積分botを解けるだけ解く, その2

実数$a$に対し, 主値積分
$$\begin{array}{r}I(a):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2}}{\sinh \pi (x+a)}dx\end{array}$$
を考える. $\frac{1}{\sinh \pi (x+a)}$は$0\le \mathrm{Im}(a)\le \frac{1}{2}$において, $\mathrm{Re}(x)\to \pm \mathrm{\infty }$で一様に$0$に収束するので, 留数定理より, ちょうど$-a$における留数の半分が入っているとみなせることを考えて,
$$\begin{array}{rcl}I(a)& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi (x+i/2{)}^2}}{\sinh \pi (x+\frac{i}{2}+a)}dx+\pi iRe{s}_{x=-a}\frac{e^{i\pi x^2}}{\sinh \pi (x+a)}\\ & =& -i{e}^{-i\pi /4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2-\pi x}}{\cosh \pi (x+a)}dx+i{e}^{i\pi a^2}\end{array}$$
同様に,
$$\begin{array}{rcl}I(a)& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi (x-i/2{)}^2}}{\sinh \pi (x-\frac{i}{2}+a)}dx-\pi iRe{s}_{x=-a}\frac{e^{i\pi x^2}}{\sinh \pi (x+a)}\\ & =& i{e}^{-i\pi /4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2+\pi x}}{\cosh \pi (x+a)}dx-i{e}^{i\pi a^2}\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2+\pi x}}{\cosh \pi (x+a)}dx& =& -i{e}^{i\pi /4}I(a)+e^{i\pi a^2+i\pi /4}\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2-\pi x}}{\cosh \pi (x+a)}dx& =& i{e}^{i\pi /4}I(a)+e^{i\pi a^2+i\pi /4}\end{array}$$
これより,
$$\begin{array}{rcl}{e}^{i\pi /4}& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i\pi x^2}dx\\ & =& \frac{1}{2}(e^{\pi a}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2+\pi x}}{\cosh \pi (x+a)}dx+e^{-\pi a}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2-\pi x}}{\cosh \pi (x+a)}dx)\\ & =& \frac{1}{2}(-i{e}^{i\pi /4+\pi a}I(a)+e^{i\pi a^2+i\pi /4+\pi a}+i{e}^{i\pi /4-\pi a}I(a)+e^{i\pi a^2+i\pi /4-\pi a})\\ & =& e^{i\pi /4}(-iI(a)\sinh \pi a+e^{i\pi a^2}\cosh \pi a)\end{array}$$
よって,
$$\begin{array}{r}I(a)=\frac{i(1-e^{i\pi a^2}\cosh \pi a)}{\sinh \pi a}\end{array}$$
虚部を考えることにより, 主値積分,

$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \pi x^2}{\sinh (x+a)}dx& =& \frac{1-\cos \pi a^2\cosh \pi a}{\sinh \pi a}\end{array}$$
を得て, 両辺は$a\to 0$において,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\sin \pi x^2(\frac{1}{\sinh \pi x}-\frac{\pi a\cosh \pi x}{{\sinh }^2\pi x}+O(a^2))dx& =& -\frac{\pi }{2}a+O(a^2)\end{array}$$
より,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \pi x^2\cosh \pi x}{{\sinh }^2\pi x}dx& =& \frac{1}{2}\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{\sin \pi x^2\cosh \pi x}{{\sinh }^2\pi x}dx& =& \frac{1}{4}\end{array}$$
が示された.

$$\begin{array}{rcl}& & {\int }_0^1\frac{x^2\ln \frac{1}{x}}{(1-x^4)({\pi }^2+{(\ln \frac{1+x}{1-x}+2{\tan }^{-1}\frac{1}{x})}^2)}dx\\ & =& -{\int }_0^1\frac{x^2\ln x}{(1-x^4)({\pi }^2+{(\pi +2{\tanh }^{-1}x-2{\tan }^{-1}x)}^2)}dx\\ & =& -\frac{1}{4}{\int }_0^1\frac{x^2\ln x}{(1-x^4)({\left(\frac{\pi }{2}\right)}^2+{(\frac{\pi }{2}+{\tanh }^{-1}x-{\tan }^{-1}x)}^2)}dx\\ & =& -\frac{1}{2\pi }\mathrm{Im}\left({\int }_0^1\frac{x^2\ln x}{(1-x^4)(\frac{\pi }{2}(1-i)+{\tanh }^{-1}x-{\tan }^{-1}x)}dx\right)\\ & =& -\frac{1}{2\pi }\mathrm{Im}((1+i){\int }_0^1\frac{x^2\ln x}{(1-x^4)(\pi +(1+i)({\tanh }^{-1}x-{\tan }^{-1}x))}dx)\\ & =& -\frac{1}{4\pi }\mathrm{Im}\left({[\ln (\pi +(1+i)({\tanh }^{-1}x-{\tan }^{-1}x))\ln x]}_0^1\right)\\ & +& \frac{1}{4\pi }\mathrm{Im}\left({\int }_0^1\frac{\ln (\pi +(1+i)({\tanh }^{-1}x-{\tan }^{-1}x))}{x}dx\right)\\ & =& \frac{1}{4\pi }{\int }_0^1{\tan }^{-1}\left(\frac{{\tanh }^{-1}x-{\tan }^{-1}x}{\pi +{\tanh }^{-1}-{\tan }^{-1}x}\right)\frac{dx}{x}\\ & =& \frac{1}{32}\ln \frac{{\pi }^2}{8}\end{array}$$

まず,
$$\begin{array}{rcl}\arctan \frac{2r\sin x}{1-r^2}& =& \mathrm{Im}(\ln (1-r^2+2ir\sin x))\\ & =& \mathrm{Im}(\ln ((1-r{e}^{-ix})(1+r{e}^{ix})))\\ & =& \mathrm{Im}(\ln (1-r{e}^{-ix})+\ln (1+r{e}^{ix}))\\ & =& \mathrm{Im}(-\sum _{0<n}\frac{r^n}{n}{e}^{-inx}+\sum _{0<n}\frac{(-1{)}^{n-1}{r}^n}{n}{e}^{inx}))\\ & =& \sum _{0<n}\frac{r^n}{n}\sin nx+\sum _{0<n}\frac{(-1{)}^{n-1}{r}^n}{n}\sin nx\\ & =& 2\sum _{0<n}\frac{r^{2n+1}}{2n+1}\sin (2n+1)x\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_0^{\pi /2}\arctan \frac{2r\sin x}{1-r^2}dx& =& 2\sum _{0\le n}{\int }_0^{\pi /2}\frac{r^{2n+1}}{2n+1}\sin (2n+1)xdx\\ & =& 2\sum _{0\le n}\frac{r^{2n+1}}{(2n+1{)}^2}\\ & =& 2{\chi }_2(r)\end{array}$$
また, 三角関数の直交性より,
$$\begin{array}{rcl}& & {\int }_0^{\pi /2}\arctan \frac{2r\sin x}{1-r^2}\arctan \frac{2s\sin x}{1-s^2}dx\\ & =& \frac{1}{4}{\int }_0^{2\pi }\arctan \frac{2r\sin x}{1-r^2}\arctan \frac{2s\sin x}{1-s^2}dx\\ & =& \sum _{0\le n,m}{\int }_0^{2\pi }\frac{r^{2n+1}{s}^{2m+1}}{(2n+1)(2m+1)}\sin (2n+1)x\sin (2m+1)xdx\\ & =& \pi \sum _{0\le n,m}\frac{r^{2n+1}{s}^{2m+1}}{(2n+1)(2m+1)}{\delta }_{n,m}\\ & =& \pi \sum _{0\le n}\frac{(rs{)}^{2n+1}}{(2n+1{)}^2}\\ & =& \pi {\chi }_2(rs)\end{array}$$
ここで, ${\delta }_{n,m}$はKroneckerのデルタである.

$$\begin{array}{rcl}& & {\int }_{1/N}^1(\lfloor \frac{\alpha }{x}\rfloor -\alpha \lfloor \frac{1}{x}\rfloor )dx\\ & =& {\int }_{1/N}^1\lfloor \frac{\alpha }{x}\rfloor dx-\alpha {\int }_{1/N}^1\lfloor \frac{1}{x}\rfloor dx\\ & =& \alpha {\int }_{\alpha }^{\alpha N}\frac{\lfloor x\rfloor }{x^2}dx-\alpha {\int }_1^N\frac{\lfloor x\rfloor }{x^2}dx\\ & =& \alpha {\int }_N^{\alpha N}\frac{\lfloor x\rfloor }{x^2}dx\\ & =& \alpha {\int }_N^{\alpha N}\frac{1}{x}dx-\alpha {\int }_N^{\alpha N}\frac{x-\lfloor x\rfloor }{x^2}dx\\ & =& \alpha \ln \alpha -\alpha {\int }_N^{\alpha N}\frac{x-\lfloor x\rfloor }{x^2}dx\end{array}$$
ここで, 第2項の積分は
$$\begin{array}{rcl}{\int }_1^{\mathrm{\infty }}\frac{x-\lfloor x\rfloor }{x^2}dx& \le & {\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dx\\ & =& 1\end{array}$$
が収束することにより, $N\to \mathrm{\infty }$で$0$に収束する.

$$\begin{array}{rcl}I& =& {\int }_0^{\mathrm{\infty }}\frac{\sqrt[\varphi ]{x}\arctan x}{(1+x^{\varphi }{)}^2}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{\varphi -1}\arctan x}{(1+x^{\varphi }{)}^2}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{-\varphi -1}\arctan \frac{1}{x}}{(1+x^{-\varphi }{)}^2}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x^{\varphi -1}\arctan \frac{1}{x}}{(1+x^{\varphi }{)}^2}dx\end{array}$$
よって,
$$\begin{array}{rcl}I& =& \frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x^{\varphi -1}(\arctan x+\arctan \frac{1}{x})}{(1+x^{\varphi }{)}^2}dx\\ & =& \frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{x^{\varphi -1}}{(1+x^{\varphi }{)}^2}dx\\ & =& \frac{\pi }{4\varphi }{\int }_0^{\mathrm{\infty }}\frac{1}{(1+x{)}^2}dx\\ & =& \frac{\pi }{4\varphi }\end{array}$$

置換$\tan x\to x^2$により,
$$\begin{array}{rcl}{\int }_0^{\pi /2}\frac{\arctan \sqrt{\tan x}}{\tan x}dx& =& 2{\int }_0^{\mathrm{\infty }}\frac{\arctan x}{x(1+x^4)}dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan x}{x(1+x^4)}dx\end{array}$$
ここで, $0<\mathrm{Im}(a)$として,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{(x-a)(1+b^2{x}^2)}dx& =& \frac{1}{1+a^2{b}^2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(\frac{1}{x-a}-\frac{b^2(x+a)}{1+b^2{x}^2})dx\\ & =& \frac{1}{1+a^2{b}^2}\underset{A\to -\mathrm{\infty },B\to \mathrm{\infty }}{lim}{[\ln (x-a)-\frac{1}{2}\ln (1+b^2{x}^2)-ab\arctan bx]}_A^B\\ & =& \frac{1}{1+a^2{b}^2}\pi (i-ab)\\ & =& -\frac{\pi }{i+ab}\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan bx}{x(x-a)}dx& =& \frac{\pi }{a}\ln (1-iab)+C(a)\end{array}$$
となる$a$に依存する関数があるが, $b=0$として, $C(a)=0$である. 両辺の共役を考えることにより, $a$を実数でない複素数としたとき,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan bx}{x(x-a)}dx& =& \{\begin{array}{l}-\frac{\pi }{a}\ln (1-iab)(0<\mathrm{Im}(a))\\ -\frac{\pi }{a}\ln (1+iab)(0>\mathrm{Im}(a))\end{array}\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan x}{x(1+x^4)}dx& =& -\frac{1}{4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\arctan x}{x}(\frac{e^{i\pi /4}}{x-e^{i\pi /4}}+\frac{e^{-i\pi /4}}{x-e^{-i\pi /4}}+\frac{e^{3i\pi /4}}{x-e^{3i\pi /4}}+\frac{e^{-3i\pi /4}}{x-e^{-3i\pi /4}})dx\\ & =& \frac{\pi }{4}(\ln (1-i{e}^{i\pi /4})+\ln (1+i{e}^{-i\pi /4})+\ln (1-i{e}^{3i\pi /4})+\ln (1+i{e}^{3i\pi /4}))\\ & =& \frac{\pi }{2}\ln (2+\sqrt{2})\end{array}$$

まず, 与えられた式は,
$$\begin{array}{r}{\int }_0^1\frac{x{\tanh }^{-1}x}{(1-s^2{x}^2)\sqrt{1-x^2}}dx=\frac{\pi {\tanh }^{-1}s}{2s\sqrt{1-s^2}}\end{array}$$
と書き直せるので, これを示します.
$$\begin{array}{rcl}& & {\int }_0^1\frac{x{\tanh }^{-1}x}{(1-s^2{x}^2)\sqrt{1-x^2}}dx\\ & =& \sum _{0\le n}\frac{1}{2n+1}{\int }_0^1{x}^{2n+2}(1-x^2{)}^{-1/2}(1-s^2{x}^2{)}^{-1}dx\\ & =& \sum _{0\le n,m}\frac{s^{2m}}{2n+1}{\int }_0^1{x}^{2n+2m+2}(1-x^2{)}^{-1/2}dx\\ & =& \frac{1}{2}\sum _{0\le n,m}\frac{s^{2m}}{2n+1}{\int }_0^1{x}^{n+m+1/2}(1-x{)}^{-1/2}dx\\ & =& \frac{1}{2}\sum _{0\le n,m}\frac{s^{2m}}{2n+1}\frac{\mathrm{\Gamma }(n+m+\frac{3}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+m+2)}\\ & =& \frac{\pi }{2}\sum _{0\le n,m}\frac{{\left(\frac{1}{2}\right)}_{n+m+1}}{(n+m+1)!(2n+1)}{s}^{2m}\\ & =& \frac{\pi }{2}\sum _{0\le m}{s}^{2m}\sum _{n=m+1}^{\mathrm{\infty }}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n-2m-1)}\end{array}$$
一方,
$$\begin{array}{rcl}\frac{\pi {\tanh }^{-1}s}{2s\sqrt{1-s^2}}& =& \frac{\pi }{2}\sum _{0\le n,m}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2m+1)}{s}^{2n+2m}\\ & =& \frac{\pi }{2}\sum _{0\le m}{s}^{2m}\sum _{n=0}^m\frac{{\left(\frac{1}{2}\right)}_n}{n!(2m-2n+1)}\end{array}$$
より,
$$\begin{array}{r}\sum _{n=m+1}^{\mathrm{\infty }}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n-2m-1)}=\sum _{n=0}^m\frac{{\left(\frac{1}{2}\right)}_n}{n!(2m-2n+1)}\end{array}$$
を示せばよいが, これはGaussの超幾何定理より,
$$\begin{array}{rcl}& & \sum _{n=m+1}^{\mathrm{\infty }}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n-2m-1)}-\sum _{n=0}^m\frac{{\left(\frac{1}{2}\right)}_n}{n!(2m-2n+1)}\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n-2m-1)}\\ & =& -\frac{2}{2m+1}{}_2{F}_1[\begin{array}{c}\frac{1}{2},-\frac{1}{2}-m\\ \frac{1}{2}-m\end{array};1]\\ & =& 0\end{array}$$
より従う.

$|b|<|a|$とする.
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{J}_n(bx)x^n{e}^{-ax}dx& =& \frac{2^n}{b^n}\sum _{0\le m}\frac{(-1{)}^m}{m!(n+m)!}{\int }_0^{\mathrm{\infty }}{\left(\frac{bx}{2}\right)}^{2m+2n}{e}^{-ax}dx\\ & =& \frac{2^n}{a{b}^n}\sum _{0\le m}\frac{(-1{)}^m(2n+2m)!}{m!(n+m)!}{\left(\frac{b}{2a}\right)}^{2n+2m}\\ & =& \frac{(2b{)}^n}{a^{2n+1}}\sum _{0\le m}\frac{(-1{)}^m{\left(\frac{1}{2}\right)}_{n+m}}{m!}{\left(\frac{b}{a}\right)}^{2m}\\ & =& \frac{(2b{)}^n{\left(\frac{1}{2}\right)}_n}{a^{2n+1}}\sum _{0\le m}\frac{{(\frac{1}{2}+n)}_n}{m!}{(-\frac{b^2}{a^2})}^m\\ & =& \frac{(2b{)}^n\mathrm{\Gamma }(\frac{1}{2}+n)}{\sqrt{\pi }{a}^{2n+1}}{(1+\frac{b^2}{a^2})}^{-1/2-n}\\ & =& \frac{(2b{)}^n\mathrm{\Gamma }(\frac{1}{2}+n)}{\sqrt{\pi }(a^2+b^2{)}^{1/2+n}}\end{array}$$