積分botを解けるだけ解く, その5

まず, 第1種完全楕円積分, $K\left(\frac{1}{\sqrt{2}}\right)$を求めます. $x\to \sqrt{1-x^2}$の置換により,
$$\begin{array}{rcl}K\left(\frac{1}{\sqrt{2}}\right)& =& {\int }_0^1\frac{1}{\sqrt{1-x^2}\sqrt{1-\frac{1}{2}{x}^2}}dx\\ & =& {\int }_0^1\frac{x}{x\sqrt{\frac{1+x^2}{2}}\sqrt{1-x^2}}dx\\ & =& \sqrt{2}{\int }_0^1\frac{1}{\sqrt{1-x^4}}dx\\ & =& \frac{1}{2\sqrt{2}}{\int }_0^1{x}^{-3/4}(1-x{)}^{-1/2}dx\\ & =& \frac{1}{2\sqrt{2}}\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}\\ & =& \frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}{4\sqrt{\pi }}\end{array}$$
ここで, $k^{\prime }=\sqrt{1-k^2}$として, Legendreの関係式,
$$\begin{array}{r}K(k)E(k^{\prime })+E(k)K(k^{\prime })-K(k)K(k^{\prime })=\frac{\pi }{2}\end{array}$$
に$k=\frac{1}{\sqrt{2}}$を代入すると,
$$\begin{array}{r}2K\left(\frac{1}{\sqrt{2}}\right)E\left(\frac{1}{\sqrt{2}}\right)-K{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{\pi }{2}\end{array}$$
より,
$$\begin{array}{rcl}\sqrt{2}E\left(\frac{1}{\sqrt{2}}\right)& =& \frac{\sqrt{2}\pi }{4K\left(\frac{1}{\sqrt{2}}\right)}+\frac{1}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}}\right)\\ & =& \frac{\pi \sqrt{2\pi }}{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}+\frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}{4\sqrt{2\pi }}\end{array}$$

Eulerの変換公式,
$$\begin{array}{r}\sum _{0\le n}\frac{(a,b{)}_n}{(c{)}_{n}n!}{z}^n=(1-z{)}^{c-a-b}\sum _{0\le n}\frac{(c-a,c-b{)}_n}{(c{)}_{n}n!}{z}^n\end{array}$$
の両辺を$a$で偏微分して,
$$\begin{array}{r}\sum _{0\le n}\frac{(a,b{)}_n}{(c{)}_{n}n!}{z}^n\sum _{k=0}^{n-1}\frac{1}{a+k}=-(1-z{)}^{c-a-b}(\ln (1-z)\sum _{0\le n}\frac{(c-a,c-b{)}_n}{(c{)}_{n}n!}{z}^n+\sum _{0\le n}\frac{(c-a,c-b{)}_n}{(c{)}_{n}n!}{z}^n\sum _{k=0}^{n-1}\frac{1}{c-a+k})\end{array}$$
ここで, $c=a+b$として, 右辺第2項を移項して,
$$\begin{array}{r}\sum _{0\le n}\frac{(a,b{)}_n}{(c{)}_{n}n!}{z}^n\sum _{k=0}^{n-1}(\frac{1}{a+k}+\frac{1}{b+k})=-\ln (1-z)\sum _{0\le n}\frac{(c-a,c-b{)}_n}{(c{)}_{n}n!}{z}^n\end{array}$$
ここで, $a=b=\frac{1}{2}$として,
$$\begin{array}{r}2\sum _{0\le n}\frac{{\left(\frac{1}{2}\right)}_n^2}{n{!}^2}{z}^n\sum _{k=0}^{n-1}\frac{1}{2k+1}=-\frac{1}{2}\ln (1-z){}_2{F}_1[\begin{array}{c}\frac{1}{2},\frac{1}{2}\\ 1\end{array};z]\end{array}$$
また, Euler積分表示
$$\begin{array}{r}\sum _{0\le n}\frac{(a,b{)}_n}{(c{)}_{n}n!}{z}^n=\frac{\mathrm{\Gamma }(c)}{\mathrm{\Gamma }(b)\mathrm{\Gamma }(c-b)}{\int }_0^1{x}^{b-1}(1-x{)}^{c-b-1}(1-zx{)}^{-a}dx\end{array}$$
の両辺を$a$で偏微分して, $a=b=\frac{1}{2},c=1$とすると,
$$\begin{array}{r}2\sum _{0\le n}\frac{{\left(\frac{1}{2}\right)}_n^2}{n{!}^2}{z}^n\sum _{k=0}^{n-1}\frac{1}{2k+1}=-\frac{1}{\pi }{\int }_0^1\frac{\ln (1-zx)}{\sqrt{x(1-x)(1-zx)}}dx\end{array}$$
これより, 上の式を合わせて,
$$\begin{array}{r}{\int }_0^1\frac{\ln (1-zx)}{\sqrt{x(1-x)(1-zx)}}dx=\frac{\pi }{2}\ln (1-z){}_2{F}_1[\begin{array}{c}\frac{1}{2},\frac{1}{2}\\ 1\end{array};z]\end{array}$$
与えられた積分は,
$$\begin{array}{rcl}{\int }_0^1\frac{\ln (1-x)}{\sqrt{x-x^3}}dx& =& {\int }_0^1\frac{\ln (1-x^2)-\ln (1+x)}{\sqrt{x(1-x)(1+x)}}dx\end{array}$$
ここで,
$$\begin{array}{rcl}{\int }_0^1\frac{\ln (1-x^2)}{\sqrt{x(1-x)(1+x)}}dx& =& \underset{a\to \frac{1}{2}}{lim}\frac{\partial }{\partial a}{\int }_0^1{x}^{-1/2}(1-x^2{)}^{a-1}dx\\ & =& \frac{1}{2}\underset{a\to \frac{1}{2}}{lim}\frac{\partial }{\partial a}\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }(a)}{\mathrm{\Gamma }(a+\frac{1}{4})}\\ & =& \frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}(\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{4}\right))\\ & =& \frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}{4\sqrt{2\pi }}(2\ln 2-\pi )\end{array}$$
と,
$$\begin{array}{rcl}{\int }_0^1\frac{\ln (1+x)}{\sqrt{x(1-x)(1-x)}}dx& =& \frac{\pi }{2}\ln 2{}_2{F}_1[\begin{array}{c}\frac{1}{2},\frac{1}{2}\\ 1\end{array};-1]\\ & =& \frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}{4\sqrt{2\pi }}\ln 2\end{array}$$
より,
$$\begin{array}{r}{\int }_0^1\frac{\ln (1-x)}{\sqrt{x-x^3}}dx=\frac{\mathrm{\Gamma }{\left(\frac{1}{4}\right)}^2}{4\sqrt{2\pi }}(\ln 2-\pi )\end{array}$$

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$z$について微分して,
$$\begin{array}{r}\psi (z)=\ln z-\frac{1}{2z}-2{\int }_0^{\mathrm{\infty }}\frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx\end{array}$$
これより,
$$\begin{array}{r}2z{\int }_0^{\mathrm{\infty }}\frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx=z\ln z-\frac{1}{2}-z\psi (z)\end{array}$$
$z$について, $0$から$a$まで積分して,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x\ln (1+\frac{a^2}{x^2})}{e^{2\pi x}-1}dx& =& {\int }_0^a(z\ln z-\frac{1}{2}-z\psi (z))dz\\ & =& -\frac{1}{4}{a}^2(1-2\ln a)-\frac{a}{2}-{\int }_0^{a}z\psi (z)dz\\ & =& -\frac{1}{4}{a}^2(1-2\ln a)-\frac{a}{2}-a\ln \mathrm{\Gamma }(a)+{\int }_0^a\ln \mathrm{\Gamma }(z)dz\\ & =& -\frac{1}{4}{a}^2(1-2\ln a)-\frac{a}{2}(1+2\ln \mathrm{\Gamma }(a))+\ln K(a)+\frac{a}{2}\ln 2\pi +\frac{a(1-a)}{2}\end{array}$$

$\tan x\to x$の置換により,
$$\begin{array}{rcl}{\int }_0^{\pi /2}{(x-\arctan (\frac{1-a}{1+a}\tan x))}^{2}dx& =& {\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}{(\arctan x-\arctan \left(\frac{1-a}{1+a}x\right))}^{2}dx\\ & =& {\int }_0^{\pi /2}{(x-\arctan (\frac{1-a}{1+a}\tan x))}^{2}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}{(\arctan x-\arctan \left(\frac{1-a}{1+a}x\right))}^{2}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{1}{1+x^2}{\int }_{\frac{1-a}{1+a}}^1{\int }_{\frac{1-a}{1+a}}^1\frac{x^2}{(1+y^2{x}^2)(1+z^2{x}^2)}dydzdx\\ & =& {\int }_{\frac{1-a}{1+a}}^1{\int }_{\frac{1-a}{1+a}}^1{\int }_0^{\mathrm{\infty }}\frac{x^2}{(1+x^2)(1+y^2{x}^2)(1+z^2{x}^2)}dxdydz\end{array}$$
最後の変形はFubiniの定理による. ここで, 部分分数分解などを用いて,
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{x^2}{(1+x^2)(1+y^2{x}^2)(1+z^2{x}^2)}dx=\frac{\pi }{2}\frac{1}{(1+y)(1+z)(y+z)}\end{array}$$
であることと, $y\to \frac{1-y}{1+y},z\to \frac{1-z}{1+z}$の置換により,
$$\begin{array}{rcl}& & {\int }_{\frac{1-a}{1+a}}^1{\int }_{\frac{1-a}{1+a}}^1{\int }_0^{\mathrm{\infty }}\frac{x^2}{(1+x^2)(1+y^2{x}^2)(1+z^2{x}^2)}dxdydz\\ & =& \frac{\pi }{2}{\int }_{\frac{1-a}{1+a}}^1{\int }_{\frac{1-a}{1+a}}^1\frac{1}{(1+y)(1+z)(y+z)}dydz\\ & =& \frac{\pi }{2}{\int }_0^a{\int }_0^a\frac{1}{\frac{2}{1+y}\frac{2}{1+z}(\frac{1-y}{1+y}+\frac{1-z}{1+z})}\frac{2dy}{(1+y{)}^2}\frac{2dz}{(1+z{)}^2}\\ & =& \frac{\pi }{4}{\int }_0^a{\int }_0^a\frac{1}{1-yz}dydz\\ & =& \frac{\pi }{4}\sum _{0<n}{\int }_0^a{\int }_0^a(yz{)}^{n-1}dydz\\ & =& \frac{\pi }{4}\sum _{0<n}\frac{a^{2n}}{n^2}\\ & =& \frac{\pi }{4}{\mathrm{Li}}_2(a^2)\end{array}$$

この前の記事( https://mathlog.info/articles/1901 ) の命題3において, $r=s=\beta$とすると,
$$\begin{array}{rcl}{\int }_0^1\frac{{\arctan }^{2}x}{\sqrt{1-x^2}}dx& =& {\int }_0^1{\arctan }^2(\sin x)dx\\ & =& \pi {\chi }_2({\beta }^2)\end{array}$$

よく知られたMellin変換,
$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx=\mathrm{\Gamma }(s)\zeta (s)\end{array}$$
より,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}{x}^{s-1}(1-\coth \pi x)dx& =& -2{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^{2\pi x}-1}dx\\ & =& -2(2\pi {)}^{-s}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{e^x-1}dx\\ & =& -2^{1-s}{\pi }^{-s}\mathrm{\Gamma }(s)\zeta (s)\end{array}$$

まず, 置換$x\to \frac{1}{x}$により,
$$\begin{array}{rcl}{\int }_0^1\ln \ln \frac{1+\sqrt{1-x^2}}{x}dx& =& {\int }_1^{\mathrm{\infty }}\frac{\ln \ln (x+\sqrt{x^2-1})}{x^2}dx\end{array}$$
さらに$x\to \cosh x$と置換して,
$$\begin{array}{r}{\int }_1^{\mathrm{\infty }}\frac{\ln \ln (x+\sqrt{x^2-1})}{x^2}dx={\int }_0^{\mathrm{\infty }}\frac{\ln x\sinh x}{{\cosh }^{2}x}dx\end{array}$$
ここで, Mellin変換により,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x^s\sinh x}{{\cosh }^{2}x}dx& =& s{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{\cosh x}dx\\ & =& 2\mathrm{\Gamma }(s+1)\beta (s)\end{array}$$
よって, 特殊値,
$$\begin{array}{rcl}\beta (0)& =& \frac{1}{2}\\ {\beta }^{\prime }(0)& =& \ln \frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2\mathrm{\Gamma }\left(\frac{3}{4}\right)}\end{array}$$
を用いて,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{\ln x\sinh x}{{\cosh }^{2}x}dx& =& \underset{s\to 0}{lim}\frac{\partial }{\partial s}{\int }_0^{\mathrm{\infty }}\frac{x^s\sinh x}{{\cosh }^{2}x}dx\\ & =& 2\underset{s\to 0}{lim}\frac{\partial }{\partial s}\mathrm{\Gamma }(s+1)\beta (s)\\ & =& 2\underset{s\to 0}{lim}\mathrm{\Gamma }(s+1)(\psi (s+1)\beta (s)+{\beta }^{\prime }(s))\\ & =& -2\gamma \beta (0)+2{\beta }^{\prime }(0)\\ & =& -\gamma -2\ln \frac{2\mathrm{\Gamma }\left(\frac{3}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{4}\right)}\end{array}$$

Fubiniの定理より,
$$\begin{array}{rcl}{\int }_0^{2\pi }\frac{\arctan (2{\cos }^{2}x)}{{\cos }^{2}x}dx& =& 4{\int }_0^{\pi /2}\frac{\arctan (2{\cos }^{2}x)}{{\cos }^{2}x}dx\\ & =& 4{\int }_0^{\pi /2}{\int }_0^2\frac{1}{1+y^2{\cos }^{4}x}dydx\\ & =& 4{\int }_0^2{\int }_0^{\pi /2}\frac{1}{1+y^2{\cos }^{4}x}dxdy\\ & =& 4{\int }_0^2{\int }_0^{\mathrm{\infty }}\frac{1}{1+\frac{y^2}{(1+x^2)}}\frac{1}{1+x^2}dxdy\\ & =& 4{\int }_0^2{\int }_0^{\mathrm{\infty }}\frac{1+x^2}{(1+x^2{)}^2+y^2}dxdy\\ & =& 4\mathrm{Re}\left({\int }_0^2{\int }_0^{\mathrm{\infty }}\frac{1}{1+iy+x^2}dxdy\right)\\ & =& 2\pi \mathrm{Re}\left({\int }_0^2\frac{1}{\sqrt{1+iy}}dy\right)\\ & =& 4\pi \mathrm{Re}(-i(\sqrt{1+2i}-1))\\ & =& 4\pi \mathrm{Im}\sqrt{1+2i}\\ & =& 4\pi \sqrt{\frac{\sqrt{5}-1}{2}}\\ & =& \frac{4\pi }{\sqrt{\varphi }}\end{array}$$