$\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:

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- Feb 14th 2011, 05:59 PMRandom Variabledefinite integral #4
$\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__: - Feb 14th 2011, 09:24 PMKrizalid
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 $\displaystyle \dfrac18(2\pi^2\ln2-7\zeta(3)).$

- Feb 15th 2011, 09:53 AMRandom Variable
- Feb 15th 2011, 10:27 AMKrizalid
$\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

$\displaystyle \displaystyle\ln (\sin x)=-\sum\limits_{j=1}^{\infty }{\frac{\cos (2jx)}{j}}-\ln 2.$ - Feb 15th 2011, 02:46 PMtopsquark
- Feb 15th 2011, 02:52 PMKrizalid
You mean power series for contangent? I don't know it.

Perhaps it's better you to share your solution. :D - Feb 15th 2011, 03:39 PMtopsquark
- Feb 15th 2011, 09:03 PMCaptainBlack
It lurks about halfway down >>this<< page.

CB - Feb 18th 2011, 06:18 PMchiph588@
$\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} $

$\displaystyle \displaystyle \implies \frac1x-\pi\cot(\pi x) = 2\sum_{n=1}^\infty \zeta(2n)x^{2n-1} \quad\quad \text{(for } |x|<1\text{)}$