0}\frac{d^2}{dw^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(w+1)}&=&\lim_{w\to0}\frac{\Gamma(s)\Gamma(s+w+1)}{\Gamma(s+w+1)}(\psi_0^2(w+1)-2\psi_0(w+1)\psi_0(s+w+1)\\ &&+\psi_0^2(s+w+1)+\psi_1(w+1)-\psi_1(s+w+1))\\ &=&\frac{\Gamma(s)}{6\Gamma(s+1)}(6\gamma^2+\pi^2+12\gamma\psi_0(s+1)+6\psi_0^2(s+1)-6\psi_1(s+1)). \end{eqnarray*} Note $$ \frac{\Gamma(s)}{6\Gamma(s+1)}=\frac{1}{6s}+\mathcal{O}(s^3)$$ and $$ 6\gamma^2+\pi^2+12\gamma\psi_0(s+1)+6\psi_0^2(s+1)-6\psi_1(s+1)=-6\psi_2(1)s-\frac{\pi^4}{30}s^2+(\pi^2\psi_2(1)-\psi_4(1))s^3+\mathcal{O}(s^3)$$ and hence \begin{eqnarray*} &&\frac{\Gamma(s)}{6\Gamma(s+1)}(6\gamma^2+\pi^2+12\gamma\psi_0(s+1)+6\psi_0^2(s+1)-6\psi_1(s+1))\\ &=&-\psi_2(1)-\frac{\pi^4}{180}s+\frac{1}{6}(\pi^2\psi_2(1)-\psi_4(1))s^2+\mathcal{O}(s^3). \end{eqnarray*} Thus \begin{eqnarray} \lim_{s\to0}\lim_{w\to0}\frac{d^2}{ds^2}\frac{d^2}{dw^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(w+1)} &=&\lim_{s\to0}\frac{d^2}{ds^2}\frac{\Gamma(s)}{6\Gamma(s+1)}(6\gamma^2+\pi^2+12\gamma\psi_0(s+1)+6\psi_0^2(s+1)-6\psi_1(s+1))\\ &=&\frac{1}{3}(\pi^2\psi_2(1)-\psi_4(1)). \end{eqnarray} So $$ \frac{1}{16}\lim_{w\to0}\lim_{s\to0}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}=\frac{1}{48}(\pi^2\psi_2(1)-\psi_4(1)). $$ Finally note $$ \zeta(2)=\frac{\pi^2}{6},\psi_2(1)=-2\zeta(3),\psi_4(1)=-24\zeta(5) $$ and hence $$ \frac{1}{16}\lim_{w\to0}\lim_{s\to0}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}=\frac{1}{48}(\pi^2\psi_2(1)-\psi_4(1))=\frac{1}{4}(2\zeta(5)-\zeta(2)\zeta(3)). $$

Answer 3

Here is another way to solve the integral. Let $$ \mathcal{I}=\int_0^{\Large\frac\pi2}\frac{\ln^2(\cos x)\ln^2(\sin x)}{\cos x\sin x}\ dx. $$ Multiplying $\,\mathcal{I}\,$ by $\,\dfrac{2\sin x\cos x}{2\sin x\cos x}\,$ and setting $\,t=\sin^2x\,$ yield \begin{align} \frac1{32}\int_0^1\frac{\ln^2(1-t)\ln^2t}{(1-t)\ t}\ dt&=\frac1{32}\left[\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt+\color{blue}{\underbrace{\color{black}{\int_0^1\frac{\ln^2(1-t)\ln^2t}{1-t}\ dt}}_{\color{red}{x\ \mapsto\ 1-x}}}\right]\\ &=\frac1{16}\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt. \end{align} The latter integral can be evaluated using IBP by setting $$u=\ln^2(1-t)\ \color{red}{\Rightarrow}\ du=-\dfrac{2\ln(1-t)}{1-t}\quad \text{and}\quad dv=\dfrac{\ln^2t}{t}\ dt\ \color{red}{\Rightarrow}\ v=\dfrac13\ln^3t.$$ Hence \begin{align} \frac1{16}\int_0^1\frac{\ln^2(1-t)\ln^2t}{t}\ dt&=\frac1{16}\left[\left.\frac13\ln^3t\ln^2(1-t)\right|_{t=0}^1+\frac23\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt\right]\\ &=\frac1{24}\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt. \end{align} The latter integral has been evaluated in my other answer (click the link below). \begin{align} \color{blue}{\int\frac{\ln^3x\ln (1-x)}{1-x}\ dx}=&\ -\mathbf{H}_{1}(x)\ln^3x+\operatorname{Li}_2(x)\ln^3x+3\,\mathbf{H}_{2}(x)\ln^2x-3\operatorname{Li}_3(x)\ln^2x\\&\ -6\,\mathbf{H}_{3}(x)\ln x+6\operatorname{Li}_4(x)\ln x+6\,\mathbf{H}_{4}(x)-6\operatorname{Li}_5(x), \end{align} where $\displaystyle\mathbf{H}_{k}(x)=\sum_{n=0}^\infty\frac{H_nx^n}{n^k}$ and $$ \mathbf{H}_{k}(1)=\frac{(k+2)}2\zeta(k+1)-\frac12\sum_{m=1}^{k-2}\zeta(k-m)\zeta(m+1)\quad;\quad\text{for}\ k\in\mathbb{Z}\ge2. $$

Therefore \begin{align} \int_0^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx=6\,\mathbf{H}_{4}(1)-6\operatorname{Li}_5(1)=12\zeta(5)-6\zeta(2)\zeta(3). \end{align} Alternatively, we can also use the following technique \begin{align} \int_0^1\frac{\ln^3x\ln (1-x)}{1-x}\ dx&=-\int_0^1\sum_{n=1}^\infty H_nx^n\ln^3x\ dx\\ &=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^3x\ dx\\ &=\sum_{n=1}^\infty\frac{3!\ H_n}{(n+1)^4}\tag1\\ &=6\sum_{n=1}^\infty\left[\frac{H_n}{n^4}-\frac1{n^5}\right]\tag2\\ &=6\bigg[3\zeta(5)-\zeta(2)\zeta(3)-\zeta(5)\bigg]\\ &=6\bigg[2\zeta(5)-\zeta(2)\zeta(3)\bigg].\\ \end{align} Thus $$ I=\frac1{24}\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt=\color{blue}{\frac14\bigg[2\zeta(5)-\zeta(2)\zeta(3)\bigg]}.\tag{Q.E.D.} $$

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Notes :

$\displaystyle[1]\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

$\displaystyle[2]\ \ H_{n+1}-H_n=\frac1{n+1}$

Answer 4

\begin{align} I&=\int_0^{\pi/2}\frac{\ln^2\cos x\ln^2\sin x}{\cos x\sin x}\ dx\overset{\sin x=u}{=}\frac1{4}\int_0^1\frac{\ln^2(1-x^2)\ln^2x}{x(1-x^2)}\ dx\\ &=\frac1{32}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\ &=\frac1{32}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\frac1{32}\underbrace{\int_0^1\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{\large{1-x\ \mapsto\ x}}\\ &=\frac1{16}\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx\\ &=\frac18\sum_{n=1}^{\infty}\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2x\ dx=\frac14\sum_{n=1}^{\infty}\left(\frac{H_n}{n^4}-\frac{1}{n^5}\right)\\ &=\frac14\left(3\zeta(5)-\zeta(2)\zeta(3)-\zeta(5)\right)\\ &=\frac14\left(2\zeta(5)-\zeta(2)\zeta(3)\right) \end{align}

Answer 5

Let $x=\sin ^2 \theta$, then $$ \begin{aligned} \int_0^1 \frac{\ln ^2(1-x) \ln ^2 x}{x} d x&=\int_0^1 \frac{\ln ^2\left(\cos ^2 \theta\right) \ln ^2\left(\sin ^2 \theta\right)}{\sin ^2 \theta} 2 \sin \theta d \theta\\ &=32 \int_0^{\frac{\pi}{2}} \frac{\ln ^2(\cos \theta) \ln ^2(\sin \theta)}{\tan \theta} d \theta\\ &=\left.32 \cdot \frac{1}{2} \frac{\partial^4}{\partial a^2 \partial b^2} B\left(\frac{a}{2}, \frac{b}{2}+1\right)\right| _{{a \rightarrow 0}, {b \rightarrow 0}}\\ &=\lim _{a \rightarrow 0} \lim _{b \rightarrow 1}\left[\frac{\partial^{4}}{\partial a^2\partial b^2} B(a, b)\right] \end{aligned} $$