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

Question

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)

Could help that $\displaystyle\frac{d}{dx}(\ln(\sin x)) = \frac{1}{\tan x}$ ? — user 1591719, 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 1591719, Aug 30 '12 at 20:21

score 35 · Accepted Answer · edited Aug 31 '12 at 09:52

35

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)

edited Aug 31 '12 at 09:52

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answered Aug 31 '12 at 07:04

user 1591719

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It is a nice answer. – Mhenni Benghorbal Aug 31 '12 at 13:37
@user78416 sure, wait 30 seconds. – user 1591719 May 30 '13 at 21:01
@user78416 see robjohn's answer here http://math.stackexchange.com/questions/330057/how-to-evaluate-i-displaystyle-int-0-pi-2x2-ln-sin-x-ln-cos-xdx – user 1591719 May 30 '13 at 21:02
Brilliant, thanks very much – user78416 May 30 '13 at 21:05
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 :)? – user78416 May 31 '13 at 00:39
@user78416 my fastest solution uses a clever beta function manipulation, but if you want something elementarily, combine Fourier series with double sums as robjohn did in the first answer. I didn't specifically try to find an elementary proof. – user 1591719 May 31 '13 at 06:53
Can you show me why is this true? $\ln(\cos x)=\frac{\ln(1-e^{2u})}{2}$ – user010010001 Mar 30 '14 at 15:44
@user137425 $$\ln(\sin x)=u \Rightarrow \sin(x)=e^u \Rightarrow \cos^2(x)=1-e^{2u} \Rightarrow \log(\cos(x))=\frac{\ln(1-e^{2u})}{2}$$ – user 1591719 Mar 30 '14 at 18:13
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. – pshmath0 Sep 18 '14 at 11:14

Sasha · Answer 2 · 2012-08-30T20:22:43.403

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.

I do not disagree, but I left them in the order they occurred to me. — Sasha, Aug 30 '12 at 21:11

score 10 · Answer 3 · edited Apr 13 '17 at 12:21

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$

Felix Marin · Answer 4 · 2017-07-08T21:28:54.817

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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.

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

user170231 · Answer 5 · 2022-12-16T19:52:05.427

$$\begin{align*} I &= \int_0^{\frac\pi2} \frac{\log(\cos(x)) \log(\sin(x))}{\tan(x)} \, dx \\[1ex] &= \int_0^\infty \frac{\log\left(\frac1{\sqrt{x^2+1}}\right) \log\left(\frac x{\sqrt{x^2+1}}\right)}{x} \, \frac{dx}{1+x^2} \tag{1} \\[1ex] &= \frac14 \int_0^\infty \frac{\log^2(x^2+1)-\log(x^2)\log(x^2+1)}{x} \, \frac{dx}{1+x^2} \\[1ex] &= \frac18 \int_0^\infty \frac{\log^2(x+1)-\log(x)\log(x+1)}{x(1+x)} \, dx \tag{2} \\[1ex] &= \frac18 \int_0^1 \left(\frac{\log^2(x+1)}{x} - \frac{\log(x)\log(x+1)}x\right) \, dx \tag{3} \\[1ex] &= \frac14 \int_0^1 \frac{\log(x)\log(x+1)}x \, dx - \frac12 \int_0^1 \frac{\log(x+1)\log(x)}{x+1} \, dx \tag{4} \\[1ex] &= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}n \int_0^1 x^{n-1} \log(x) \, dx + \frac12 \sum_{n=1}^\infty (-1)^n H_n \int_0^1 x^n \log(x) \, dx \tag{5} \\[1ex] &= -\frac14 \sum_{n=1}^\infty \frac{(-1)^n}{n^3} + \frac12 \sum_{n=1}^\infty \frac{(-1)^n H_n}{(n+1)^2} \tag{4} \\[1ex] &= \boxed{\frac{\zeta(3)}8} \tag{6} \end{align*}$$

$(1)$ : substitute $x\mapsto\arctan(x)$
$(2)$ : substitute $x\mapsto\sqrt x$
$(3)$ : split up the integral at $x=1$; substitute $x\mapsto\frac1x$ in the integral over $[1,\infty)$; partial fractions
$(4)$ : integrate by parts
$(5)$ : recall the power series $-\log(1-x)=\sum\limits_{n\ge1}\frac{x^n}n$ and $-\frac{\log(1-x)}{1-x}=\sum\limits_{n\ge1} H_n x^n$, where $H_n$ is the $n^{\rm th}$ harmonic number
$(6)$ : see e.g. section $7$ for a contour-integration method to evaluating the Euler sum; I'm sure there are more efficient means

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

5 Answers5

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