(0)

不定積分は
\begin{align*} \int\frac{z}{\tan\left(z\right)}dz & =-i\int z\left(1+2\Li_{0}\left(e^{2iz}\right)\right)dz\cmt{\tan\left(z\right)=-i\left(1+2\Li_{0}\left(e^{2iz}\right)\right)}\\ & =-i\int\left(z+2z\Li_{0}\left(e^{2iz}\right)\right)dz\\ & =-i\left\{ \frac{z^{2}}{2}+\frac{2}{2i}z\Li_{1}\left(e^{2iz}\right)-\frac{2}{2i}\int\Li_{1}\left(e^{2iz}\right)dz\right\} \\ & =-i\left\{ \frac{z^{2}}{2}-iz\Li_{1}\left(e^{2iz}\right)-\frac{2}{\left(2i\right)^{2}}\Li_{2}\left(e^{2iz}\right)\right\} \\ & =-\frac{i}{2}z^{2}-z\Li_{1}\left(e^{2iz}\right)-\frac{i}{2}\Li_{2}\left(e^{2iz}\right) \end{align*} となるので、
\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{x}{\tan\left(x\right)}dx & =-\left[\frac{i}{2}x^{2}+x\Li_{1}\left(e^{2ix}\right)+\frac{i}{2}\Li_{2}\left(e^{2ix}\right)\right]_{0}^{\frac{\pi}{2}}\\ & =-\left\{ \frac{i\pi^{2}}{8}+\frac{\pi}{2}\Li_{1}\left(-1\right)+\frac{i}{2}\Li_{2}\left(-1\right)-\frac{i}{2}\Li_{2}\left(1\right)\right\} \\ & =-\left\{ \frac{i\pi^{2}}{8}-\frac{\pi}{2}\log2+\frac{i}{2}\left(-\eta\left(2\right)\right)-\frac{i}{2}\zeta\left(2\right)\right\} \\ & =-\left\{ \frac{i\pi^{2}}{8}-\frac{\pi}{2}\log2-\frac{i}{2}\zeta\left(2\right)\left(1-\frac{2}{2^{2}}\right)-\frac{i}{2}\zeta\left(2\right)\right\} \\ & =-\left\{ \frac{i\pi^{2}}{8}-\frac{\pi}{2}\log2-\frac{i}{2}\zeta\left(2\right)\left(2-\frac{2}{2^{2}}\right)\right\} \\ & =-\left\{ \frac{i\pi^{2}}{8}-\frac{\pi}{2}\log2-\frac{i}{2}\cdot\frac{\pi^{2}}{6}\cdot\frac{3}{2}\right\} \\ & =\frac{\pi}{2}\log2 \end{align*} となる。

(0)-2

\begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{x}{\tan x}dx & =\int_{0}^{\frac{\pi}{2}}\frac{x\cos x}{\sin x}dx\\ & =\left[x\log\left(\sin x\right)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\log\left(\sin x\right)dx\\ & =-\lim_{x\rightarrow0}\frac{\log\left(\sin x\right)}{x^{-1}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left\{ \log\left(\sin x\right)+\log\left(\sin\left(\frac{\pi}{2}-x\right)\right)\right\} dx\\ & =-\lim_{x\rightarrow0}\frac{\cos x}{-x^{-2}\sin x}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left\{ \log\left(\sin x\right)+\log\left(\cos x\right)\right\} dx\\ & =\lim_{x\rightarrow0}\frac{x^{2}}{\sin x}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left\{ \log\left(\sin x\cos x\right)\right\} dx\\ & =\lim_{x\rightarrow0}\frac{2x}{\cos x}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left\{ \log\left(\frac{1}{2}\sin\left(2x\right)\right)\right\} dx\\ & =-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left\{ \log\left(\frac{1}{2}\right)+\log\left(\sin\left(2x\right)\right)\right\} dx\\ & =-\frac{1}{2}\log\left(\frac{1}{2}\right)\frac{\pi}{2}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\log\left(\sin\left(2x\right)\right)dx\\ & =\frac{\pi}{4}\log2-\frac{1}{4}\int_{0}^{\pi}\log\left(\sin y\right)dy\cmt{2x=y}\\ & =\frac{\pi}{4}\log2-\frac{1}{4}\left\{ \int_{0}^{\frac{\pi}{2}}\log\left(\sin y\right)dy+\int_{\frac{\pi}{2}}^{\pi}\log\left(\sin y\right)dy\right\} \\ & =\frac{\pi}{4}\log2-\frac{1}{4}\left\{ \int_{0}^{\frac{\pi}{2}}\log\left(\sin y\right)dy-\int_{\frac{\pi}{2}}^{0}\log\left(\sin\left(\pi-t\right)\right)dt\right\} \cmt{y=\pi-t}\\ & =\frac{\pi}{4}\log2-\frac{1}{4}\left\{ \int_{0}^{\frac{\pi}{2}}\log\left(\sin y\right)dy+\int_{0}^{\frac{\pi}{2}}\log\left(\sin t\right)dt\right\} \\ & =\frac{\pi}{4}\log2-\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\log\left(\sin y\right)dy\\ & =2\frac{\pi}{4}\log2\\ & =\frac{\pi}{2}\log2 \end{align*}