$\displaystyle \displaystyle \int^{\pi/2}_{0} x^{2} \cot x \ dx $
I have a solution using contour integration. You could either try to replicate that solution, or preferably find a simpler approach.
Spoiler:
$\displaystyle \displaystyle \int^{\pi/2}_{0} x^{2} \cot x \ dx $
I have a solution using contour integration. You could either try to replicate that solution, or preferably find a simpler approach.
Spoiler:
Integrate by parts and the integral equals $\displaystyle -2\displaystyle\int_0^{\frac\pi2}x\ln(\sin x)\,dx,$ now use $\displaystyle \displaystyle\ln (\sin x)=-\ln 2-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}},$ and that gives the answer $\displaystyle \dfrac18(2\pi^2\ln2-7\zeta(3)).$
$\displaystyle \begin{aligned}
\ln (\cos x)&=\text{Re}\ln \left( \frac{1+{{e}^{2ix}}}{2{{e}^{2ix}}} \right) \\
& =\text{Re}\left( \ln \left( 1+{{e}^{2ix}} \right)-2ix-\ln 2 \right) \\
& =\text{Re}\left( -\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}{{e}^{2ijx}}}{j}} \right)-\ln 2 \\
& =-\sum\limits_{j=1}^{\infty }{\frac{{{(-1)}^{j}}\cos (2jx)}{j}}-\ln 2, \\
\end{aligned}$
now put $\displaystyle x\mapsto\dfrac\pi2-x$ and get
$\displaystyle \displaystyle\ln (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\ln 2.$
$\displaystyle \displaystyle \frac{\sin\pi x}{\pi x} = \prod_{n=1}^\infty \left(1-\frac{x^2}{n^2}\right) \implies \log(\pi x) - \log(\sin\pi x) = \sum_{n=1}^\infty \frac{\zeta(2n)}{n}x^{2n} $
$\displaystyle \displaystyle \implies \frac1x-\pi\cot(\pi x) = 2\sum_{n=1}^\infty \zeta(2n)x^{2n-1} \quad\quad \text{(for } |x|<1\text{)}$