積分botを解けるだけ解く, その1

積分路1

上のような積分路$C$を考えます.

$$\begin{array}{rcl}{\int }_C{x}^{a-1}(1-x{)}^{b-1}dx& =& (1-e^{2\pi ia}){\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx+i{\int }_0^{2\pi }{e}^{iax}(1-e^{ix}{)}^{b-1}dx\\ & =& -2i{e}^{i\pi a}B(a,b)\sin \pi a-e^{-i\pi b/2}{\int }_0^{2\pi }{e}^{i(a+(b-1)/2)x}{(2\sin \frac{x}{2})}^{b-1}dx\end{array}$$

一方, $\ln$の主値は上の曲線内で$1$価正則に取れるので, この積分は$0$になります. よって,

$$\begin{array}{rcl}{\int }_0^{2\pi }{e}^{i(a+(b-1)/2)x}{(2\sin \frac{x}{2})}^{b-1}dx& =& 2e^{i\pi (a+(b-1)/2)}B(a,b)\sin \pi a\\ & =& \frac{2\pi \mathrm{\Gamma }(b)e^{i\pi (a+(b-1)/2)}}{\mathrm{\Gamma }(1-a)\mathrm{\Gamma }(a+b)}\end{array}$$

$a\mapsto \frac{1-b+c}{2}$と置き換えて,
$$\begin{array}{rcl}{\int }_0^{2\pi }{e}^{icx/2}{(2\sin \frac{x}{2})}^{b-1}dx& =& \frac{2\pi \mathrm{\Gamma }(b)e^{i\pi c/2}}{\mathrm{\Gamma }\left(\frac{1-c+b}{2}\right)\mathrm{\Gamma }\left(\frac{1+c+b}{2}\right)}\\ {\int }_0^{\pi }{e}^{icx}{\sin }^{b-1}xdx& =& \frac{2^{1-b}\pi \mathrm{\Gamma }(b)e^{i\pi c/2}}{\mathrm{\Gamma }\left(\frac{1-c+b}{2}\right)\mathrm{\Gamma }\left(\frac{1+c+b}{2}\right)}\end{array}$$
この虚部を考えることにより, 証明ができました.

$$\begin{array}{r}\ln \frac{1+ax}{1-ax}=2\sum _{0\le n}\frac{(ax{)}^{2n+1}}{2n+1}\end{array}$$
なので, 項別に積分していくと,
$$\begin{array}{rcl}{\int }_0^1\ln \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}& =& \sum _{0\le n}\frac{2}{2n+1}{a}^{2n+1}{\int }_0^1{x}^{2n}(1-x^2{)}^{-1/2}dx\\ & =& \sum _{0\le n}\frac{a^{2n+1}}{2n+1}{\int }_0^1{x}^{n-1/2}(1-x{)}^{-1/2}dx\\ & =& \sum _{0\le n}\frac{a^{2n+1}}{2n+1}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\frac{1}{2}+n)}{\mathrm{\Gamma }(n+1)}\\ & =& \pi \sum _{0\le n}\frac{{\left(\frac{1}{2}\right)}_n}{n!(2n+1)}{a}^{2n+1}\\ & =& \pi \arcsin a\end{array}$$
となって証明ができました.

$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\ln (2-2\cos x^2)dx& =& 2{\int }_0^{\mathrm{\infty }}\ln (2-2\cos x^2)dx\\ & =& 4{\int }_0^{\mathrm{\infty }}\ln |2\sin \frac{x^2}{2}|dx\\ & =& 2{\int }_0^{\mathrm{\infty }}{x}^{-1/2}\ln |2\sin \frac{x}{2}|dx\end{array}$$
$0<\mathrm{Re}s$として,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{x}^{s-1}\ln (1-e^{-ix})dx& =& -\sum _{0<n}\frac{1}{n}{\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-inx}dx\\ & =& -i^{-s}\mathrm{\Gamma }(s)\sum _{0<n}\frac{1}{n^{s+1}}\\ & =& -i^{-s}\mathrm{\Gamma }(s)\zeta (s+1)\end{array}$$
$s=\frac{1}{2}$として, これの実部をとることにより,
$$\begin{array}{r}2{\int }_0^{\mathrm{\infty }}{x}^{-1/2}\ln |2\sin \frac{x}{2}|dx=-\sqrt{2\pi }\zeta \left(\frac{3}{2}\right)\end{array}$$
を得ます.

$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{x}^{s-1}(\cos x^2-\cos x)dx& =& \frac{1}{2}\mathrm{\Gamma }\left(\frac{s}{2}\right)\cos \frac{\pi s}{4}-\mathrm{\Gamma }(s)\cos \frac{\pi s}{2}\\ & =& \frac{1}{2}(\frac{2}{s}-\gamma +O(s))(1+O(s^2))-(\frac{1}{s}-\gamma +O(s))(1+O(s^2))\\ & =& \frac{\gamma }{2}+O(s)\end{array}$$
ここで, $s\to +0$とすればよい.

まず, 少し準備をします.
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{1}{(1+x^2)(1+a^2{x}^2)}dx& =& \frac{1}{1-a}{\int }_0^{\mathrm{\infty }}(\frac{1}{1+x^2}-\frac{a^2}{1+a^2{x}^2})dx\\ & =& \frac{\pi }{2(1+a)}\end{array}$$
となります. ここから, ある定数$C$があって,
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{\arctan ax}{x(1+x^2)}dx=\frac{\pi }{2}\ln (1+a)+C\end{array}$$
となりますが, $a=0$として, $C=0$を得ます.
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{\arctan ax}{x(1+x^2)}dx=\frac{\pi }{2}\ln (1+a)\end{array}$$
これを用いて, 与えられた積分を考えていきます.

$$\begin{array}{rcl}& & {\int }_0^{\pi /2}\frac{1}{\tan x}\ln \left(\frac{1+(\tan x+\sinh y{)}^2{\cosh }^{2}y}{1+(\tan x-\sinh y{)}^2{\cosh }^{2}y}\right)dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{x(1+x^2)}\ln \left(\frac{1+(x+\sinh y{)}^2{\cosh }^{2}y}{1+(x-\sinh y{)}^2{\cosh }^{2}y}\right)dx\\ & =& 2{\int }_0^{\mathrm{\infty }}\frac{1}{x(1+x^2)}\ln \left|\frac{1+i((\sinh y+x)\cosh y)}{1+i((\sinh y-x)\cosh y)}\right|dx\\ & =& 2\mathrm{Re}{\int }_0^{\mathrm{\infty }}\frac{1}{x(1+x^2)}\ln \left(\frac{1+i((\sinh y+x)\cosh y)}{1+i((\sinh y-x)\cosh y)}\right)dx\\ & =& -4\mathrm{Re}(i{\int }_0^{\mathrm{\infty }}\frac{1}{x(1+x^2)}\arctan \left(\frac{x\cosh y}{1+i\sinh y\cosh y}\right)dx)\\ & =& 4\mathrm{Im}(\frac{\pi }{2}\ln (1+\frac{\cosh y}{1+i\sinh y\cosh y}))\\ & =& 2\pi \arctan \left(\frac{\sinh y{\cosh }^{2}y}{1+\cosh y+{\sinh }^{2}y{\cosh }^{2}y}\right)\end{array}$$
となりました.

Binetの第2公式
$$\begin{array}{r}\ln \mathrm{\Gamma }(z)=\frac{1}{2}\ln 2\pi +(z-\frac{1}{2})\ln z-z+2{\int }_0^{\mathrm{\infty }}\frac{\arctan \frac{x}{z}}{e^{2\pi x}-1}dx\end{array}$$
を二階微分すると,
$$\begin{array}{r}{\psi }^{\prime }(z)=\frac{1}{z}+\frac{1}{2z^2}+4z{\int }_0^{\mathrm{\infty }}\frac{x}{(z^2+x^2{)}^2(e^{2\pi x}-1)}dx\end{array}$$
よって,
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{x}{(z^2+x^2{)}^2(e^{2\pi x}-1)}dx=\frac{1}{4z}({\psi }^{\prime }(z)-\frac{1}{z}-\frac{1}{2z^2})\end{array}$$
これを用いて,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2{)}^2\tanh \frac{\pi x}{2}}dx& =& {\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2{)}^2}dx+2{\int }_0^{\mathrm{\infty }}\frac{x}{(1+x^2{)}^2(e^{\pi x}-1)}dx\\ & =& {[-\frac{1}{2(1+x^2)}]}_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x}{{({\left(\frac{1}{2}\right)}^2+x^2)}^2(e^{2\pi x}-1)}dx\\ & =& \frac{1}{2}+\frac{1}{4}{\psi }^{\prime }\left(\frac{1}{2}\right)-1\\ & =& \frac{{\pi }^2}{8}-\frac{1}{2}\end{array}$$

$\frac{f(x)}{x}$は偶関数だから,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{f(x)}{x}dx& =& \frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{f(x)}{x}dx\\ & =& \frac{1}{2}\sum _{n\in \mathbb{Z}}{\int }_{nT}^{(n+1)T}\frac{f(x)}{x}dx\\ & =& \frac{1}{2}\sum _{n\in \mathbb{Z}}{\int }_0^T\frac{f(x)}{x+nT}dx\\ & =& \frac{1}{4}\sum _{n\in \mathbb{Z}}{\int }_0^{T}f(x)(\frac{1}{x-nT}+\frac{1}{x+nT})dx\end{array}$$
ここで, $\pi \cot \pi z$の部分分数展開,
$$\begin{array}{rcl}\pi \cot \pi z& =& \frac{1}{2}\sum _{n\in \mathbb{Z}}(\frac{1}{z-n}+\frac{1}{z+n})\end{array}$$
を用いると,
$$\begin{array}{rcl}& & \frac{1}{4}\sum _{n\in \mathbb{Z}}{\int }_0^{T}f(x)(\frac{1}{x-nT}+\frac{1}{x+nT})dx\\ & =& \frac{\pi }{2T}{\int }_0^{T}f(x)\cot \frac{\pi x}{T}dx\\ & =& \frac{\pi }{2T}({\int }_0^{T/2}f(x)\cot \frac{\pi x}{T}dx+{\int }_0^{T/2}f(T-x)\cot \frac{\pi (T-x)}{T}dx)\\ & =& \frac{\pi }{T}{\int }_0^{T/2}f(x)\cot \frac{\pi x}{T}dx\end{array}$$
となって, 示すことができました.

$0<a$としておきます. $t=\tan x$と置換して,
$$\begin{array}{rcl}{\int }_0^{\pi /4}\frac{{\tan }^{a}x-2{\sin }^{2}x}{\sin 2x\ln \tan x}dx& =& {\int }_0^1\frac{t^a-2\frac{t^2}{1+t^2}}{\frac{2t}{1+t^2}\ln t}\frac{dt}{1+t^2}\\ & =& \frac{1}{2}{\int }_0^1(\frac{t^{a-1}}{\ln t}-\frac{2t}{(1+t^2)\ln t})dt\\ & =& \frac{1}{2}{\int }_0^1\frac{t^{a-1}-1}{\ln t}dt+\frac{1}{2}{\int }_0^1(\frac{1}{\ln t}-\frac{2t}{(1+t^2)\ln t})dt\end{array}$$
初項はFrullani integral
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{f(ax)-f(bx)}{x}dx=(f(\mathrm{\infty })-f(0))\ln \frac{a}{b}\end{array}$$
より,
$$\begin{array}{rcl}\frac{1}{2}{\int }_0^1\frac{t^{a-1}-1}{\ln t}dt& =& -\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-ax}-e^{-x}}{x}dx\\ & =& \frac{1}{2}\ln a\end{array}$$
第2項は
$$\begin{array}{r}\frac{1}{2}{\int }_0^1(\frac{1}{\ln t}-\frac{2t}{(1+t^2)\ln t})dt=-\frac{1}{2}{\int }_0^{\mathrm{\infty }}(\frac{e^{-x}}{x}-\frac{2e^{-2x}}{x(1+e^{-2x})})dx\end{array}$$
ここで, Mellin変換,
$$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}{x}^{s-1}(e^{-x}-\frac{2e^{-2x}}{1+e^{-2x}})dx\\ & =& \mathrm{\Gamma }(s)-2{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}{e}^{-2x}}{1+e^{-2x}}dx\\ & =& \mathrm{\Gamma }(s)-2^{1-s}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x+1}dx\\ & =& \mathrm{\Gamma }(s)(1-2^{1-s}\eta (s))\end{array}$$
において, $s\to 0$とすることで, $\eta (0)=\frac{1}{2},{\eta }^{\prime }(0)=\ln \frac{\pi }{2}$を用いて,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}(\frac{e^{-x}}{x}-\frac{2e^{-2x}}{x(1+e^{-2x})})dx& =& \underset{s\to 0}{lim}\mathrm{\Gamma }(s)(1-2^{1-s}\eta (s))\\ & =& -\underset{s\to 0}{lim}{2}^{1-s}\eta (s)(-\ln 2+\frac{{\eta }^{\prime }(s)}{\eta (s)})\\ & =& \ln 2-\ln \frac{\pi }{2}\end{array}$$
よって, 求める積分は
$$\begin{array}{rcl}\frac{1}{2}(\ln a-\ln 2+\ln \frac{\pi }{2})& =& \frac{1}{2}(\ln \frac{a}{2}+\ln \frac{\pi }{2})\end{array}$$
となり, $a=2e$とすればよい.