How to calculate $\int_{0}^{\frac{\pi}{2}}{x(x-\pi)\cot(x)dx}$

Question

I know how to calculate $\int_{0}^{\frac{\pi}{2}}{x\cot(x)dx}$

But when it adds $(x-\pi)$, I don't know how to do.

Answer 1

Integration by parts tells that

$$ \int_{0}^{\frac{\pi}{2}} x(x-\pi)\cot x \, dx = \underbrace{\left[ x(x-\pi) \log\sin x \right]_{0}^{\frac{\pi}{2}}}_{=0} - \int_{0}^{\frac{\pi}{2}} (2x-\pi) \log\sin x \, dx. $$

Now notice that

$$ -\log\sin x = -\log\left| \frac{1 - e^{2ix}}{2}\right| = \log 2 + \sum_{n=1}^{\infty} \frac{\cos (2nx)}{n} $$

So we have

\begin{align*} \int_{0}^{\frac{\pi}{2}} x(x-\pi)\cot x \, dx &= \int_{0}^{\frac{\pi}{2}} (2x-\pi) \left( \log 2 + \sum_{n=1}^{\infty} \frac{\cos (2nx)}{n} \right) \, dx \\ &= -\frac{\pi^2}{4}\log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n - 1}{2n^3} \\ &= -\frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3). \end{align*}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\pi/2}x\pars{x - \pi}\cot\pars{x}\,\dd x}} \stackrel{\mrm{IBP}}{=} -\int_{0}^{\pi/2}\pars{2x - \pi}\ln\pars{\sin\pars{x}}\,\dd x \\[5mm] \stackrel{x\ \mapsto\ \pi/2 - x}{=}&\ 2\int_{0}^{\pi/2}x\ln\pars{\cos\pars{x}}\,\dd x = \left.2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2} \bracks{-\ic\ln\pars{z}}\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left. -2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2} \ln\pars{z}\ln\pars{1 + z^{2} \over 2z}\,{\dd z \over z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}&\ 2\,\Re\int_{1}^{\epsilon} \bracks{\ln\pars{y} + {\pi \over 2}\,\ic} \bracks{\ln\pars{1 - y^{2} \over 2y} - {\pi \over 2}\,\ic} \,{\ic\,\dd y \over \ic y} \\[2mm] &\ + 2\,\Re\int_{\pi/2}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta} \bracks{-\ln\pars{2\epsilon} - \ic\theta}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} + 2\,\Re\int_{\epsilon}^{1} \ln\pars{x}\ln\pars{1 + x^{2} \over 2x}\,{\dd x \over x} \\[1cm] =&\ -2\int_{\epsilon}^{1} \bracks{\ln\pars{y}\ln\pars{1 - y^{2} \over 2y} + {\pi^{2} \over 4}} \,{\dd y \over y} - 2\int_{0}^{\pi/2}\bracks{\ln\pars{2} + 2\ln\pars{\epsilon}}\theta\,\dd\theta \\[2mm] &\ + 2\int_{\epsilon}^{1} \ln\pars{x}\ln\pars{1 + x^{2} \over 2x}\,{\dd x \over x} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{\large +}}{\sim}&\ -2\int_{0}^{1}{\ln\pars{y}\ln\pars{1 - y^{2}} \over y}\,\dd y - 2\int_{\epsilon}^{1}{-\ln\pars{y}\ln\pars{2y} + \pi^{2}/4 \over y}\,\dd y - \bracks{\ln\pars{2} + 2\ln\pars{\epsilon}}\,{\pi^{2} \over 4} \\[2mm] &\ 2\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over x}\,\dd x - 2\int_{\epsilon}^{1}{\ln\pars{x}\ln\pars{2x} \over x}\,\dd x \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}&\ -\,{\pi^{2} \over 4}\,\ln\pars{2} - {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi + {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 + \xi} \over \xi}\,\dd\xi \\[1cm] = &\ -\,{\pi^{2} \over 4}\,\ln\pars{2} - {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi + \left[{1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi^{2}} \over \xi}\,\dd\xi\right. \\[2mm] &\ \left. - {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi\right] \\[1cm] = &\ -\,{\pi^{2} \over 4}\,\ln\pars{2} - {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi + \left[{1 \over 8}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi\right. \\[2mm] &\ \left. - {1 \over 2}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi\right] \\[1cm] = &\ -\,{\pi^{2} \over 4}\,\ln\pars{2} - {7 \over 8}\int_{0}^{1}{\ln\pars{\xi}\ln\pars{1 - \xi} \over \xi}\,\dd\xi = -\,{\pi^{2} \over 4}\,\ln\pars{2} + {7 \over 8}\int_{0}^{1}\ln\pars{\xi}\,\mrm{Li}_{2}'\pars{\xi}\,\dd\xi \\[5mm] = &\ -\,{\pi^{2} \over 4}\,\ln\pars{2} - {7 \over 8}\int_{0}^{1}\overbrace{\mrm{Li}_{2}\pars{\xi} \over \xi} ^{\ds{\mrm{Li}_{3}'\pars{\xi}}}\,\dd\xi = -\,{\pi^{2} \over 4}\,\ln\pars{2} - {7 \over 8}\mrm{Li}_{3}\pars{1} \\[5mm] = &\ \bbx{-\,{\pi^{2} \over 4}\,\ln\pars{2} - {7 \over 8}\,\zeta\pars{3}} \approx -2.7621 \end{align}