How can I find $\int_0^{\pi/2}x\cot x\,dx$? Can $\displaystyle\int_0^{\pi/2}x\cot x\,dx$ be found using elementary functions? If so how could I possibly do it? Is there any other way to calculate above definite integral?
 A: Using integration by parts, we find that
$$\begin{align}I = \int_0^{\pi/2} dx \, x \, \cot{x} &= \left [x \log{\sin{x}} \right ]_0^{\pi/2} - \int_0^{\pi/2} dx \, \log{\sin{x}} \\ &= - \int_0^{\pi/2} dx \, \log{\sin{x}} \end{align} $$
Note that
$$I = - \int_0^{\pi/2} dx \, \log{\cos{x}} $$
so that
$$\begin{align} 2 I &= -\int_0^{\pi/2} dx \, \log{\sin{x}} -\int_0^{\pi/2} dx \, \log{\cos{x}}\\ &= -\int_0^{\pi/2} dx \, \log{(\sin{x} \cos{x})} \\ &= -\int_0^{\pi/2} dx \, \log{\frac{\sin{2x}}{2}} \\ &= \frac{\pi}{2} \log{2} - \int_0^{\pi/2} dx \, \log{\sin{2 x}}\\ &= \frac{\pi}{2} \log{2} - \frac12 \int_0^{\pi} du \, \log{\sin{u}}\\ &= \frac{\pi}{2} \log{2} + I\end{align} $$
Therefore,
$$I = \int_0^{\pi/2} dx \, x \, \cot{x} = \frac{\pi}{2} \log{2} $$
A: Using the Riemann-Lebesgue lemma, one can show that $$ \int_{a}^{b} f(x) \cot (x) \ dx = 2 \sum_{n=1}^{\infty} \int_{a}^{b} f(x) \sin(2nx) \ dx$$
Then $$\int^{\pi /2}_{0} x \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{0}^{\pi /2} x \sin(2nx) \ dx$$
$$ = 2 \sum_{n=1}^{\infty} \Big( \frac{\sin(2nx)}{4n^{2}} - \frac{x}{2n} \cos(2nx) \Big|^{\pi/2}_{0} \Big)$$
$$ = -\frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} = \frac{\pi \ln 2}{2}$$
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x&=
\overbrace{\left.x\ln\pars{\sin\pars{x}}\right\vert_{0}^{\pi/2}}^{\ds{0}}\
-\
\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x
\\[3mm]&=-\,\half\bracks{%
\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x
+ \int_{0}^{\pi/2}\ln\pars{\sin\pars{{\pi \over 2} - x}}\,\dd x}
\\[3mm]&=-\,\half\int_{0}^{\pi/2}\ln\pars{\sin\pars{2x} \over 2}\,\dd x
=-\,{1 \over 4}\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\,\dd x
+ {1 \over 4}\,\pi\ln\pars{2}
\\[3mm]&=-\,{1 \over 2}\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\,\dd x
+ {1 \over 4}\,\pi\ln\pars{2}
=\half\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x + {1 \over 4}\,\pi\ln\pars{2}
\end{align}
$$
\color{#00f}{\large\int_{0}^{\pi/2}x\cot\pars{x}\,\dd x = \half\,\pi\ln\pars{2}}
$$
A: \begin{equation}
\int_0^\frac{\pi}{2}x\cot xdx=\int_0^\frac{\pi}{2}xd\ln(\sin(x))=xd\ln(\sin(x))\bigg|^\frac{\pi}{2}_0-\int_0^\frac{\pi}{2}\ln(\sin(x))dx\\
=\frac{\pi}{2}\ln2
\end{equation}
For the last step, see Last Step.
