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

With $t\in\Complex$ and

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

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