$\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\sin^2 \left ( x- \frac{\pi}{4} \right )} \;dx$

Evaluate the following integral:

$$\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\sin^2 \left ( x- \frac{\pi}{4} \right )}  \, {\rm d}x$$

Solution


We have successively:

\begin{align*}
\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\sin^2 \left ( x- \frac{\pi}{4} \right )} \, {\rm d}x  &= \int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \right )^2} \, {\rm d}x\\
 &= 2\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \sin x  - \cos x \right )^2} \, {\rm d}x\\
 &= 2 \int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \frac{\sin x}{ \cos x} -1 \right )^2 \cos^2 x} \, {\rm d}x\\
 &= 2\int_{0}^{\pi/2} \frac{\log^2 \left ( \tan x \right )}{\left ( \tan x - 1 \right )^2} \sec^2 x \, {\rm d}x\\
 &\!\!\!\!\! \overset{u=\tan x}{=\! =\! =\! =\!} \; 2 \int_{0}^{\infty} \frac{\log^2 u}{\left ( u-1 \right )^2} \, {\rm d}u \\
 &= 2 \left [ \int_{0}^{1} \frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d} u + \int_{1}^{\infty} \frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d}u \right ] \\
 &=4 \int_{0}^{1}\frac{\log^2 u}{\left ( 1-u \right )^2} \, {\rm d}u \\
 &= 4 \int_{0}^{1} \log^2 u \sum_{n=1}^{\infty} n u^{n-1} \, {\rm d}u \\
 &= 4 \sum_{n=1}^{\infty} n \int_{0}^{1} u^{n-1 }\log^2 u \, {\rm d}u \\
 &= 4 \cdot 2 \sum_{n=1}^{\infty} \frac{1}{n^2} \\
 &= 8 \zeta(2) \\
 &=\frac{4\pi^2}{3}
\end{align*}