\(\epsilon\rightarrow+0,R\rightarrow\infty\)とする。
\(C_{1}\)を\(x:\epsilon\rightarrow R\)で実軸上を\(x\)として、\(C_{2}\)を\(\theta:0\rightarrow\pi\)で半径\(R\)を一定で角度を反時計回りに\(Re^{i\theta}\)として、\(C_{3}\)を\(x:-R\rightarrow\epsilon\)で実軸上を\(x\)として、\(C_{4}\)を\(\theta:\pi\rightarrow0\)で半径\(\epsilon\)を一定で角度を時計回りに\(\epsilon e^{i\theta}\)とする。
このとき\(C=C_{1}+C_{2}+C_{3}+C_{4}\)とすると\(C\)は閉曲線となる。

(1)

経路\(C,C_{1},C_{2},C_{3},C_{4}\)について積分をしていく。
\begin{align*} \int_{C_{1}}\frac{\log z}{z^{2}+1}dz & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{\epsilon}^{R}\frac{\log x}{x^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx \end{align*} \begin{align*} \left|\int_{C_{2}}\frac{\log z}{z^{2}+1}dz\right| & =\left|\lim_{R\rightarrow\infty}\int_{0}^{\pi}\frac{\log\left(Re^{i\theta}\right)}{R^{2}e^{2i\theta}+1}iRe^{i\theta}d\theta\right|\\ & \leq\lim_{R\rightarrow\infty}\int_{0}^{\pi}\left|\frac{\log\left(Re^{i\theta}\right)}{R^{2}e^{2i\theta}+1}iRe^{i\theta}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{R\rightarrow\infty}\left|\frac{\ln\left|R\right|+i\theta}{Re^{2i\theta}+\frac{1}{R}}\right|d\theta\\ & =\int_{0}^{\pi}0d\theta\\ & =0 \end{align*} となるので、
\[ \int_{C_{2}}\frac{\log z}{z^{2}+1}dz=0 \] となる。
\begin{align*} \int_{C_{3}}\frac{\log z}{z^{2}+1}dz & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{-R}^{-\epsilon}\frac{\log x}{x^{2}+1}dx\\ & =-\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{R}^{\epsilon}\frac{\log\left(-x\right)}{\left(-x\right)^{2}+1}dx\cmt{x\rightarrow-x}\\ & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{\epsilon}^{R}\frac{\log x+i\pi}{\left(-x\right)^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+i\pi\int_{0}^{\infty}\frac{1}{x^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+i\pi\left[\tan^{\bullet}x\right]_{0}^{\infty}\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+i\pi\left(\frac{\pi}{2}-0\right)\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+i\frac{\pi^{2}}{2} \end{align*} \begin{align*} \left|\int_{C_{4}}\frac{\log z}{z^{2}+1}dz\right| & =\left|\lim_{\epsilon\rightarrow0}\int_{\pi}^{0}\frac{\log\left(\epsilon e^{i\theta}\right)}{\epsilon^{2}e^{2i\theta}+1}i\epsilon e^{i\theta}d\theta\right|\\ & \leq\lim_{\epsilon\rightarrow+0}\int_{0}^{\pi}\left|\frac{\log\left(\epsilon e^{i\theta}\right)}{\epsilon^{2}e^{2i\theta}+1}i\epsilon e^{i\theta}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|\frac{\ln\left|\epsilon\right|+i\theta}{\epsilon e^{2i\theta}+\frac{1}{\epsilon}}\right|d\theta\\ & =\int_{0}^{\pi}0d\theta\\ & =0 \end{align*} \begin{align*} \int_{C}\frac{\log z}{z^{2}+1}dz & =\frac{1}{2i}\int_{C}\left(\frac{\log z}{z-i}-\frac{\log z}{z+i}\right)dz\\ & =\frac{1}{2i}\int_{C}\frac{\log z}{z-i}dz\\ & =\frac{2\pi i}{2i}\res\left(\frac{\log z}{z-i},i\right)\\ & =\pi\left[\log z\right]_{z=i}\\ & =\pi\log e^{i\frac{\pi}{2}}\\ & =i\frac{\pi^{2}}{2} \end{align*} これらより、
\[ \int_{C}\frac{\log z}{z^{2}+1}dz=\int_{C_{1}}\frac{\log z}{z^{2}+1}dz+\int_{C_{2}}\frac{\log z}{z^{2}+1}dz+\int_{C_{3}}\frac{\log z}{z^{2}+1}dz+\int_{C_{4}}\frac{\log z}{z^{2}+1}dz \] なので、
\[ i\frac{\pi^{2}}{2}=\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+0+\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+i\frac{\pi^{2}}{2}+0 \] となり、
\[ \int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx=0 \] となる。

(2)

経路\(C,C_{1},C_{2},C_{3},C_{4}\)について積分をしていく。
\begin{align*} \int_{C_{1}}\frac{\log^{2}z}{z^{2}+1}dz & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{\epsilon}^{R}\frac{\log^{2}x}{x^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log^{2}x}{x^{2}+1}dx \end{align*} \begin{align*} \left|\int_{C_{2}}\frac{\log^{2}z}{z^{2}+1}dz\right| & =\left|\lim_{R\rightarrow\infty}\int_{0}^{\pi}\frac{\log^{2}\left(Re^{i\theta}\right)}{R^{2}e^{2i\theta}+1}iRe^{i\theta}d\theta\right|\\ & \leq\lim_{R\rightarrow\infty}\int_{0}^{\pi}\left|\frac{\log^{2}\left(Re^{i\theta}\right)}{R^{2}e^{2i\theta}+1}iRe^{i\theta}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{R\rightarrow\infty}\left|\frac{\left(\ln\left|R\right|+i\theta\right)^{2}}{Re^{2i\theta}+\frac{1}{R}}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{R\rightarrow\infty}\left|\frac{2\ln\left|R\right|}{Re^{2i\theta}}\right|d\theta\\ & =\int_{0}^{\pi}0d\theta\\ & =0 \end{align*} となるので、
\[ \int_{C_{2}}\frac{\log z}{z^{2}+1}dz=0 \] となる。
\begin{align*} \int_{C_{3}}\frac{\log^{2}z}{z^{2}+1}dz & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{-R}^{-\epsilon}\frac{\log^{2}x}{x^{2}+1}dx\\ & =-\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{R}^{\epsilon}\frac{\log^{2}\left(-x\right)}{\left(-x\right)^{2}+1}dx\cmt{x\rightarrow-x}\\ & =\lim_{\epsilon\rightarrow+0}\lim_{R\rightarrow\infty}\int_{\epsilon}^{R}\frac{\left(\log x+i\pi\right)^{2}}{\left(-x\right)^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log^{2}x}{x^{2}+1}dx+2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx-\pi^{2}\int\frac{1}{x^{2}+1}dx\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx-\pi^{2}\left[\tan^{\bullet}x\right]_{0}^{\infty}\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx-\pi^{2}\left(\frac{\pi}{2}-0\right)\\ & =\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx-\frac{\pi^{3}}{2} \end{align*} \begin{align*} \left|\int_{C_{4}}\frac{\log^{2}z}{z^{2}+1}dz\right| & =\left|\lim_{\epsilon\rightarrow0}\int_{\pi}^{0}\frac{\log^{2}\left(\epsilon e^{i\theta}\right)}{\epsilon^{2}e^{2i\theta}+1}i\epsilon e^{i\theta}d\theta\right|\\ & \leq\lim_{\epsilon\rightarrow+0}\int_{0}^{\pi}\left|\frac{\log^{2}\left(\epsilon e^{i\theta}\right)}{\epsilon^{2}e^{2i\theta}+1}i\epsilon e^{i\theta}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|\frac{\left(\ln\left|\epsilon\right|+i\theta\right)^{2}}{\epsilon e^{2i\theta}+\frac{1}{\epsilon}}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|\frac{2\ln\left|\epsilon\right|}{-\frac{1}{\epsilon^{2}}\epsilon}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|\frac{2\ln\left|\epsilon\right|}{\frac{1}{\epsilon}}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|\frac{2}{-\frac{1}{\epsilon^{2}}\epsilon}\right|d\theta\\ & =\int_{0}^{\pi}\lim_{\epsilon\rightarrow+0}\left|2\epsilon\right|d\theta\\ & =\int_{0}^{\pi}0d\theta\\ & =0 \end{align*} \begin{align*} \int_{C}\frac{\log^{2}z}{z^{2}+1}dz & =\frac{1}{2i}\int_{C}\left(\frac{\log^{2}z}{z-i}-\frac{\log^{2}z}{z+i}\right)dz\\ & =\frac{1}{2i}\int_{C}\frac{\log^{2}z}{z-i}dz\\ & =\frac{2\pi i}{2i}\res\left(\frac{\log^{2}z}{z-i},i\right)\\ & =\pi\left[\log^{2}z\right]_{z=i}\\ & =\pi\log^{2}e^{i\frac{\pi}{2}}\\ & =\pi\left(i\frac{\pi}{2}\right)^{2}\\ & =-\frac{\pi^{3}}{4} \end{align*} これらより、
\[ \int_{C}\frac{\log z}{z^{2}+1}dz=\int_{C_{1}}\frac{\log z}{z^{2}+1}dz+\int_{C_{2}}\frac{\log z}{z^{2}+1}dz+\int_{C_{3}}\frac{\log z}{z^{2}+1}dz+\int_{C_{4}}\frac{\log z}{z^{2}+1}dz \] なので、
\[ -\frac{\pi^{3}}{4}=\int_{0}^{\infty}\frac{\log^{2}x}{x^{2}+1}dx+0+\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx+2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx-\frac{\pi^{3}}{2}+0 \] となり、
\begin{align*} \int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx & =\frac{1}{2}\left(-\frac{\pi^{3}}{4}+\frac{\pi^{3}}{2}-2i\pi\int_{0}^{\infty}\frac{\log x}{x^{2}+1}dx\right)\\ & =\frac{1}{2}\left(-\frac{\pi^{3}}{4}+\frac{\pi^{3}}{2}\right)\\ & =\frac{\pi^{3}}{8} \end{align*} となる。