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Prove that:

$$\int_{0}^{\pi/2} \theta^2 \cot \theta \, {\rm d}\theta= \frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3)$$

I like to use the relation

$$\int_{a}^{b}p(x)\cot(x)dx=2\sum_{k=1}^{\infty}\int_{a}^{b}p(x)\sin(2kx)dx$$ when doing integrals that involve the product of a polynomial and cot

There are also similar relations for csc and so forth. If this were csc instead of cot, we would use $\sin[(2k+1)x]$ instead.

In this case $p(x)=x^{2}$

So, we can say, $$2\sum_{k=1}^{\infty}\int_{0}^{\frac{\pi}{2}}x^{2}\sin(2kx)dx$$

$$=\frac{\pi^{2}}{4}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k^{3}}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k^{3}}$$

These sums are familiar and we have:

$$\frac{\pi^{2}}{4}\log(2)-\frac{3}{8}\zeta(3)-\frac{1}{2}\zeta(3)$$

$$\frac{\pi^{2}}{4}\log(2)-\frac{7}{8}\zeta(3)$$

Tolaso J Kos wrote:Prove that:

$$\int_{0}^{\pi/2} \theta^2 \cot \theta \, {\rm d}\theta= \frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3)$$

$$I=\int\limits_{0}^{\frac{\pi }{2}}{x^{2}\cot xdx}=\left( x^{2}\ln \sin x \right)\left| _{\left( \frac{\pi }{2},0 \right)} \right.-\int\limits_{0}^{\frac{\pi }{2}}{2x\ln \sin xdx}=\int\limits_{0}^{\frac{\pi }{2}}{2x\left( \ln 2-\ln \left( 2\sin x \right) \right)dx}$$

$$=\frac{\pi ^{2}}{4}\ln 2-\int\limits_{0}^{\frac{\pi }{2}}{2x\ln \left( 2\sin x \right)dx}=\frac{\pi ^{2}}{4}\ln 2-\int\limits_{0}^{\frac{\pi }{2}}{2x\sum\limits_{n=1}^{+\infty }{\frac{\cos 2nx}{n}dx}}=\frac{\pi ^{2}}{4}\ln 2-\sum\limits_{n=1}^{+\infty }{\frac{1}{n}\int\limits_{0}^{\frac{\pi }{2}}{2x\cos 2nxdx}}$$

$$=\frac{\pi ^{2}}{4}\ln 2-\sum\limits_{n=0}^{+\infty }{\frac{1}{\left( 2n+1 \right)^{3}}}=\frac{\pi ^{2}}{4}\ln 2-\frac{7}{8}\zeta \left( 3 \right)$$

Regards