# §22.12 Expansions in Other Trigonometric Series and Doubly-Infinite Partial Fractions: Eisenstein Series

With $t\in\mathbb{C}$ and

 22.12.1 $\tau=i{K^{\prime}}\left(k\right)/K\left(k\right),$
 22.12.2 $2Kk\operatorname{sn}\left(2Kt,k\right)=\sum_{n=-\infty}^{\infty}\frac{\pi}{% \sin\left(\pi(t-(n+\frac{1}{2})\tau)\right)}=\sum_{n=-\infty}^{\infty}\left(% \sum_{m=-\infty}^{\infty}\frac{(-1)^{m}}{t-m-(n+\frac{1}{2})\tau}\right),$
 22.12.3 $2iKk\operatorname{cn}\left(2Kt,k\right)=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n% }\pi}{\sin\left(\pi(t-(n+\frac{1}{2})\tau)\right)}=\sum_{n=-\infty}^{\infty}% \left(\sum_{m=-\infty}^{\infty}\frac{(-1)^{m+n}}{t-m-(n+\frac{1}{2})\tau}% \right),$
 22.12.4 $2iK\operatorname{dn}\left(2Kt,k\right)=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n% }\frac{\pi}{\tan\left(\pi(t-(n+\frac{1}{2})\tau)\right)}=\lim_{N\to\infty}\sum% _{n=-N}^{N}(-1)^{n}\left(\lim_{M\to\infty}\sum_{m=-M}^{M}\frac{1}{t-m-(n+\frac% {1}{2})\tau}\right).$

The double sums in (22.12.2)–(22.12.4) are convergent but not absolutely convergent, hence the order of the summations is important. Compare §20.5(iii).

 22.12.5 $\displaystyle 2Kk\operatorname{cd}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{\pi}{\sin\left(\pi(t+\frac{1}{2}-% (n+\frac{1}{2})\tau)\right)}=\sum_{n=-\infty}^{\infty}\left(\sum_{m=-\infty}^{% \infty}\frac{(-1)^{m}}{t+\frac{1}{2}-m-(n+\frac{1}{2})\tau}\right),$ 22.12.6 $\displaystyle-2iKkk^{\prime}\operatorname{sd}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}\pi}{\sin\left(\pi(t+% \frac{1}{2}-(n+\frac{1}{2})\tau)\right)}=\sum_{n=-\infty}^{\infty}\left(\sum_{% m=-\infty}^{\infty}\frac{(-1)^{m+n}}{t+\frac{1}{2}-m-(n+\frac{1}{2})\tau}% \right),$ 22.12.7 $\displaystyle 2iKk^{\prime}\operatorname{nd}\left(2Kt,k\right)$ $\displaystyle=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n}\frac{\pi}{\tan\left(\pi% (t+\frac{1}{2}-(n+\frac{1}{2})\tau)\right)}=\lim_{N\to\infty}\sum_{n=-N}^{N}(-% 1)^{n}\lim_{M\to\infty}\left(\sum_{m=-M}^{M}\frac{1}{t+\frac{1}{2}-m-(n+\frac{% 1}{2})\tau}\right),$ 22.12.8 $\displaystyle 2K\operatorname{dc}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{\pi}{\sin\left(\pi(t+\frac{1}{2}-% n\tau)\right)}=\sum_{n=-\infty}^{\infty}\left(\sum_{m=-\infty}^{\infty}\frac{(% -1)^{m}}{t+\frac{1}{2}-m-n\tau}\right),$ 22.12.9 $\displaystyle 2Kk^{\prime}\operatorname{nc}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}\pi}{\sin\left(\pi(t+% \frac{1}{2}-n\tau)\right)}=\sum_{n=-\infty}^{\infty}\left(\sum_{m=-\infty}^{% \infty}\frac{(-1)^{m+n}}{t+\frac{1}{2}-m-n\tau}\right),$ 22.12.10 $\displaystyle-2Kk^{\prime}\operatorname{sc}\left(2Kt,k\right)$ $\displaystyle=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n}\frac{\pi}{\tan\left(\pi% (t+\frac{1}{2}-n\tau)\right)}=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n}\left(% \lim_{M\to\infty}\sum_{m=-M}^{M}\frac{1}{t+\frac{1}{2}-m-n\tau}\right),$ 22.12.11 $\displaystyle 2K\operatorname{ns}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{\pi}{\sin\left(\pi(t-n\tau)\right% )}=\sum_{n=-\infty}^{\infty}\left(\sum_{m=-\infty}^{\infty}\frac{(-1)^{m}}{t-m% -n\tau}\right),$ 22.12.12 $\displaystyle 2K\operatorname{ds}\left(2Kt,k\right)$ $\displaystyle=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}\pi}{\sin\left(\pi(t-n% \tau)\right)}=\sum_{n=-\infty}^{\infty}\left(\sum_{m=-\infty}^{\infty}\frac{(-% 1)^{m+n}}{t-m-n\tau}\right),$ 22.12.13 $\displaystyle 2K\operatorname{cs}\left(2Kt,k\right)$ $\displaystyle=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n}\frac{\pi}{\tan\left(\pi% (t-n\tau)\right)}=\lim_{N\to\infty}\sum_{n=-N}^{N}(-1)^{n}\left(\lim_{M\to% \infty}\sum_{m=-M}^{M}\frac{1}{t-m-n\tau}\right).$