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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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    Finally, you get $\frac{1}{8} \zeta(3)$.2012-08-30

4 Answers 4

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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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    That's the first integral in a long time I've seen where the Taylor series has been used to expand an improper integral! Nice to see.2014-09-18
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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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    Nice solution. Good technique.2012-08-31
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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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$\ds{\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}} \over \tan\pars{x}}\,\dd x:\ {\large ?}}$

\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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    @RandomVariable Thanks. I'm ${\rm Li}$ fan.2014-07-17