logがある積分の解法まとめ

$$\begin{array}{rl}I& ={\int }_0^{1}x\log xdx\\ & ={[\frac{x^2}{2}\log x]}_0^1-\frac{1}{2}{\int }_0^{1}xdx\\ & =\frac{1}{2}\underset{x\to 0}{lim}{x}^2\log x-\frac{1}{4}\\ & =-\frac{1}{4}\end{array}$$
$$\begin{array}{rl}\underset{x\to 0}{lim}{x}^2\log x& =\underset{x\to 0}{lim}\frac{\log x}{\frac{1}{x^2}}\\ & =\underset{x\to 0}{lim}\frac{\frac{1}{x}}{-\frac{2}{x^3}}\\ & =-\frac{1}{2}\underset{x\to 0}{lim}{x}^2\\ & =0\end{array}$$

$$\begin{array}{rl}I& ={\int }_1^e\frac{\log x}{x}dx\\ & ={\int }_0^{1}tdt(\log x\mapsto t)\\ & =1\end{array}$$

$$\log (1-x)=-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}(|x|<1)$$
$$\log (1+x)=-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^n}{n}(|x|<1)$$
$$\log (2\sin \frac{x}{2})=-\sum _{n=1}^{\mathrm{\infty }}\frac{\cos nx}{n}(0<x<\frac{\pi }{2})$$
$$\log (2\cos \frac{x}{2})=-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n\cos nx}{n}(0<x<\frac{\pi }{2})$$
$$\log \tan \frac{x}{2}=-2\sum _{n=1}^{\mathrm{\infty }}\frac{\cos (2n+1)x}{2n+1}(0<x<\frac{\pi }{2})$$

$$\begin{array}{rl}I& ={\int }_0^1\frac{\log (1+x)}{\sqrt{x}}dx\\ & =-{\int }_0^1\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{x}^n}{n\sqrt{x}}dx\\ & =-\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n}{\int }_0^1{x}^{n-\frac{1}{2}}dx\\ & =-2\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n(2n+1)}\\ & =-2(\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n}-2\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1})\\ & =-2(\log 2+\frac{\pi }{2}-2)\\ & =-2\log 2-\pi +4\end{array}$$

$$\begin{array}{rl}I& ={\int }_0^{\frac{\pi }{2}}\log (2\sin \frac{x}{2})\log (2\cos \frac{x}{2})dx\\ & =\frac{1}{2}{\int }_0^{\pi }\log (2\sin \frac{x}{2})\log (2\cos \frac{x}{2})dx\\ & =\frac{1}{2}\sum _{n,m>0}\frac{(-1{)}^n}{nm}{\int }_0^{\pi }\cos mx\cos nxdx\\ & =\frac{\pi }{4}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n^2}\\ & =-\frac{{\pi }^3}{48}\end{array}$$

$$\begin{array}{rl}I& ={\int }_0^{\mathrm{\infty }}{e}^{-x}\log xdx\\ & ={\frac{d}{ds}{\int }_0^{\mathrm{\infty }}{x}^s{e}^{-x}dx|}_{s=1}\\ & ={\mathrm{\Gamma }}^{\prime }(1)\\ & =\mathrm{\Gamma }(1)\psi (1)\\ & =-\gamma \end{array}$$

$$I={\int }_0^1\frac{x^2-x}{\log x}dx$$
$$f(t)={\int }_0^1\frac{x^t}{\log x}$$
$$\begin{array}{rl}{f}^{\prime }(t)& ={\int }_0^1{x}^t\\ & =\frac{1}{t+1}\end{array}$$
$$\begin{array}{rl}f(2)-f(1)& ={\int }_1^2{f}^{\prime }(t)dt\\ & ={\int }_1^2\frac{1}{t+1}dt\\ & =\log \frac{3}{2}\end{array}$$

$$I={\int }_0^{\mathrm{\infty }}\frac{x\log x}{x^4+1}dx$$
$$f(z)=\frac{z\log z}{z^4+1}$$
積分経路
$\log 0$を回避するために$0$は避けています。
${\displaystyle C_2,C_4}$は評価して極限とばすと$0$になります。
確認してみてください。
全体の積分を$C$としておきます。
それぞれ計算していきます。

$$\begin{array}{rl}{\int }_{C_1}& ={\int }_{\epsilon }^{R}f(z)dz\\ & =I(\epsilon \to 0,R\to \mathrm{\infty })\end{array}$$
$$\begin{array}{rl}{\int }_{C_3}& ={\int }_{iR}^{i\epsilon }f(z)dz\\ & =-i{\int }_{\epsilon }^{R}f(ix)dx\\ & ={\int }_{\epsilon }^R\frac{x\log ix}{x^4+1}dx\\ & ={\int }_{\epsilon }^R\frac{x\log x}{x^4+1}dx+\frac{i\pi }{2}{\int }_{\epsilon }^R\frac{x}{x^4+1}dx\\ & =I+\frac{{\pi }^2}{8}i(\epsilon \to 0,R\to \mathrm{\infty })\end{array}$$
$$\begin{array}{rl}{\oint }_C& =2\pi i\underset{z=e^{\frac{i\pi }{4}}}{\mathrm{Res}}\frac{z\log z}{z^4+1}\\ & =2\pi i\underset{z\to e^{\frac{i\pi }{4}}}{lim}\frac{z\log z(z-e^{\frac{i\pi }{4}})}{z^4+1}\\ & =2\pi i\underset{z\to e^{\frac{i\pi }{4}}}{lim}\frac{(2z-e^{\frac{i\pi }{4}})\log z-e^{\frac{i\pi }{4}}+z}{4z^3}\\ & =\frac{{\pi }^2}{8}i\end{array}$$
$${\oint }_C={\int }_{C_1}+{\int }_{C_2}+{\int }_{C_3}+{\int }_{C_4}$$
$$\frac{{\pi }^2}{8}i=I+I+\frac{{\pi }^2}{8}i$$
$$I=0$$