𝐭𝐚𝐧𝐡𝝅𝒏, 𝐜𝐬𝐜𝐡𝝅𝒏 を含む級数

次の等式を証明する。

$1$の証明

$\begin{array}{rl}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\tanh \pi n}{n(n^2-z^2)}}& =\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n(n^2-z^2)}{\int }_0^{\mathrm{\infty }}\frac{\sin 2\pi nx}{\sinh \pi x}dx\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n(n^2-z^2)}\sum _{m=0}^{\mathrm{\infty }}{\int }_m^{m+1}\frac{\sin 2\pi nx}{\sinh \pi x}dx\\ & =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n(n^2-z^2)}{\int }_0^1\frac{\sin 2\pi n(x+m)}{\sinh \pi (x+m)}dx\\ & =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n(n^2-z^2)}{\int }_0^1\frac{\sin 2\pi nx}{\sinh \pi (x+m)}dx\\ & =2\sum _{m=0}^{\mathrm{\infty }}{\int }_0^1\frac{2}{e^{\pi (x+m)}-e^{-\pi (x+m)}}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin 2\pi nx}{n(n^2-z^2)}dx\\ & =4\sum _{m,n=0}^{\mathrm{\infty }}{\int }_0^1{e}^{-\pi (x+m)(2n+1)}\frac{\pi }{2z^2}(2x-\frac{\sin 2\pi zx}{\tan \pi z}-1+\cos 2\pi zx)dx\\ & =\frac{2\pi }{z^2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{1-e^{-\pi (2n+1)}}{\int }_0^1{e}^{-\pi (2n+1)x}(2x-\frac{\sin 2\pi zx}{\tan \pi z}-1+\cos 2\pi zx)dx\\ & =\frac{2{\pi }^3}{z^2}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{1-e^{-\pi (2n+1)}}\frac{1}{{\pi }^2(2n+1{)}^2({\pi }^2(2n+1{)}^2+(2\pi z{)}^2)}(2(1-e^{-\pi (2n+1)})(4z^2+(2n+1{)}^2(1-\frac{\pi z}{\tan \pi z}))-4\pi z^2(2n+1)(1+e^{-\pi (2n+1)}))\\ & =-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{2})({(n+\frac{1}{2})}^2+z^2)\tanh \pi (n+\frac{1}{2})}+\frac{16}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2((2n+1{)}^2+(2z{)}^2)}+\frac{4}{\pi z^2}(1-\frac{\pi z}{\tan \pi z})\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2+(2z{)}^2}\\ & =-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{2})({(n+\frac{1}{2})}^2+z^2)\tanh \pi (n+\frac{1}{2})}+\frac{16}{\pi }\frac{\pi (\pi z-\tanh \pi z)}{32z^3}+\frac{4}{\pi z^2}(1-\frac{\pi z}{\tan \pi z})\frac{\pi \tanh \pi z}{8z}\\ & =-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+\frac{1}{2})({(n+\frac{1}{2})}^2+z^2)\tanh \pi (n+\frac{1}{2})}+\frac{\pi }{2z^2}(1-\frac{\tanh \pi z}{\tan \pi z})\end{array}$

$2$の証明

　$1$の場合とほとんど同じ流れなので略証明を示す。

$\begin{array}{rl}{\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n^2-z^2)\cosh \pi n}}& =\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^2-z^2}{\int }_0^{\mathrm{\infty }}\frac{\cos 2\pi nx}{\cosh \pi x}dx\\ & =\sum _{m=0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n^2-z^2}{\int }_0^1\frac{\cos 2\pi nx}{\cosh \pi (x+m)}dx\\ & =4\sum _{m=0}^{\mathrm{\infty }}{\int }_0^1\frac{e^{-\pi (x+m)}}{1+e^{-2\pi (x+m)}}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos 2\pi nx}{n^2-z^2}dx\\ & =\frac{2}{z^2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{1-e^{-\pi (2n+1)}}{\int }_0^1{e}^{-\pi (2n+1)x}(1-\pi z\sin 2\pi zx-\frac{\pi z\cos 2\pi zx}{\tan \pi z})dx\\ & =\frac{2}{z^2}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(\frac{1}{\pi (2n+1)}-\frac{1}{\pi }\frac{2n+1}{(2n+1{)}^2+(2z{)}^2}\frac{\pi z}{\tan \pi z}-\frac{2z^2}{((2n+1{)}^2+(2z{)}^2)\tanh \pi (n+\frac{1}{2})})\\ & =\frac{1}{2z^2}(1-\frac{\pi z}{\tan \pi z\cosh \pi z})-\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{({(n+\frac{1}{2})}^2+z^2)\tanh \pi (n+\frac{1}{2})}\end{array}$

$3$の証明

　元となった積分は

$\begin{array}{r}{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{\sin 2\pi nx}{e^{2\pi x}-1}dx=\frac{1}{4}(\frac{1}{\tanh \pi n}-\frac{1}{\pi n})}\end{array}$

であり，これに対して${\displaystyle \frac{1}{n(n^2-z^2)}}$をかけて和をとる。

$4$の証明

　元となった積分は

$\begin{array}{r}{\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x\cos 2\pi nx}{\sinh \pi x}dx=\frac{1}{4{\cosh }^2\pi n}}\end{array}$

であり，これに対して${\displaystyle \frac{1}{n^2-z^2}}$をかけて和をとる。