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I need to solve $$ \int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx $$

I tried to use symmetric properties of the trigonometric functions as is commonly used to compute $$ \int_0^{\Large\frac\pi2}\ln\sin x\ dx = -\frac{\pi}{2}\ln2 $$ but never succeeded. (see this for example)

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  • $\begingroup$ Could help that $\displaystyle\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\tan x}$ ? $\endgroup$ – user 1357113 Aug 30 '12 at 19:41
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    $\begingroup$ With the substitution $\ln(\sin x)=u$ you get $\displaystyle \frac{1}{2} \int_{-oo}^{0} \ln(1-e^{2u}) u \ du$. Then you may use Taylor expansion for $\ln(1-e^{2u})$ and integrate term by term. Does it help? $\endgroup$ – user 1357113 Aug 30 '12 at 20:21
  • $\begingroup$ Finally, you get $\frac{1}{8} \zeta(3)$. $\endgroup$ – user 1357113 Aug 30 '12 at 20:28
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Let's start out with the substitution $ \displaystyle \ln(\sin x) = u $ and get: $$\ln(\cos x)=\frac{\ln(1-e^{2u})}{2}$$ $$\displaystyle\frac{1}{\tan x} \ dx =du$$ that further yields $$\int_0^{\pi/2}\frac{(\ln{\sin x})(\ln{\cos x})}{\tan x}dx= \frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du$$

According to Taylor expansion we have $$\ln(1-e^{2u})= \sum_{k=1}^{\infty} (-1)^{2k+1} \frac{e^{2 k u}}{k}$$ then $$\frac{1}{2} \int_{-\infty}^{0} \ln(1-e^{2u}) u \ du=$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \int_{-\infty}^{0} u e^{2ku} \ du =$$ $$\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{2k+1}}{k} \frac{-1}{4k^2} = \frac{1}{8} \sum_{k=1}^{\infty} \frac{1}{k^3}=\frac{1}{8} \zeta(3).$$

Remark: the value of $\zeta(3)\approx1.2020569$ is called Apéry's Constant - see here.

Q.E.D. (Chris)

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  • $\begingroup$ It is a nice answer. $\endgroup$ – Mhenni Benghorbal Aug 31 '12 at 13:37
  • $\begingroup$ @user78416 sure, wait 30 seconds. $\endgroup$ – user 1357113 May 30 '13 at 21:01
  • $\begingroup$ @user78416 see robjohn's answer here math.stackexchange.com/questions/330057/… $\endgroup$ – user 1357113 May 30 '13 at 21:02
  • $\begingroup$ Brilliant, thanks very much $\endgroup$ – user78416 May 30 '13 at 21:05
  • $\begingroup$ Hello Chris, I was querying as to whether there is elementary solution without the use of contour integration/ the beta function as I have not delved into such topics yet :)? $\endgroup$ – user78416 May 31 '13 at 0:39
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Let $u = \sin^2(x)$. Then $\frac{\mathrm{d}x}{\tan(x)} = \frac{\cos(x)}{\sin(x)} \mathrm{d}x = \frac{d\sin(x)}{\sin(x)} = \frac{1}{2}\frac{\mathrm{d}u}{u}$, $\ln(\sin(x)) = \frac{1}{2}\ln(u)$ and $\ln(\cos(x)) = \frac{1}{2} \ln(1-u)$: $$ \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \int_0^{1} \frac{\ln u}{u} \cdot \ln(1-u) \mathrm{d} u = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \int_0^1 u^{s-1} (1-u)^{t-1} \mathrm{d}u \right|_{s\to 0^+,t=1} = \left.\frac{1}{8} \frac{\mathrm{d}^2}{\mathrm{d} s \mathrm{d} t} \frac{\Gamma(s) \Gamma(t)}{\Gamma(s+t)} \right|_{s\to 0^+,t=1} $$ First differentiate with respect to $t$ and substitute $t=1$: $$ \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\Gamma(s)}{\Gamma(s+1)}\left( \psi(1) - \psi(s+1)\right) \right|_{s\to 0^+} = \left.\frac{1}{8} \frac{\mathrm{d}}{\mathrm{d} s} \frac{\left( \psi(1) - \psi(s+1)\right) }{s} \right|_{s\to 0^+} $$ Using Taylor series expansion for the digamma function $\psi(s)$ we have: $$ \frac{\left( \psi(1) - \psi(s+1)\right) }{s} = -\zeta(2) + \zeta(3) s + \mathcal{o}(s) $$ Hence the value of the integral is: $$ \int_0^{\pi/2} \frac{\ln(\sin(x)) \ln(\cos(x))}{\tan(x)} \mathrm{d} x = \frac{1}{8} \zeta(3) $$


Alternatively you could use $$\frac{\ln(1-u)}{u} = -\sum_{k=0}^\infty \frac{u^k}{k+1}$$ and integrate term-wise: $$\int_0^1 u^k \ln(u) \mathrm{d} u \stackrel{u=\exp(-t)}{=} \int_0^\infty t \exp(-t(k+1)) \mathrm{d} t = -\frac{1}{(k+1)^2}$$ which yields the result immediately.

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  • $\begingroup$ It looks better the alternative with the series. $\endgroup$ – user 1357113 Aug 30 '12 at 21:07
  • $\begingroup$ I do not disagree, but I left them in the order they occurred to me. $\endgroup$ – Sasha Aug 30 '12 at 21:11
  • $\begingroup$ Nice solution. Good technique. $\endgroup$ – Mhenni Benghorbal Aug 31 '12 at 13:34
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Rewrite the integral as $$ \int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx=\int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{\sqrt{1-\sin^2 x}}}{\sin x}\cdot\cos x\ dx. $$ Set $t=\sin x\ \color{red}{\Rightarrow}\ dt=\cos x\ dx$, then we obtain \begin{align} \int_0^{\Large\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx&=\frac12\int_0^1\frac{\ln t\ \ln(1-t^2)}{t}\ dt\\ &=-\frac12\int_0^1\ln t\sum_{n=1}^\infty\frac{t^{2n}}{nt}\ dt\tag1\\ &=-\frac12\sum_{n=1}^\infty\frac{1}{n}\int_0^1t^{2n-1}\ln t\ dt\\ &=\frac12\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{1}{(2n)^2}\tag2\\ &=\large\color{blue}{\frac{\zeta(3)}{8}}. \end{align}


Notes :

$[1]\ \ $Use Maclaurin series for natural logarithm: $\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\ $ for $|x|<1$.

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

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

$\ds{\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}} \over \tan\pars{x}}\,\dd x:\ {\large ?}}$

\begin{align} &\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \tan\pars{x}}\,\dd x \,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\, \int_{0}^{1}{\ln\pars{t}\ln\pars{\root{1 - t^{2}}} \over t\,/\root{1 - t^{2}}}\, {\dd t \over \root{1 - t^{2}}} \\[5mm]= &\ \half\int_{0}^{1}{\ln\pars{t}\ln\pars{1 - t^{2}} \over t}\,\dd t = \half\int_{0}^{1}{\ln\pars{t^{1/2}}\ln\pars{1 - t}\over t^{1/2}}\,\half\,t^{-1/2}\,\dd t \\[5mm] = &\ {1 \over 8}\int_{0}^{1}{\ln\pars{t}\ln\pars{1 - t} \over t}\,\dd t =-\,{1 \over 8}\int_{0}^{1}{{\rm Li}_{1}\pars{t} \over t}\,\ln\pars{t}\,\dd t =-\,{1 \over 8}\int_{0}^{1}{\rm Li}_{2}'\pars{t}\ln\pars{t}\,\dd t \\[5mm]&={1 \over 8}\int_{0}^{1}{{\rm Li}_{2}\pars{t} \over t}\,\dd t ={1 \over 8}\int_{0}^{1}{\rm Li}_{3}'\pars{t}\,\dd t ={1 \over 8}\,{\rm Li}_{3}\pars{1} = \bbox[10px,border:1px dotted navy]{\ds{\zeta\pars{3}}} \approx 0.1503 \end{align}

$\ds{{\rm Li}_{\rm s}\pars{z}}$ is a PolyLogarithm Function: I used some properties of them as reported in the cited link. $\ds{\zeta\pars{s}}$ is the Riemann Zeta Function.

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  • $\begingroup$ Very nice answer. +1 $\endgroup$ – Random Variable Jul 17 '14 at 14:25
  • $\begingroup$ @RandomVariable Thanks. I'm ${\rm Li}$ fan. $\endgroup$ – Felix Marin Jul 17 '14 at 20:05

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