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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?

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    $\begingroup$ What is so nice about it? $\endgroup$
    – Kaster
    Jan 30, 2013 at 2:48
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    $\begingroup$ It is interesting, in that neither of $\zeta(3),\ \zeta(5)$ are "known", in the sense that $\zeta(2)=\pi^2/6$ is known. $\endgroup$
    – coffeemath
    Jan 31, 2013 at 9:03
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    $\begingroup$ Please, make titles more informative. $\endgroup$
    – Pedro
    May 28, 2013 at 18:55
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    $\begingroup$ I edited again the tags deleting the tag "harmonic numbers" because it is completely unrelated to the question despite the fact that some solutions can be stated in terms of harmonic numbers $\endgroup$
    – Masacroso
    Feb 16, 2020 at 14:39
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    $\begingroup$ @AliShather the tags are for the question, not for particular answers. I called for moderation about this particular matter, let the moderators decide if this is correct or not, to me is clearly not $\endgroup$
    – Masacroso
    Feb 16, 2020 at 16:27

4 Answers 4

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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.

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  • $\begingroup$ Thx for the answering ! I think since there are a lots of zeta involves. Maybe we use the series at the start? :) $\endgroup$
    – Ryan
    Feb 1, 2013 at 23:40
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    $\begingroup$ @Ryan:You are welcome. Note that, you can get different forms of solutions for this integral. $\endgroup$ Feb 2, 2013 at 3:11
  • $\begingroup$ :You may hav a look at this question too :) math.stackexchange.com/questions/291850/… $\endgroup$
    – Ryan
    Feb 2, 2013 at 14:26
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    $\begingroup$ @MhenniBenghorbal Nice and concise! (+1) $\endgroup$ May 28, 2013 at 19:12
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    $\begingroup$ @MhenniBenghorbal, how exactly did you take the beta function limit (excellent answer, by the way)? $\endgroup$
    – Meow
    Dec 9, 2013 at 18:16
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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 $$ I=\frac1{24}\int_0^1\frac{\ln(1-t)\ln^3t}{1-t}\ dt=\color{blue}{\frac14\bigg[2\zeta(5)-\zeta(2)\zeta(3)\bigg]}.\tag{Q.E.D.} $$


Notes :

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

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

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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 $$ \frac{1}{16}\lim_{w\to0}\lim_{s\to0}\frac{d^2}{dw^2}\frac{d^2}{ds^2}\frac{\Gamma(s)\Gamma(w+1)}{\Gamma(s+w+1)}=\frac{1}{48}(\pi^2\psi_2(1)-\psi_4(1))=\frac{1}{4}(2\zeta(5)-\zeta(2)\zeta(3)). $$

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\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}

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