積分botを解けるだけ解く, その4

途中で, $t=\tan x$の置換をします. また, $\varphi (\varphi -1)=1$であることを用いることができます.
$$\begin{array}{rcl}{\int }_0^{2\pi }\frac{x}{\varphi -{\cos }^{2}x}dx& =& \frac{1}{2}({\int }_0^{2\pi }\frac{x}{\varphi -{\cos }^{2}x}dx+{\int }_0^{2\pi }\frac{2\pi -x}{\varphi -{\cos }^{2}x}dx)\\ & =& \pi {\int }_0^{2\pi }\frac{1}{\varphi -{\cos }^{2}x}dx\\ & =& 4\pi {\int }_0^{\pi /2}\frac{1}{\varphi -{\cos }^{2}x}dx\\ & =& 4\pi {\int }_0^{\mathrm{\infty }}\frac{1}{\varphi (1+t^2)-1}dt\\ & =& \frac{4\pi }{\sqrt{\varphi (\varphi -1)}}{\int }_0^{\mathrm{\infty }}\frac{1}{1+t^2}dt\\ & =& 2{\pi }^2\end{array}$$

変数変換$x_k\to \frac{x_k}{a_k}$により,
$$\begin{array}{r}{\int }_{{\mathbb{R}}^n}\frac{\sin \sum _{k=1}^n{x}_k}{\sum _{k=1}^n{x}_k}\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k={\pi }^n\end{array}$$
を示せばよい. まず, $0<a<1$に対し, $x_k\to -x_k$の置換により,
$$\begin{array}{rcl}{\int }_{{\mathbb{R}}^n}\sin \left(a\sum _{k=1}^n{x}_k\right)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k& =& -{\int }_{{\mathbb{R}}^n}\sin \left(a\sum _{k=1}^n{x}_k\right)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k=0\end{array}$$
ここで, $n$に関する帰納法により, $0<a<1$に対し,
$$\begin{array}{r}{\int }_{{\mathbb{R}}^n}\cos \left(a\sum _{k=1}^n{x}_k\right)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k={\pi }^n\end{array}$$
であることを示す. $n-1$まで成立するとすると,
$$\begin{array}{rcl}& & {\int }_{{\mathbb{R}}^n}\cos \left(a\sum _{k=1}^n{x}_k\right)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k\\ & =& {\int }_{{\mathbb{R}}^n}\cos (a\sum _{k=1}^{n-1}{x}_k+a{x}_n)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k\\ & =& {\int }_{{\mathbb{R}}^n}\cos \left(a\sum _{k=1}^{n-1}{x}_k\right)\cos a{x}_n\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k-{\int }_{{\mathbb{R}}^n}\sin \left(a\sum _{k=1}^{n-1}{x}_k\right)\sin a{x}_n\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin x_n\cos a{x}_n}{x_n}d{x}_n{\int }_{{\mathbb{R}}^{n-1}}\cos \left(a\sum _{k=1}^{n-1}{x}_k\right)\prod _{k=1}^{n-1}\frac{\sin x_k}{x_k}d{x}_k\\ & -& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin x_n\sin a{x}_n}{x_n}d{x}_n{\int }_{{\mathbb{R}}^{n-1}}\sin \left(a\sum _{k=1}^{n-1}{x}_k\right)\prod _{k=1}^{n-1}\frac{\sin x_k}{x_k}d{x}_k\\ & =& \pi \cdot {\pi }^{n-1}-0\\ & =& {\pi }^n\end{array}$$
最後の変形は帰納法の仮定と,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin x\cos ax}{x}dx& =& \frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin (1+a)x+\sin (1-a)x}{x}dx\\ & =& \pi \end{array}$$
であることを用いた. よって,
$$\begin{array}{r}{\int }_{{\mathbb{R}}^n}\cos \left(a\sum _{k=1}^n{x}_k\right)\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k={\pi }^n\end{array}$$
において, $a$を$0$から$1$まで積分し, 積分順序の交換をして,
$$\begin{array}{r}{\int }_{{\mathbb{R}}^n}\frac{\sin \sum _{k=1}^n{x}_k}{\sum _{k=1}^n{x}_k}\prod _{k=1}^n\frac{\sin x_k}{x_k}d{x}_k={\pi }^n\end{array}$$
を得る.

$$\begin{array}{rcl}& & \frac{d}{da}\arctan \frac{\sqrt{\sqrt{1+a^4}-1}}{a}\\ & =& \frac{1}{1+\frac{\sqrt{1+a^4}-1}{a^2}}\left(\frac{\frac{1}{2\sqrt{\sqrt{1+a^4}-1}}\frac{4a^3}{2\sqrt{1+a^4}}a-\sqrt{\sqrt{1+a^4}-1}}{a^2}\right)\\ & =& \frac{1}{a^2+\sqrt{1+a^4}-1}(\frac{a^4}{\sqrt{1+a^4}\sqrt{\sqrt{1+a^4}-1}}-\sqrt{\sqrt{1+a^4}-1})\\ & =& \frac{1}{a^2+\sqrt{1+a^4}-1}\frac{\sqrt{1+a^4}-1}{\sqrt{1+a^4}\sqrt{\sqrt{1+a^4}-1}}\\ & =& \frac{1}{\sqrt{2(\sqrt{1+a^4}-1)(\sqrt{1+a^4}+a^2)}}\sqrt{\frac{\sqrt{1+a^4}-1}{1+a^4}}\\ & =& \frac{1}{\sqrt{2}}\sqrt{\frac{\sqrt{1+a^2}-a^2}{1+a^4}}\end{array}$$
より, 示された.

まず,
$$\begin{array}{rcl}I& :=& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2}}{1+2\cosh \frac{2\pi x}{\sqrt{3}}}dx\end{array}$$
と置く, 被積分関数は$x=\pm \frac{i}{\sqrt{3}}$に極をもつが, 虚部が$\pm \frac{i}{\sqrt{3}}$の直線上で主値積分を考えることができる.
積分路1
上のような積分路で積分をすると, 主値積分を考えるので, $x=\frac{i}{\sqrt{3}}$における留数の半分が含まれているとみなせて,
$$\begin{array}{rcl}I& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2}}{1+2\cosh \frac{2\pi x}{\sqrt{3}}}dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi (x+i/\sqrt{3}{)}^2}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}+\frac{2\pi i}{3})}dx+\pi iRe{s}_{x=\frac{i}{3}}\frac{e^{i\pi x^2}}{1+2\cosh \frac{2\pi x}{\sqrt{3}}}\\ & =& e^{-i\pi /3}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2-2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}+\frac{2\pi i}{3})}dx+\frac{1}{2}{e}^{-i\pi /3}\\ & =& e^{-i\pi /3}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2+2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}-\frac{2\pi i}{3})}dx+\frac{1}{2}{e}^{-i\pi /3}\end{array}$$
よって,
$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2-2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}+\frac{2\pi i}{3})}dx& =& e^{i\pi /3}I-\frac{1}{2}\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x^2+2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}-\frac{2\pi i}{3})}dx& =& e^{i\pi /3}I-\frac{1}{2}\end{array}$$
ここで, $y=e^{-2\pi x/\sqrt{3}},\omega =e^{2\pi i/3}$とすると,
$$\begin{array}{rcl}& & \frac{1}{1+2\cosh \frac{2\pi x}{\sqrt{3}}}=\frac{1}{1+y^{-1}+y}=\frac{y(1-y)}{1-y^3}\\ & & \frac{e^{-2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}+\frac{2\pi i}{3})}=\frac{y}{1+\omega y^{-1}+{\omega }^{2}y}=\frac{{\omega }^2{y}^2(1-{\omega }^{2}y)}{1-y^3}\\ & & \frac{e^{2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}-\frac{2\pi i}{3})}=\frac{y^{-1}}{1+{\omega }^2{y}^{-1}+\omega y}=\frac{\omega (1-\omega y)}{1-y^3}\end{array}$$
よって,
$$\begin{array}{r}\omega \frac{y(1-y)}{1-y^3}+{\omega }^2\frac{{\omega }^2{y}^2(1-{\omega }^{2}y)}{1-y^3}+{\omega }^2\frac{\omega (1-\omega y)}{1-y^3}=1\end{array}$$
より,
$$\begin{array}{rcl}{e}^{i\pi /4}& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i\pi x^2}dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i\pi x^2}(\frac{e^{2\pi i/3}}{1+2\cosh \frac{2\pi x}{\sqrt{3}}}+\frac{e^{-2\pi i/3-2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}+\frac{2\pi i}{3})}+\frac{e^{-2\pi i/3+2\pi x/\sqrt{3}}}{1+2\cosh (\frac{2\pi x}{\sqrt{3}}-\frac{2\pi i}{3})})dx\\ & =& e^{2\pi i/3}I+2e^{-2\pi i/3}(e^{i\pi /3}I-\frac{1}{2})\\ & =& (e^{2\pi i/3}+2e^{-i\pi /3})I-e^{-2\pi i/3}\end{array}$$
よって,
$$\begin{array}{rcl}I& =& \frac{e^{i\pi /4}+e^{-2\pi i/3}}{e^{2\pi i/3}+2e^{-i\pi /3}}\\ & =& e^{i\pi /3}{e}^{i\pi /4}+e^{-i\pi /3}\\ & =& \frac{1+i\sqrt{3}}{2}\frac{\sqrt{2}+i\sqrt{2}}{2}+\frac{1-i\sqrt{3}}{2}\\ & =& \frac{2+\sqrt{2}-\sqrt{6}}{4}+i\left(\frac{\sqrt{2}+\sqrt{6}-2\sqrt{3}}{4}\right)\end{array}$$
あとは, $I$の実部と虚部を比較すればよい.

まず,
$$\begin{array}{r}y=\frac{1}{1+\frac{f(x)}{f(a+b-x)}}\end{array}$$
と置いて,
$$\begin{array}{rcl}\sqrt{1+\sqrt{y}}-\sqrt{1-\sqrt{y}}& =& \frac{2\sqrt{y}}{\sqrt{1+\sqrt{y}}+\sqrt{1-\sqrt{y}}}\\ & =& \sqrt{2}\frac{\sqrt{y}}{\sqrt{1+\sqrt{1-y}}}\\ & =& \sqrt{2}\sqrt{\frac{y}{1+\sqrt{1-y}}}\\ & =& \sqrt{2}\sqrt{\frac{f(a+b-x)}{\sqrt{f(x)+f(a+b-x)}(\sqrt{f(x)+f(a+b-x)}+\sqrt{f(x)})}}\\ & =& \sqrt{2}\sqrt{\frac{\sqrt{f(x)+f(a+b-x)}-\sqrt{f(x)}}{\sqrt{f(x)+f(a+b-x)}}}\\ & =& \sqrt{2}\sqrt{1-\frac{1}{\sqrt{1+\frac{f(a+b-x)}{f(x)}}}}\end{array}$$
よって,
$$\begin{array}{r}{\int }_a^b\sqrt{1+\sqrt{y}}dx={\int }_a^b\sqrt{1-\sqrt{y}}dx+\sqrt{2}{\int }_a^b\sqrt{1-\frac{1}{\sqrt{1+\frac{f(a+b-x)}{f(x)}}}}dx\end{array}$$
この最後の項に, $x\to a+b-x$の置換をすればよい.

$x\to \frac{(1+p)x}{p+x}$の置換により,
$$\begin{array}{rcl}B(a,b)& =& {\int }_0^1{x}^{a-1}(1-x{)}^{b-1}dx\\ & =& (1+p{)}^a{p}^b{\int }_0^1{x}^{a-1}(1-x{)}^{b-1}(p+x{)}^{-a-b}dx\end{array}$$
よって,
$$\begin{array}{r}{\int }_0^1{x}^{a-1}(1-x{)}^{b-1}(p+x{)}^{-a-b}dx=\frac{B(a,b)}{(1+p{)}^a{p}^b}\end{array}$$

$$\begin{array}{r}K(x)={\int }_0^{\pi /2}\frac{1}{\sqrt{1-y^2}\sqrt{1-x^2{y}^2}}dy\end{array}$$
より,
$$\begin{array}{rcl}I(m,1)& =& {\int }_0^{1}K(x^{1/m})dx\\ & =& m{\int }_0^1{x}^{m-1}K(x)dx\\ & =& m{\int }_0^1{\int }_0^1\frac{x^{m-1}}{\sqrt{1-y^2}\sqrt{1-x^2{y}^2}}dxdy\\ & =& m{\int }_0^1{\int }_0^y\frac{x^{m-1}{y}^{-m}}{\sqrt{1-y^2}\sqrt{1-x^2}}dxdy\end{array}$$
ここで, $x\to \sin x,y\to \sin y$とすると,
$$\begin{array}{rcl}{\int }_0^1{\int }_0^y\frac{x^{m-1}{y}^{-m}}{\sqrt{1-y^2}\sqrt{1-x^2}}dxdy& =& {\int }_0^{\pi /2}{\int }_0^y\frac{{\sin }^{m-1}x}{{\sin }^{m}y}dxdy\\ & =& {\int }_0^{\pi /2}{\int }_x^{\pi /2}\frac{{\sin }^{m-1}x}{{\sin }^{m}y}dydx\end{array}$$
$m$が偶数, $m=2r$のとき, 不定積分,
$$\begin{array}{rcl}\int \frac{1}{{\sin }^{2r}x}dx& =& -\frac{2^{2r}}{r(\genfrac{}{}{0ex}{}{2r}{r})}\cos x\sum _{k=0}^{r-1}\frac{(\genfrac{}{}{0ex}{}{2k}{k})}{(2\sin x{)}^{2k+1}}\end{array}$$
を用いて,
$$\begin{array}{rcl}& & {\int }_0^{\pi /2}{\int }_x^{\pi /2}\frac{{\sin }^{2r-1}x}{{\sin }^{2r}y}dydx\\ & =& \frac{2^{2r}}{r(\genfrac{}{}{0ex}{}{2r}{r})}{\int }_0^{\pi /2}\cos x{\sin }^{2r-1}x\sum _{k=0}^{r-1}\frac{(\genfrac{}{}{0ex}{}{2k}{k})}{(2\sin x{)}^{2k+1}}dx\\ & =& \frac{2^{2r}}{r(\genfrac{}{}{0ex}{}{2r}{r})}\sum _{k=0}^{r-1}\frac{(\genfrac{}{}{0ex}{}{2k}{k})}{2^{2k+1}}{\int }_0^{\pi /2}\cos x{\sin }^{2r-2k-2}xdx\\ & =& \frac{2^{2r-1}}{r(\genfrac{}{}{0ex}{}{2r}{r})}\sum _{k=0}^{r-1}\frac{(\genfrac{}{}{0ex}{}{2k}{k})}{2^{2k}(2r-2k-1)}\end{array}$$
$m$が奇数, $m=2r+1$のとき, 不定積分,
$$\begin{array}{rcl}\int \frac{1}{{\sin }^{2r+1}x}dx& =& \frac{1}{2^{2r}}(\genfrac{}{}{0ex}{}{2r}{r})\ln \tan \frac{x}{2}-\frac{1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r})\cos x\sum _{k=1}^r\frac{1}{k(\genfrac{}{}{0ex}{}{2k}{k})}{\left(\frac{2}{\sin x}\right)}^{2k}\end{array}$$
を用いて,
$$\begin{array}{rcl}& & {\int }_0^{\pi /2}{\int }_x^{\pi /2}\frac{{\sin }^{2r}x}{{\sin }^{2r+1}y}dydx\\ & =& \frac{1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r}){\int }_0^{\pi /2}{\sin }^{2r}x\cos x\sum _{k=1}^r\frac{1}{k(\genfrac{}{}{0ex}{}{2k}{k})}{\left(\frac{2}{\sin x}\right)}^{2k}dx-\frac{1}{2^{2r}}(\genfrac{}{}{0ex}{}{2r}{r}){\int }_0^{\pi /2}{\sin }^{2r}x\ln \tan \frac{x}{2}dx\\ & =& \frac{1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=1}^r\frac{2^{2k}}{k(\genfrac{}{}{0ex}{}{2k}{k})}{\int }_0^{\pi /2}{\sin }^{2r-2k}x\cos xdx-\frac{1}{2^{2r}}(\genfrac{}{}{0ex}{}{2r}{r}){\int }_0^{\pi /2}{\sin }^{2r}x\ln \tan \frac{x}{2}dx\\ & =& \frac{1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=1}^r\frac{2^{2k}}{k(2r-2k+1)(\genfrac{}{}{0ex}{}{2k}{k})}-\frac{1}{2^{2r}}(\genfrac{}{}{0ex}{}{2r}{r}){\int }_0^{\pi /2}{\sin }^{2r}x\ln \tan \frac{x}{2}dx\end{array}$$

ここで, $\tan \frac{x}{2}\to x$の置換により,
$$\begin{array}{rcl}-{\int }_0^{\pi /2}{\sin }^{2r}x\ln \tan \frac{x}{2}dx& =& -{\int }_0^1{\left(\frac{2x}{1+x^2}\right)}^{2r+1}\frac{\ln x}{x}dx\\ & =& {\int }_0^{\mathrm{\infty }}\frac{x}{{\cosh }^{2r+1}x}dx\end{array}$$
ここで, $\mathrm{Re}(s)$を十分大きくとって, Mellin変換を考える.
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{{\cosh }^{2r+1}x}dx& =& 2^{2r+1}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}{e}^{-(2r+1)x}}{(1+e^{-2x}{)}^{2r+1}}dx\\ & =& 2^{2r+1}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k(\genfrac{}{}{0ex}{}{2r+k}{k}){\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-(2r+2k+1)x}dx\\ & =& \frac{2^{2r+1}}{(2r)!}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\frac{(k+1)\cdots (k+2r)}{(2r+2k+1{)}^s}\\ & =& \frac{2^{2r+1}}{(2r)!}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^{k-r}\frac{(k+1-r)(k+2-r)\cdots (k+r)}{(2k+1{)}^s}\\ & =& \frac{2^{2r+1}}{(2r)!}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^{k-r}}{(2k+1{)}^s}\prod _{i=0}^{r-1}({(k+\frac{1}{2})}^2-{(i+\frac{1}{2})}^2)\\ & =& \frac{2^{2r+1}{\left(\frac{1}{2}\right)}_n^2}{(2r)!}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(2k+1{)}^s}\prod _{i=0}^{r-1}(1-\frac{{(2k+1)}^2}{{(2i+1)}^2})\\ & =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(2k+1{)}^s}\prod _{i=0}^{r-1}(1-\frac{{(2k+1)}^2}{{(2i+1)}^2})\end{array}$$
ここで, 自然数の組$(k_1,\dots ,k_a)$に対し,
$$\begin{array}{r}{t}_n(k_1,\dots ,k_a):=\sum _{0\le n_1<n_2<\cdots <n_a<n}\frac{1}{(2n_1+1{)}^{k_1}\cdots (2n_a+1{)}^{k_a}}\end{array}$$
と定義すると,
$$\begin{array}{rcl}\prod _{i=0}^{r-1}(1-\frac{{(2k+1)}^2}{{(2i+1)}^2})& =& \sum _{j=0}^r(-1{)}^j{t}_r(\{2{\}}^j)(2k+1{)}^{2j}\end{array}$$
ここで, $\{2{\}}^j$は$2$が$j$個並んだものを表し, $t_n(\{2{\}}^0)=t_n(\varnothing )=1$とする. これを先ほどの式に代入して,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x^{s-1}}{{\cosh }^{2r+1}x}dx& =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(2k+1{)}^s}\sum _{j=0}^r(-1{)}^j{t}_r(\{2{\}}^j)(2k+1{)}^{2j}\\ & =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{j=0}^r(-1{)}^j{t}_r(\{2{\}}^j)\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(2k+1{)}^{s-2j}}\\ & =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{j=0}^r(-1{)}^j{t}_r(\{2{\}}^j)\beta (s-2j)\end{array}$$
ここで, $\beta (s)$はDirichletのベータ関数である. 上の式において, 解析接続により, $s=2$を代入して,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x}{{\cosh }^{2r+1}x}dx& =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{j=0}^r{t}_r(\{2{\}}^j)\beta (2-2j)\\ & =& \frac{1}{2^{2r-1}}(\genfrac{}{}{0ex}{}{2r}{r})(\beta (2)-\sum _{j=0}^{r-1}(-1{)}^j{t}_r(\{2{\}}^{j+1})\beta (-2j))\end{array}$$
ここで, Euler数$E_j$を用いて, $\beta (-2j)=\frac{E_{2j}}{2}$と表せることを用いて,
$$\begin{array}{rcl}{\int }_0^{\mathrm{\infty }}\frac{x}{{\cosh }^{2r+1}x}dx& =& \frac{1}{2^{2r}}(\genfrac{}{}{0ex}{}{2r}{r})(2\beta (2)-\sum _{j=0}^{r-1}(-1{)}^j{t}_r(\{2{\}}^{j+1})E_{2j})\end{array}$$
よって, これより,
$$\begin{array}{rcl}& & {\int }_0^{\pi /2}{\int }_x^{\pi /2}\frac{{\sin }^{2r}x}{{\sin }^{2r+1}y}dydx\\ & =& \frac{1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=1}^r\frac{2^{2k}}{k(2r-2k+1)(\genfrac{}{}{0ex}{}{2k}{k})}+\frac{1}{2^{4r}}{(\genfrac{}{}{0ex}{}{2r}{r})}^2(2\beta (2)-\sum _{j=0}^{r-1}(-1{)}^j{t}_r(\{2{\}}^{j+1})E_{2j})\end{array}$$
まとめると, $m$を自然数として,
$$\begin{array}{rcl}I(2r,1)& =& \frac{2^{2r}}{(\genfrac{}{}{0ex}{}{2r}{r})}\sum _{k=0}^{r-1}\frac{(\genfrac{}{}{0ex}{}{2k}{k})}{2^{2k}(2r-2k-1)}\\ I(2r+1,1)& =& \frac{2r+1}{2^{2r+1}}(\genfrac{}{}{0ex}{}{2r}{r})\sum _{k=1}^r\frac{2^{2k}}{k(2r-2k+1)(\genfrac{}{}{0ex}{}{2k}{k})}+\frac{2r+1}{2^{4r}}{(\genfrac{}{}{0ex}{}{2r}{r})}^2(2\beta (2)-\sum _{j=0}^{r-1}(-1{)}^j{t}_r(\{2{\}}^{j+1})E_{2j})\end{array}$$
よって, $I(m,1)$をすべて求めることができた.

$$\begin{array}{rcl}{\int }_0^{\pi /2}{J}_0(\cos x)dx& =& \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^{2n}n{!}^2}{\int }_0^{\pi /2}{\cos }^{2n}xdx\\ & =& \frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^{2n}n{!}^2}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(n+\frac{1}{2})}{\mathrm{\Gamma }(n+1)}\\ & =& \frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{\left(\frac{1}{2}\right)}_n}{2^{2n}n{!}^3}\\ & =& \frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n{!}^2}(\genfrac{}{}{0ex}{}{2n}{n}){(-\frac{1}{16})}^n\end{array}$$
一方,
$$\begin{array}{rcl}{\left(J_0\left(\frac{1}{2}\right)\right)}^2& =& {\left(\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n{!}^2}{(-\frac{1}{16})}^n\right)}^2\\ & =& \sum _{n=0}^{\mathrm{\infty }}{(-\frac{1}{16})}^n\sum _{k=0}^n\frac{1}{k{!}^2(n-k){!}^2}\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n{!}^2}{(-\frac{1}{16})}^n\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2\end{array}$$
ここで, Vandermondeの恒等式より,
$$\begin{array}{r}\sum _{k=0}^n{(\genfrac{}{}{0ex}{}{n}{k})}^2=(\genfrac{}{}{0ex}{}{2n}{n})\end{array}$$
だから,
$$\begin{array}{rcl}{\left(J_0\left(\frac{1}{2}\right)\right)}^2& =& \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n{!}^2}(\genfrac{}{}{0ex}{}{2n}{n}){(-\frac{1}{16})}^n\end{array}$$
よって, 示された.