# Evaluating $\int_0^{\frac\pi2}\frac{\ln{(\sin x)}\ \ln{(\cos x})}{\tan x}\ dx$

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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Could help that $\displaystyle\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\tan x}$ ? – user 1618033 Aug 30 '12 at 19:41
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? – user 1618033 Aug 30 '12 at 20:21
Finally, you get $\frac{1}{8} \zeta(3)$. – user 1618033 Aug 30 '12 at 20:28

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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+1 I love this answer – dot dot Aug 31 '12 at 9:50
It is a nice answer. – Mhenni Benghorbal Aug 31 '12 at 13:37
@user78416 sure, wait 30 seconds. – user 1618033 May 30 '13 at 21:01
@user78416 see robjohn's answer here math.stackexchange.com/questions/330057/… – user 1618033 May 30 '13 at 21:02
Brilliant, thanks very much – user78416 May 30 '13 at 21:05

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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It looks better the alternative with the series. – user 1618033 Aug 30 '12 at 21:07
I do not disagree, but I left them in the order they occurred to me. – Sasha Aug 30 '12 at 21:11
Nice solution. Good technique. – Mhenni Benghorbal Aug 31 '12 at 13:34

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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\begin{align}&\overbrace{\color{#66f}{\large\int_{0}^{\pi/2}{\ln\pars{\sin\pars{x}}\ln\pars{\cos\pars{x}}\over \tan\pars{x}}\,\dd x}} ^{\ds{\mbox{Set}\ \sin\pars{x} \equiv t\ \imp\ 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}=\color{#66f}{\large{1 \over 8}\,\zeta\pars{3}} \approx 0.1503 \end{align}

$\ds{{\rm Li}_{\rm s}\pars{z}}$ is a PolyLogarithm Function and we used well known properties of them as reported in the cited link.

$\ds{\zeta\pars{s}}$ is the Riemann Zeta Function.

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