# How may one show that $\int_{0}^{\pi/2}{\ln\cos x\over \tan x}\cdot\ln\left({\ln\sin x\over \ln \cos x}\right)\mathrm dx={\pi^2\over 4!}?$

Motivated from Integral contest, a slight variation of it

$$\int_{0}^{\pi/2}{\ln\cos x\over \tan x}\cdot\ln\left({\ln\sin x\over \ln \cos x}\right)\mathrm dx={\pi^2\over 4!}\tag1$$

Making an attempt:

$$u={\ln\sin x\over \ln \cos x}\implies du={\cot x\ln\cos x+\tan x \ln\sin x\over \ln^2\cos x}dx\tag2$$

$$u=\ln\cos x\implies du=-\tan x dx\tag3$$

$$u=\ln\sin x\implies du=\cot x dx\tag4$$

Using $(4)$, then $(1)$ becomes

$$\int_{0}^{\infty}\ln\sqrt{1-e^{-2u}}\ln\left({\ln\sqrt{1-e^{-2u}}\over u}\right)\mathrm du\tag5$$

Recall $$\ln(1-x)=\sum_{n=1}^{\infty}{(-1)^n\over n}x^n\tag6$$

Then $(5)$ becomes

$$\sum_{n=1}^{\infty}{(-1)^n\over 2n}\color{red}{\int_{0}^{1}e^{-2un}\ln\left({\ln\sqrt{1-e^{-2u}}\over u}\right)\mathrm du}\tag7$$

The red part rewrite as

$$I_1-I_2={1\over 2}\int_{0}^{1}e^{-2un}\ln\left(\ln(1-e^{-2un})\right)\mathrm du-\int_{0}^{1}e^{-2un}\ln u\mathrm du\tag8$$

$$I_1-\color{blue}{I_2}={1\over 2}\int_{0}^{1}e^{-2un}\ln\left[\ln(1-e^{-2un})\right]\mathrm du-\color{blue}{{1\over 4}[Ei(-2n)-\gamma -\ln(2n)]}\tag9$$

Where Ei is the Exponential integral Ei

As for $I_1$ seems very difficult.

How can we prove $(1)?$

• have you proved that this integral does converge? – Dr. Sonnhard Graubner Apr 15 '17 at 6:32

Consider the integral

$$I=\int_{0}^{\pi/2}{\log(\cos x)\over \tan x}\cdot\log {|\log(\sin x)|\over |\log (\cos x)|}\mathrm dx$$

Then by separating and then using the substitution $x \to \pi/2-x$

$$I = \int^{\pi/2}_0(\cot(x)\log(\cos x)-\tan(x)\log(\sin x))\log| \log (\sin x)|\,dx$$

By noting that

$$\int \cot(x)\log(\cos x)-\tan(x)\log(\sin x)\,dx = \log(|\sin x|)\log(|\cos x|)+C$$

Using integration by parts

$$I = -\int^{\pi/2}_0\cot(x)\log(\cos x)\,dx$$

Let $\cos(x) =t$

\begin{align}I =- \int^{1}_0\frac{t\log(t)}{1-t^2}\,dt &= -\frac{1}{2}\int^{1}_0\frac{\log(1-t^2)}{t}\,dt\\ &= -\frac{1}{2}\int^{1}_0\frac{\log(1+t)}{t}\,dt-\frac{1}{2}\int^{1}_0\frac{\log(1+t)}{t}\,dt\\ &=\frac{\mathrm{Li}_2(1)+\mathrm{Li}_2(-1)}{2}\\ &=\frac{1}{2}\left[\frac{\pi^2}{6}-\frac{\pi^2}{12}\right]\\ &=\frac{\pi^2}{24} \end{align}

• How did you get to $$I=-\frac 12\int\limits_0^1 dt\,\frac {\log(1-t^2)}t$$? DId you substitute $u=1-t^2$? – Frank W May 2 '18 at 15:32
• @FrankW, yes ... – Zaid Alyafeai May 2 '18 at 16:44
• Then shouldn’t the natural log become $\sqrt{u-1}$? – Frank W May 2 '18 at 17:13
• @FrankW, it will be $\log(\sqrt{1-u}) = \frac{1}{2} \log(1-u)$ – Zaid Alyafeai May 2 '18 at 21:43
• I’m aware of that. But your natural logarithmic argument is $1-t^2$ instead of $1-u$. Also, when you differentiate, it becomes $du=-2t\, dt$ so $dt=-\tfrac {du}{2t}$ – Frank W May 2 '18 at 22:05