ๆฌกใฎ็ญๅผใ่จผๆใใใ
$\rm{THEOREM}$
${\large 1}.\qquad\BA\D
\sum_{n=1}^\infty\frac{\tanh\pi n}{n(n^2-z^2)}+\sum_{n=0}^\infty\frac{1}{\L(n+\frac{1}{2}\R)\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)}=\frac{\pi}{2z^2}\L(1-\frac{\tanh\pi z}{\tan\pi z}\R)
\EA$
${\large 2}.\qquad\BA\D
\sum_{n=1}^\infty\frac{1}{(n^2-z^2)\cosh\pi n}+\sum_{n=0}^\infty \frac{(-1)^n}{\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)}=\frac{1}{2z^2}\L(1-\frac{\pi z}{\tan\pi z\cosh\pi z}\R)
\EA$
${\large 3}.\qquad\BA\D
\sum_{n=1}^\infty\frac{n}{(n^4-z^4)\tanh\pi n}=\frac{\pi}{4z^2}\L(\frac{1}{\pi^2z^2}-\frac{1}{\tan\pi z\tanh\pi z}\R)
\EA$
${\large 4}.\qquad\BA\D
&\sum_{n=1}^\infty\frac{1}{(n^2-z^2)\cosh^2\pi n}+\sum_{n=0}^\infty\L(\frac{1}{\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh^2\pi\L(n+\frac{1}{2}\R)}+\frac{2n+1}{\pi\L(\L(n+\frac{1}{2}\R)^2+z^2\R)^2\tanh\pi\L(n+\frac{1}{2}\R)}\R)\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\frac{\pi\tanh\pi z}{2z}+\frac{1}{2z^2}-\frac{\pi}{2z\tan\pi z\cosh^2\pi z}
\EA$
$\qquad\BA\D \sum_{n=1}^\infty \frac{\tanh\pi n}{n(n^2-z^2)} &=\sum_{n=1}^\infty \frac{2}{n(n^2-z^2)}\int_0^\infty \frac{\sin2\pi nx}{\sinh\pi x}dx\\ &=\sum_{n=1}^\infty \frac{2}{n(n^2-z^2)}\sum_{m=0}^\infty \int_m^{m+1} \frac{\sin2\pi nx}{\sinh\pi x}dx\\ &=\sum_{m=0}^\infty\sum_{n=1}^\infty \frac{2}{n(n^2-z^2)} \int_0^1 \frac{\sin2\pi n(x+m)}{\sinh\pi (x+m)}dx\\ &=\sum_{m=0}^\infty\sum_{n=1}^\infty \frac{2}{n(n^2-z^2)} \int_0^1 \frac{\sin2\pi nx}{\sinh\pi (x+m)}dx\\ &=2\sum_{m=0}^\infty\int_0^1\frac{2}{e^{\pi(x+m)}-e^{-\pi(x+m)}}\sum_{n=1}^\infty\frac{\sin2\pi nx}{n(n^2-z^2)}dx\\ &=4\sum_{m,n=0}^\infty \int_0^1 e^{-\pi(x+m)(2n+1)}\frac{\pi}{2z^2}\L(2x-\frac{\sin2\pi zx}{\tan\pi z}-1+\cos2\pi zx\R)dx\\ &=\frac{2\pi}{z^2}\sum_{n=0}^\infty\frac{1}{1-e^{-\pi(2n+1)}}\int_0^1 e^{-\pi(2n+1)x}\L(2x-\frac{\sin2\pi zx}{\tan\pi z}-1+\cos2\pi zx\R)dx\\ &=\frac{2\pi^3}{z^2}\sum_{n=0}^\infty\frac{1}{1-e^{-\pi(2n+1)}}\frac{1}{\pi^2(2n+1)^2\L(\pi^2(2n+1)^2+(2\pi z)^2\R)}\L(2(1-e^{-\pi(2n+1)})\L(4z^2+(2n+1)^2\L(1-\frac{\pi z}{\tan\pi z}\R)\R)-4\pi z^2(2n+1)(1+e^{-\pi(2n+1)})\R)\\ &=-\sum_{n=0}^\infty \frac{1}{\L(n+\frac{1}{2}\R)\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)}+\frac{16}{\pi}\sum_{n=0}^\infty\frac{1}{(2n+1)^2((2n+1)^2+(2z)^2)}+\frac{4}{\pi z^2}\L(1-\frac{\pi z}{\tan\pi z}\R)\sum_{n=0}^\infty\frac{1}{(2n+1)^2+(2z)^2}\\ &=-\sum_{n=0}^\infty \frac{1}{\L(n+\frac{1}{2}\R)\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)} +\frac{16}{\pi}\frac{\pi(\pi z-\tanh\pi z)}{32z^3} +\frac{4}{\pi z^2}\L(1-\frac{\pi z}{\tan\pi z}\R)\frac{\pi\tanh\pi z}{8z}\\ &=-\sum_{n=0}^\infty \frac{1}{\L(n+\frac{1}{2}\R)\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)} +\frac{\pi}{2z^2}\L(1-\frac{\tanh\pi z}{\tan\pi z}\R) \EA$
ใ$1$ใฎๅ ดๅใจใปใจใใฉๅใๆตใใชใฎใง็ฅ่จผๆใ็คบใใ
$\qquad\BA\D \sum_{n=1}^\infty \frac{1}{(n^2-z^2)\cosh\pi n} &=\sum_{n=1}^\infty \frac{2}{n^2-z^2}\int_0^\infty \frac{\cos2\pi nx}{\cosh\pi x}dx\\ &=\sum_{m=0}^\infty\sum_{n=1}^\infty \frac{2}{n^2-z^2}\int_0^1 \frac{\cos2\pi nx}{\cosh\pi(x+m)}dx\\ &=4\sum_{m=0}^\infty \int_0^1 \frac{e^{-\pi(x+m)}}{1+e^{-2\pi(x+m)}}\sum_{n=1}^\infty\frac{\cos2\pi nx}{n^2-z^2}dx\\ &=\frac{2}{z^2}\sum_{n=0}^\infty\frac{(-1)^n}{1-e^{-\pi(2n+1)}}\int_0^1 e^{-\pi(2n+1)x}\L(1-\pi z\sin2\pi zx-\frac{\pi z\cos2\pi zx}{\tan\pi z}\R)dx\\ &=\frac{2}{z^2}\sum_{n=0}^\infty(-1)^n\L(\frac{1}{\pi(2n+1)}-\frac{1}{\pi}\frac{2n+1}{(2n+1)^2+(2z)^2}\frac{\pi z}{\tan\pi z}-\frac{2z^2}{((2n+1)^2+(2z)^2)\tanh\pi\L(n+\frac{1}{2}\R)}\R)\\ &=\frac{1}{2z^2}\L(1-\frac{\pi z}{\tan\pi z\cosh\pi z}\R)-\sum_{n=0}^\infty \frac{(-1)^n}{\L(\L(n+\frac{1}{2}\R)^2+z^2\R)\tanh\pi\L(n+\frac{1}{2}\R)} \EA$
ใๅ ใจใชใฃใ็ฉๅใฏ
$\qquad\BA\D \int_0^\infty \frac{\sin 2\pi nx}{e^{2\pi x}-1}dx=\frac{1}{4}\L(\frac{1}{\tanh\pi n}-\frac{1}{\pi n}\R) \EA$
ใงใใ๏ผใใใซๅฏพใใฆ$\D\frac{1}{n(n^2-z^2)}$ใใใใฆๅใใจใใ
ใๅ ใจใชใฃใ็ฉๅใฏ
$\qquad\BA\D \int_0^\infty \frac{x\cos 2\pi nx}{\sinh\pi x}dx=\frac{1}{4\cosh^2\pi n} \EA$
ใงใใ๏ผใใใซๅฏพใใฆ$\D\frac{1}{n^2-z^2}$ใใใใฆๅใใจใใ