ラマヌジャンの公式集：級数、積分

　$\mathrm{Re}(x+y)>-1$なる複素数$x,y,z$に対し
$$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(-1{)}^n}{z+n}\frac{1}{\mathrm{\Gamma }(x+1-k)\mathrm{\Gamma }(y+1+k)}=\frac{\pi }{\sin \pi z}\frac{1}{\mathrm{\Gamma }(x+z+1)\mathrm{\Gamma }(y-z+1)}$$

　整数$m$に対し
$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+z^2+z^4/n^2}& =\frac{\pi }{2z\sqrt{3}}\frac{\sinh \left(\pi z\sqrt{3}\right)-\sqrt{3}\sin \left(\pi z\right)}{\cosh \left(\pi z\sqrt{3}\right)-\cos \left(\pi z\right)}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+(2m{)}^2+(2m{)}^4/n^2}& =\frac{1}{12m^2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+3m^2}\end{array}$$

　正整数$n$に対し
$$\begin{array}{rlrl}\sum _{k=1}^{\mathrm{\infty }}\frac{{\sin }^{2n+1}(kx)}{k}& =\frac{\pi }{2^{2n+1}}(\genfrac{}{}{0ex}{}{2n}{n})& & (0<x<\frac{\pi }{n+\frac{1}{2}})\\ \sum _{k=1}^{\mathrm{\infty }}\frac{{\sin }^{2n+2}(kx)}{k^2}& =\frac{\pi x}{2^{2n+1}}(\genfrac{}{}{0ex}{}{2n}{n})& & (0\le x\le \frac{\pi }{n+1})\end{array}$$

　非負整数$m,n$に対し
$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2n+1}x}{x}{\cos }^{2m}xdx& ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2n+2}x}{x^2}{\cos }^{2m}xdx\\ & =\frac{\pi }{2^{2m+2n+1}}\frac{(2m)!(2n)!}{m!n!(m+n)!}\\ \\ {\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2n+1}x}{x}\cos \left(2mx\right)dx& ={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2n+2}x}{x^2}\cos \left(2mx\right)dx\\ & =(-1{)}^m\frac{\pi }{2^{2n+1}}(\genfrac{}{}{0ex}{}{2n}{n-m})\end{array}$$

　偶奇の異なる正整数$m,n$に対し$l=\lfloor (m+n)/2\rfloor$とおくと
$$\frac{1}{2}+\sum _{k=1}^l{\cos }^{2n}\frac{\pi k}{m+n}=\frac{m+n}{2^{2n+1}}(\genfrac{}{}{0ex}{}{2n}{n})$$
また$0\le m\le (n-1)/2,n>0$なる偶奇の異なる整数$m,n$に対し$l=\lfloor (n-m)/2\rfloor$とおくと
$$\frac{1}{2}+\sum _{k=1}^l{\cos }^{2n}\frac{\pi k}{n-m}=\frac{n-m}{2^{2n+1}}((\genfrac{}{}{0ex}{}{2n}{n})+2(\genfrac{}{}{0ex}{}{2n}{m}))$$

$$f(z)=\sum _n\frac{A_n}{{\alpha }_n-z},g(z)=\sum _n\frac{B_n}{{\beta }_n-z}$$
および$xy\ne 0$に対し
$$f(x)g(y)=\sum _n\frac{A_{n}g({\alpha }_{n}y/x)}{{\alpha }_n-x}+\sum _n\frac{B_{n}f({\beta }_{n}x/y)}{{\beta }_n-y}$$

　$|\theta |,|\phi |<\pi$なる実数$\theta ,\phi$と$x/y$が純虚数とならない複素数$x,y$に対し
$$\begin{array}{rl}{\pi }^{2}xy\frac{\cos \left(\theta x\right)\cosh \left(\phi y\right)}{\sin \left(\pi x\right)\sinh \left(\pi y\right)}=1& +2\pi xy\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\cos \left(n\phi \right)}{n^2+y^2}\frac{\cosh \left(n\theta x/y\right)}{\sinh \left(\pi nx/y\right)}\\ & -2\pi xy\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\cos \left(n\theta \right)}{n^2-x^2}\frac{\cosh \left(n\phi y/x\right)}{\sinh \left(\pi ny/x\right)}\end{array}$$
また$|\theta |,|\phi |<\pi /2$のとき
$$\begin{array}{rl}\frac{\pi }{4}\frac{\sin \left(\theta x\right)\sinh \left(\phi y\right)}{\cos \left(\pi x/2\right)\cosh \left(\pi y/2\right)}=& y^2\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\sin \left(n\phi \right)}{n^2+y^2}\frac{\sinh \left(n\theta x/y\right)}{n\cosh \left(\pi nx/2y\right)}\\ & +x^2\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\sin \left(n\theta \right)}{n^2-x^2}\frac{\sinh \left(n\phi y/x\right)}{n\cosh \left(\pi ny/2x\right)}\end{array}$$
$$\begin{array}{rl}\frac{\pi }{4}\frac{\cos \left(\theta x\right)\sinh \left(\phi y\right)}{\sin \left(\pi x/2\right)\cosh \left(\pi y/2\right)}=\frac{\phi y}{2x}& +y^2\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\sin \left(n\phi \right)}{n^2+y^2}\frac{\cosh \left(n\theta x/y\right)}{n\sinh \left(\pi nx/2y\right)}\\ & -x^2\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{\cos \left(2n\theta \right)}{(2n{)}^2-x^2}\frac{\sinh \left(2n\phi y/x\right)}{n\cosh \left(\pi ny/x\right)}\end{array}$$

$$\begin{array}{rl}& {\pi }^{2}xy\cot \left(\pi x\right)\cosh \left(\pi y\right)\\ =& 1+2\pi xy\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi nx/y\right)}{n^2+y^2}-2\pi xy\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi ny/x\right)}{n^2-x^2}\\ \\ & {\pi }^{2}xy\csc \left(\pi x\right)\mathrm{csch}(\pi y)\\ =& 1+2\pi xy\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\mathrm{csch}(\pi nx/y)}{n^2+y^2}-2\pi xy\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\mathrm{csch}(\pi ny/x)}{n^2-x^2}\\ \\ & \frac{\pi }{4}\tan \left(\pi x/2\right)\tanh \pi y/2\\ =& y^2\sum _{n=0}^{\mathrm{\infty }}\frac{\tanh ((2n+1)\pi x/2y)}{(2n+1)((2n+1{)}^2+y^2)}+x^2\sum _{n=0}^{\mathrm{\infty }}\frac{\tanh ((2n+1)\pi y/2x)}{(2n+1)((2n+1{)}^2-x^2)}\\ \\ & \frac{\pi }{4}\sec \left(\pi x/2\right)\mathrm{sech}(\pi y/2)\\ =& \sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{n\mathrm{sech}(\pi nx/2y)}{n^2+y^2}+\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{n\mathrm{sech}(\pi ny/2x)}{n^2-x^2}\\ \\ & \frac{\pi }{4}\cot \left(\pi x/2\right)\mathrm{sech}(\pi y/2)\\ =& \frac{1}{2x}-y\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\coth \left(\pi nx/2y\right)}{n^2+y^2}-x\sum _{n=1}^{\mathrm{\infty }}\frac{\coth \left(\pi ny/x\right)}{(2n{)}^2-x^2}\end{array}$$

$$\begin{array}{rl}{\pi }^2{z}^2\cot \left(\pi z\right)\coth \left(\pi z\right)& =1-4\pi z^4\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi n\right)}{n^4-z^4}\\ {\pi }^2{z}^2\frac{\cosh \left(\pi z\sqrt{2}\right)+\cos \left(\pi z\sqrt{2}\right)}{\cosh \left(\pi z\sqrt{2}\right)-\cos \left(\pi z\sqrt{2}\right)}& =1+4\pi z^4\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi n\right)}{n^4+z^4}\\ {\pi }^2{z}^2\csc \left(\pi z\right)\mathrm{csch}(\pi z)& =1-4\pi z^4\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\coth \left(\pi n\right)}{n^4-z^4}\\ \frac{2{\pi }^2{z}^2}{\cosh \left(\pi z\sqrt{2}\right)-\cos \left(\pi z\sqrt{2}\right)}& =1+4\pi z^4\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n\coth \left(\pi n\right)}{n^4+z^4}\\ \frac{\pi }{8z^2}\tan \left(\pi z/2\right)\tanh \left(\pi z/2\right)& =\sum _{n=0}^{\mathrm{\infty }}\frac{(2n+1)\tanh ((2n+1)\pi /2)}{(2n+1{)}^4-z^4}\\ \frac{\pi }{8z^2}\frac{\cosh \left(\pi z/\sqrt{2}\right)-\cos \left(\pi z/\sqrt{2}\right)}{\cosh \left(\pi z/\sqrt{2}\right)+\cos \left(\pi z/\sqrt{2}\right)}& =\sum _{n=0}^{\mathrm{\infty }}\frac{(2n+1)\tanh ((2n+1)\pi /2)}{(2n+1{)}^4+z^4}\\ \frac{\pi }{8}\sec \left(\pi z/2\right)\mathrm{sech}(\pi z/2)& =\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{n^3\mathrm{sech}(\pi n/2)}{n^4-z^4}\\ \frac{\pi /4}{\cosh \left(\pi z/\sqrt{2}\right)+\cos \left(\pi z/\sqrt{2}\right)}& =\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{n^3\mathrm{sech}(\pi n/2)}{n^4+z^4}\end{array}$$

$$\begin{array}{rl}& {\pi }^{2}xy\frac{\cosh (\pi (x+y)\sqrt{2})+\cos (\pi (x-y)\sqrt{2})-\cosh (\pi (x-y)\sqrt{2})-\cos (\pi (x+y)\sqrt{2})}{(\cosh \left(\pi x\sqrt{2}\right)-\cos \left(\pi x\sqrt{2}\right))(\cosh \left(\pi y\sqrt{2}\right)-\cos \left(\pi y\sqrt{2}\right))}\\ =& 2+4\pi x{y}^3\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi nx/y\right)}{n^4+y^4}+4\pi x^{3}y\sum _{n=1}^{\mathrm{\infty }}\frac{n\coth \left(\pi ny/x\right)}{n^4+x^4}\end{array}$$

$${\int }_0^{\mathrm{\infty }}\frac{\cos \left(2nx\right)}{\cosh \left(\pi \sqrt{x}\right)+\cos \left(\pi \sqrt{x}\right)}dx=\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{k}{\cosh \left(\pi k/2\right)}{e}^{-n{k}^2}$$

$$\begin{array}{rl}\frac{\pi e^{-2\pi z}}{2z(\cosh 2\pi z-\cos 2\pi z)}=& \frac{1}{8\pi z^3}-\frac{1}{4z^2}+\frac{\pi }{4z}\\ & -\sum _{n=1}^{\mathrm{\infty }}\frac{1}{z^2+(z+n{)}^2}+4z\sum _{n=1}^{\mathrm{\infty }}\frac{n}{e^{2\pi n}-1}\frac{1}{4z^4+n^4}\\ \frac{\pi e^{-\pi z}}{4z(\cosh \pi z+\cos \pi z)}=& \frac{\pi }{8z}\\ & -\sum _{n=1}^{\mathrm{\infty }}\frac{1}{z^2+(z+2n+1{)}^2}-4z\sum _{n=1}^{\mathrm{\infty }}\frac{2n+1}{e^{(2n+1)\pi }-1}\frac{1}{4z^4+(2n+1{)}^4}\end{array}$$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{\coth \pi n}{n^3}& =\frac{7{\pi }^3}{180}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{\coth \pi n}{n^7}& =\frac{19{\pi }^7}{56700}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{\tanh ((2n+1)\pi /2)}{(2n+1{)}^3}& =\frac{{\pi }^3}{32}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{\tanh ((2n+1)\pi /2)}{(2n+1{)}^7}& =\frac{7{\pi }^7}{23040}\\ \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}\frac{\mathrm{csch}\pi n}{n^3}& =\frac{{\pi }^3}{360}\\ \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}\frac{\mathrm{csch}\pi n}{n^7}& =\frac{13{\pi }^7}{453600}\\ \sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\mathrm{sech}\pi n}{n}& =\frac{\pi }{8}\\ \sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\mathrm{sech}\pi n}{n^5}& =\frac{{\pi }^5}{768}\\ \sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{\mathrm{sech}\pi n}{n^9}& =\frac{23{\pi }^9}{1720320}\end{array}$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{\chi (n)}{n^2(e^{\pi n}-1)}+\frac{1}{8}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2\cosh \pi n}=\frac{5{\pi }^2}{96}-\frac{1}{2}{\int }_0^1\frac{\arctan x}{x}dx$$

$$\begin{array}{rl}& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+(n+1{)}^2}\frac{1}{\cosh ((2n+1)\pi )-\cosh \pi }\\ =& \frac{1}{2\sinh \pi }(\frac{1}{\pi }+\coth \pi -\frac{\pi }{2}{\tanh }^2\frac{\pi }{2})\\ \\ & \sum _{n=0}^{\mathrm{\infty }}\frac{2n+1}{25+(2n+1{)}^4/100}\frac{1}{e^{(2n+1)\pi }+1}\\ =& \frac{4689}{11890}-\frac{\pi }{8}{\coth }^2\frac{5\pi }{2}\end{array}$$

$$\phi (t)=\prod _{k=0}^{\mathrm{\infty }}\frac{1}{1+(\frac{t}{a+k}{)}^2}$$
とおくと$\alpha \beta =\pi$において
$$\sqrt{\alpha }(\frac{1}{2}+\sum _{k=1}^{\mathrm{\infty }}(\mathrm{sech}\alpha k{)}^{2a})=\frac{\mathrm{\Gamma }(a)}{\mathrm{\Gamma }(a+\frac{1}{2})}\sqrt{\beta }(\frac{1}{2}+\sum _{k=1}^{\mathrm{\infty }}\phi (\beta k))$$

　$\alpha \beta =\pi$において
$$e^{z^2/4}\sqrt{\alpha }(\frac{1}{2}+\sum _{k=1}^{\mathrm{\infty }}{e}^{-{\alpha }^2{k}^2}\cos \alpha kz)=\sqrt{\beta }(\frac{1}{2}+\sum _{k=1}^{\mathrm{\infty }}{e}^{-{\beta }^2{k}^2}\cosh \beta kz)$$

　$\alpha \beta =\pi$および$\beta t<\pi$において
$$\begin{array}{rl}\alpha \sum _{k=1}^{\mathrm{\infty }}\frac{\sinh 2\alpha kt}{e^{2{\alpha }^{2}k}-1}+\beta \sum _{k=1}^{\mathrm{\infty }}\frac{\sin 2\beta kt}{e^{2{\beta }^{2}k}-1}& =\frac{1}{4}\alpha \coth \alpha t-\frac{1}{4}\beta \cot \beta t-\frac{1}{2}t\\ 2\sum _{k=1}^{\mathrm{\infty }}\frac{\cosh 2\alpha kt}{k(e^{2{\alpha }^{2}k}-1)}-2\sum _{k=1}^{\mathrm{\infty }}\frac{\cos 2\beta kt}{k(e^{2{\beta }^{2}k}-1)}& =\frac{1}{6}({\alpha }^2-{\beta }^2)+\log \frac{\sinh \alpha t}{\cos \beta t}-t^2\end{array}$$

　整関数$\phi (z)$が良い性質を満たすとき、$\alpha \beta =\pi$において
$$\begin{array}{rl}& \sum _{k=1}^{\mathrm{\infty }}\frac{\phi (\alpha kt)+\phi (-\alpha kt)}{k(e^{2{\alpha }^{2}k}-1)}+\sum _{k=1}^{\mathrm{\infty }}\frac{\phi (\alpha kt)}{k}-\sum _{k=1}^{\mathrm{\infty }}\frac{\phi (\beta kit)+\phi (-\beta kit)}{k(e^{2{\beta }^{2}k}-1)}-\sum _{k=1}^{\mathrm{\infty }}\frac{\phi (\beta kit)}{k}\\ =& (\frac{\pi i}{2}-\frac{{\alpha }^2}{6}+\frac{{\beta }^2}{6})\phi (0)+(-\frac{\alpha t}{2}+\frac{\beta it}{2}){\phi }^{\prime }(0)-\frac{t^2}{4}{\phi }^{″}(0)\end{array}$$

　$\alpha \beta ={\pi }^2$において
$$\begin{array}{rl}\alpha \sum _{k=1}^{\mathrm{\infty }}\frac{k}{e^{2\alpha k}-1}+\beta \sum _{k=1}^{\mathrm{\infty }}\frac{k}{e^{2\beta k}-1}& =\frac{\alpha +\beta }{24}-\frac{1}{4}\\ {\left(\frac{\alpha }{\beta }\right)}^{\frac{1}{4}}\prod _{k=1}^{\mathrm{\infty }}\frac{1-e^{-2\alpha k}}{1-e^{-2\beta k}}& =e^{\frac{\alpha -\beta }{12}}\end{array}$$

$$\sum _{k=1}^{\mathrm{\infty }}\frac{k}{e^{2\alpha k}-1}=\frac{1}{24}-\frac{1}{8\pi }$$

　$[0,h]$上の連続関数$\phi$に対し
$$\psi (t)={\int }_0^h\phi (x)e^{itx}dx$$
とおくと$m$を$h/\alpha$以下の最大の奇数としたとき、$\alpha \beta =\pi /2$および$h/\alpha >m\ge 1$において
$$\alpha \sum _{k=1}^m\chi (k)\phi (\alpha k)e^{i\alpha kt}=\frac{1}{2i}\sum _{k=1}^{\mathrm{\infty }}\chi (k)(\psi (t+\beta k)-\psi (t-\beta k))$$

　$\alpha \beta =\pi /4$において
$$e^{z^2/4}\sqrt{\alpha }\sum _{k=1}^{\mathrm{\infty }}\chi (k)e^{-{\alpha }^2{k}^2}\sin \alpha kz=\sqrt{\beta }\sum _{k=1}^{\mathrm{\infty }}\chi (k)e^{-{\beta }^2{k}^2}\sinh \beta kz$$

　$\alpha \beta =\pi$および$|t|<\beta /2$において
$$\alpha (\frac{1}{4\cos \alpha t}+\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\cos \alpha kt}{e^{{\alpha }^{2}k}-1})=\beta (\frac{1}{4}+\sum _{k=1}^{\mathrm{\infty }}\frac{\cosh 2\beta kt}{e^{{\beta }^{2}k}+e^{-{\beta }^{2}k}})$$

　$\alpha \beta =\pi /2$および$0<t<\pi /2\alpha$において
$$\alpha \sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\sin \alpha kt}{\cosh {\alpha }^{2}k}=\beta \sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\sinh \beta kt}{\cosh {\beta }^{2}k}$$

　$\alpha \beta ={\pi }^2$および$n\ge 2$において
$${\alpha }^n\sum _{k=1}^{\mathrm{\infty }}\frac{k^{2n-1}}{e^{2\alpha k}-1}-(-\beta {)}^n\sum _{k=1}^{\mathrm{\infty }}\frac{k^{2n-1}}{e^{2\beta k}-1}=({\alpha }^n-(-\beta {)}^n)\frac{B_{2n}}{4n}$$

$$\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}\frac{k^5}{e^{2\pi k}-1}& =\frac{1}{504}\\ \sum _{k=1}^{\mathrm{\infty }}\frac{k^9}{e^{2\pi k}-1}& =\frac{1}{264}\\ \sum _{k=1}^{\mathrm{\infty }}\frac{k^{13}}{e^{2\pi k}-1}& =\frac{1}{24}\\ \sum _{k=1}^{\mathrm{\infty }}\frac{k^{4n+1}}{e^{2\pi k}-1}& =\frac{B_{4n+2}}{4(2n+1)}\end{array}$$

　$\alpha \beta ={\pi }^2$において
$${\alpha }^n\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{k^{2n-1}}{\cosh \left(\alpha k/2\right)}+(-\beta {)}^n\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{k^{2n-1}}{\cosh \left(\beta k/2\right)}=0$$

　$\alpha \beta ={\pi }^2/4$において
$$\begin{array}{rl}& 2\sum _{k=1}^{\mathrm{\infty }}\chi (k)\arctan e^{-\alpha k}+2\sum _{k=1}^{\mathrm{\infty }}\chi (k)\arctan e^{-\beta k}\\ =& \sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\mathrm{sech}\alpha k}{k}+\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\mathrm{sech}\alpha k}{k}\\ =& \frac{\pi }{4}\end{array}$$

$$\sum _{k=1}^{\mathrm{\infty }}\chi (k)\arctan e^{-\pi k/2}=\frac{\pi }{16}$$

　$\alpha \beta ={\pi }^2$において
$$\begin{array}{rl}& {\alpha }^{-n}(\frac{1}{2}\zeta (2n+1)+\sum _{k=1}^{\mathrm{\infty }}\frac{k^{-2n-1}}{e^{2\alpha k}-1})\\ =& (-\beta {)}^{-n}(\frac{1}{2}\zeta (2n+1)+\sum _{k=1}^{\mathrm{\infty }}\frac{k^{-2n-1}}{e^{2\beta k}-1})\\ & -2^{2n}\sum _{k=0}^{n+1}(-1{)}^k\frac{B_{2k}}{(2k)!}\frac{B_{2n+2-2k}}{(2n+2-2k)!}{\alpha }^{n+1-k}{\beta }^k\end{array}$$

　$\alpha \beta ={\pi }^2/4$において
$$\begin{array}{rl}& {\alpha }^{-n}\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\mathrm{sech}\alpha k}{k^{2n+1}}+(-\beta {)}^{-n}\sum _{k=1}^{\mathrm{\infty }}\chi (k)\frac{\mathrm{sech}\beta k}{k^{2n+1}}\\ =& \frac{\pi }{4}\sum _{k=0}^n(-1{)}^k\frac{E_{2k}}{(2k)!}\frac{E_{2n-2k}}{(2n-2k)!}{\alpha }^{n-k}{\beta }^k\end{array}$$

　$\chi$を何らかの原始的ディリクレ指標とし
$$L(s)=\sum _{n=1}^{\mathrm{\infty }}\frac{\chi (n)}{n^s}$$
とおくと$\alpha \beta ={\pi }^2$において
$$\begin{array}{rl}& {\alpha }^{-n+\frac{1}{2}}(\frac{1}{2}L(2n)+\sum _{k=1}^{\mathrm{\infty }}\frac{\chi (k)}{k^{2n}(e^{\alpha k}-1)})\\ =& (-1{)}^n\frac{{\beta }^{-n+\frac{1}{2}}}{2^{2n+1}}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2n}\cosh \beta k}\\ & +\frac{1}{4}\sum _{k=0}^n\frac{(-1{)}^k}{2^{2k}}\frac{E_{2k}}{(2k)!}\frac{B_{2n-2k}}{(2n-2k)!}{\alpha }^{n-k}{\beta }^{k+\frac{1}{2}}\end{array}$$

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin ax\cot xdx& =2a^2\sum _{n=1}^{\mathrm{\infty }}\frac{n}{a^4+4n^4}\\ {\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin ax\coth xdx& =\frac{1}{2a}+2a\sum _{n=1}^{\mathrm{\infty }}\frac{1}{a^2+(a+2n{)}^2}\\ {\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin ax(\cot x+\coth x)dx& =\frac{\pi }{2}\frac{\sinh \pi a}{\cosh \pi a-\cos \pi a}\end{array}$$

$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}\frac{x\sin 2ax}{e^{x^2}-1}dx& =\frac{\sqrt{\pi }a}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{e^{-a^2/n}}{n\sqrt{n}}\\ & =\frac{\pi }{2}+\pi \sum _{n=1}^{\mathrm{\infty }}{e}^{-2a\sqrt{n\pi }}\cos \left(2a\sqrt{n\pi }\right)\\ {\int }_0^{\mathrm{\infty }}\frac{x\sin 2ax}{e^{x^2}+e^{-x^2}}dx& =\frac{\sqrt{\pi }a}{2}\sum _{n=1}^{\mathrm{\infty }}\chi (n)\frac{e^{-a^2/n}}{n\sqrt{n}}\\ & =\frac{\pi }{2}\sum _{n=1}^{\mathrm{\infty }}\chi (n)e^{-a\sqrt{n\pi }}\sin \left(a\sqrt{n\pi }\right)\end{array}$$

$$\varphi (t)={\int }_0^{\mathrm{\infty }}\frac{\cos tx}{e^{2\pi \sqrt{x}}-1}dx$$
とおくと
$${\int }_0^{\mathrm{\infty }}\frac{\sin tx}{e^{2\pi \sqrt{x}}-1}dx=\varphi (t)-\frac{1}{2t}+\varphi \left(\frac{{\pi }^2}{t}\right)\sqrt{\frac{2{\pi }^3}{t^3}}$$

　$\psi (t)$を
$$\frac{1}{4\pi }+2\sum _{n=1}^{\mathrm{\infty }}\frac{n\cos t{n}^2}{e^{2\pi n}-1}=\varphi (t)+\psi (t)$$
によって定めると
$$\begin{array}{rl}{\int }_0^{\mathrm{\infty }}{e}^{-2a^{2}t}\psi (t)dt& =\frac{\pi }{e^{4\pi a}-2e^{2\pi a}\cos 2\pi a+1}\\ t\sqrt{\frac{t}{2}}\psi (\pi t)& =\sum _{r=1}^{\mathrm{\infty }}(r{e}^{-2\pi r^2/t}+\sum _{j=1}^{\mathrm{\infty }}{e}^{-2\pi r(r+1)/t}((2r+j)\cos \frac{\pi j(2r+j)}{t}+\sin \frac{\pi j(2r+j)}{t}))\end{array}$$

　$t=a/b>0$を既約分数とする。このとき$a,b$が共に奇数であれば
$$\varphi \left(\frac{\pi a}{b}\right)=\frac{1}{4}\sum _{r=1}^b(b-2r)\cos \left(\frac{r^2\pi a}{b}\right)-\frac{b}{4a}\sqrt{\frac{b}{a}}\sum _{r=1}^a(a-2r)\sin (\frac{\pi }{4}+\frac{r^{2}b\pi }{a})$$
が成り立ち、また$a,b$の一方が偶数であれば
$$\begin{array}{rl}\varphi \left(\frac{\pi a}{b}\right)=\frac{1}{4\pi a}\sum _{r=1}^b\sin \left(\frac{r^2\pi a}{b}\right)-\frac{1}{2}& \sum _{r=1}^{b}r(1-\frac{r}{b})\cos \left(\frac{r^2\pi a}{b}\right)\\ +\frac{b}{2a}\sqrt{\frac{b}{a}}& \sum _{r=1}^{b}r(1-\frac{r}{a})\sin (\frac{\pi }{4}+\frac{r^2\pi b}{a})\end{array}$$
が成り立つ。例えば
$$\begin{array}{c}\varphi (0)=\frac{1}{12},\varphi \left(\frac{\pi }{2}\right)=\frac{1}{4\pi },\varphi (\pi )=\frac{2-\sqrt{2}}{8},\varphi (2\pi )=\frac{1}{16},\varphi (\mathrm{\infty })=0\\ \\ \varphi \left(\frac{\pi }{5}\right)=\frac{6+\sqrt{5}}{4}-\frac{5\sqrt{10}}{8},\varphi \left(\frac{2\pi }{5}\right)=\frac{8-3\sqrt{5}}{16},\varphi \left(\frac{2\pi }{3}\right)=\frac{1}{3}-\sqrt{3}(\frac{3}{16}-\frac{1}{8\pi })\end{array}$$
となる。

　$a>0$に対し
$${\int }_0^{\mathrm{\infty }}{e}^{-2a^{2}t}f(t)dt=\pi e^{-4ap}$$
が成り立つとき
$$f(t)=\frac{p\sqrt{2\pi }}{t\sqrt{t}}{e}^{2p^2/t}$$

$$\frac{1}{p}{\int }_0^1{x}^{a-1}(1-x^{p/(1-nx)})dx=\sum _{k=0}^{\mathrm{\infty }}\frac{n^k(a+k{)}^{k-1}}{(p+a+k{)}^{k+1}}$$

$$\begin{array}{rlrl}{\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-x}\sin xdx& =2^{-\frac{s}{2}}\mathrm{\Gamma }(s)\sin \frac{\pi s}{4}& & (\mathrm{Re}(s)>-1)\\ {\int }_0^{\mathrm{\infty }}{x}^{s-1}{e}^{-x}\cos xdx& =2^{-\frac{s}{2}}\mathrm{\Gamma }(s)\cos \frac{\pi s}{4}& & (\mathrm{Re}(s)>0)\end{array}$$