Разложение мероморфных функций в ряды простейших дробей и в бесконечные произведения

$\frac{1}{e^{z}-e^{a}} = e^{-a}\left[-\frac{1}{2}+\frac{1}{z-a}+\sum_{n = 1}^{\infty } \frac{2(z-a)}{(z-a)^{2}+4 n^{2} \pi^{2}}\right]$;

$\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{(-1)^{n}}{z-n}$;

$\frac{1}{\operatorname {ch} 2 z-\cos 2 z} = \frac{1}{4 z^{2}}+\sum_{n = 1}^{\infty } \frac{n(-1)^{n}}{\operatorname {sh} \pi n} \frac{2 \pi z^{2}}{4 z^{4}+n^{4} \pi^{4}}$.

$\frac{\pi^{2}}{\cos^{2} \pi z} = \sum_{n = -\infty }^{\infty } \frac{1}{\left(z+\frac{1}{2}-n\right)^{2}}$;

$\frac{1}{\operatorname {ch} z} = \sum_{n = 1}^{\infty } \frac{(-1)^{n+1} \pi }{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}}$;

$\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}} = \sum_{n = 1}^{\infty } \frac{2 z}{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}}$;

$\frac{e^{a z}}{e^{z}-1} = \frac{1}{z}+\sum_{n = 1}^{\infty } \frac{2 z \cdot \cos 2 \pi n a-4 \pi n \sin 2 \pi n a}{z^{2}+4 n^{2} \pi^{2}}, 0 < a < 1$.

$\frac{\sin a z}{\sin \pi z} = \frac{2}{\pi } \sum_{n = 1}^{\infty } \frac{(-1)^{n} n \sin n a}{z^{2}-n^{2}},-\pi < a < \pi ;$

$\frac{\operatorname {ch} a z}{\operatorname {sh} \pi z} = \frac{1}{\pi z}+\frac{1}{\pi } \sum_{n = 1}^{\infty }(-1)^{n} \frac{2 z \cos n a}{z^{2}+n^{2}},-\pi < a < \pi ;$

$\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{e^{i(\pi n+\alpha n-\alpha z)}}{z-n},-\pi < \alpha < \pi$;

$\sum_{n = -\infty }^{\infty }(-1)^{n} \frac{\sin \alpha (z-n)}{z-n} = 0,-\pi < \alpha < \pi$;

$\frac{\pi^{2}}{\sin^{2} \pi z} = \pi \alpha +\sum_{n = -\infty }^{\infty } \frac{\cos \alpha (z-n)}{(z-n)^{2}}, 0 \leqslant \alpha < \pi$.

$\frac{e^{(\alpha +\beta ) z}-1}{\left(e^{\alpha z}-1\right)\left(e^{\beta z}-1\right)} = \left(\frac{1}{\alpha }+\frac{1}{\beta }\right) \frac{1}{z}+\sum_{k = 1}^{\infty }\left(\frac{2 \alpha z}{\alpha^{2} z^{2}+4 k^{2} \pi^{2}}+\frac{2 \beta z}{\beta^{2} z^{2}+4 k^{2} \pi^{2}}\right)$, $\alpha \beta \neq 0, \alpha \neq \beta ;$

$\operatorname {ctg}(z-a)+\operatorname {ctg} a = z \sum_{n = -\infty }^{\infty } \frac{1}{(a-\pi n)(z-a+\pi n)}, \sin a \neq 0$;

$\frac{\sin a}{\cos z-\cos a} = \sum_{n = -\infty }^{\infty } \frac{2 a}{a^{2}-(z+2 \pi n)^{2}}, \sin a \neq 0$;

$\frac{\cos a}{\sin a-\sin z} = \sum_{n = -\infty }^{\infty } \frac{\pi -2 a}{(z-a+2 \pi n)(z+a+(2 n-1) \pi )}, \cos a \neq 0$.

$\operatorname {sh} z = z \prod_{n = 1}^{\infty }\left(1+\frac{z^{2}}{n^{2} \pi^{2}}\right)$;

$\cos z = \prod_{n = 1}^{\infty }\left(1-\frac{z^{2}}{(n-(1 / 2))^{2} \pi^{2}}\right)$;

$\operatorname {th} z = z \prod_{n = 1}^{\infty } \frac{\left[1+(z / \pi n)^{2}\right]}{\left[1+\left(\frac{z}{\pi (n-(1 / 2))}\right)^{2}\right]}$;

$e^{a z}-e^{b z} = (a-b) z e^{(1 / 2)(a+b) z} \prod_{n = 1}^{\infty }\left[1+\frac{(a-b)^{2} z^{2}}{4 n^{2} \pi^{2}}\right]$;

$e^{z}-e^{a} = (z-a) e^{(z+a) / 2} \prod_{n = 1}^{\infty }\left[1+\left(\frac{z-a}{2 \pi n}\right)^{2}\right]$;

$\operatorname {ch} z-\cos z = z^{2} \prod_{n = 1}^{\infty }\left(1+\frac{z^{4}}{4 n^{4} \pi^{4}}\right)$;

$\cos \pi z-\cos \pi a = -\frac{\pi^{2}}{2}\left(z^{2}-a^{2}\right) \prod_{n = 1}^{\infty }\left[1-\left(\frac{z+a}{2 n}\right)^{2}\right] \prod_{n = 1}^{\infty }\left[1-\left(\frac{z-a}{2 n}\right)^{2}\right]$;

$\cos \pi z = \pi (2 z+1) \prod_{n = 1}^{\infty }\left(1-\frac{z(z+1)}{n^{2}-(1 / 4)}\right)$;

$\frac{\cos z-\cos a}{1-\cos a} \prod_{n = -\infty }^{\infty }\left(1-\frac{z^{2}}{(2 \pi n+a)^{2}}\right)$;

$\sin (z-a)+\sin a = \frac{z(\pi +2 a-z)}{\pi +2 a} \prod_{n \neq 0}\left\{ 1+\frac{z(\pi +2 a-z)}{2 \pi n(\pi (2 n-1)-2 a)}\right\}$;

$\frac{\sin (z-a)}{\sin a-\sin z} = -\prod_{n = 0}^{\infty } \frac{1-\frac{z(z-2 a)}{\pi^{2}(2 n+1)^{2}-a^{2}}}{1-\frac{z(z+2 a)}{\pi^{2}(2 n+1)^{2}-a^{2}}}$;

$e^{z^{2}}+e^{2 z-1} = 2 e^{(1 / 2)\left(z^{2}+2 z-1\right)} \prod_{n = 1}^{\infty }\left(1+\frac{(z-1)^{4}}{\pi^{2}(2 n-1)^{2}}\right)$;

$e^{2 \operatorname {ch} z}-e^{2 \operatorname {sh} z} = 2 e^{e^{z}-z} \prod_{n = 1}^{\infty }\left(1+\frac{e^{-2 z}}{\pi^{2} n^{2}}\right)$.