# §23.8 Trigonometric Series and Products

## §23.8(i) Fourier Series

If $q=e^{i\pi\omega_{3}/\omega_{1}}$, $\Im\left(z/\omega_{1}\right)<2\Im\left(\omega_{3}/\omega_{1}\right)$, and $z\notin\mathbb{L}$, then

 23.8.1 $\displaystyle\wp\left(z\right)+\frac{\eta_{1}}{\omega_{1}}-\frac{\pi^{2}}{4% \omega_{1}^{2}}{\csc}^{2}\left(\frac{\pi z}{2\omega_{1}}\right)$ $\displaystyle=-\frac{2\pi^{2}}{\omega_{1}^{2}}\sum_{n=1}^{\infty}\frac{nq^{2n}% }{1-q^{2n}}\cos\left(\frac{n\pi z}{\omega_{1}}\right),$ 23.8.2 $\displaystyle\zeta\left(z\right)-\frac{\eta_{1}z}{\omega_{1}}-\frac{\pi}{2% \omega_{1}}\cot\left(\frac{\pi z}{2\omega_{1}}\right)$ $\displaystyle=\frac{2\pi}{\omega_{1}}\sum_{n=1}^{\infty}\frac{q^{2n}}{1-q^{2n}% }\sin\left(\frac{n\pi z}{\omega_{1}}\right).$

## §23.8(ii) Series of Cosecants and Cotangents

When $z\notin\mathbb{L}$,

 23.8.3 $\wp\left(z\right)=-\frac{\eta_{1}}{\omega_{1}}+\frac{\pi^{2}}{4\omega_{1}^{2}}% \sum_{n=-\infty}^{\infty}{\csc}^{2}\left(\frac{\pi(z+2n\omega_{3})}{2\omega_{1% }}\right),$
 23.8.4 $\zeta\left(z\right)=\frac{\eta_{1}z}{\omega_{1}}+\frac{\pi}{2\omega_{1}}\sum_{% n=-\infty}^{\infty}\cot\left(\frac{\pi(z+2n\omega_{3})}{2\omega_{1}}\right),$

where in (23.8.4) the terms in $n$ and $-n$ are to be bracketed together (the Eisenstein convention or principal value: see Weil (1999, p. 6) or Walker (1996, p. 3)).

 23.8.5 $\eta_{1}=\frac{\pi^{2}}{2\omega_{1}}\left(\frac{1}{6}+\sum_{n=1}^{\infty}{\csc% }^{2}\left(\frac{n\pi\omega_{3}}{\omega_{1}}\right)\right),$

with similar results for $\eta_{2}$ and $\eta_{3}$ obtainable by use of (23.2.14).

## §23.8(iii) Infinite Products

 23.8.6 $\sigma\left(z\right)=\frac{2\omega_{1}}{\pi}\exp\left(\frac{\eta_{1}z^{2}}{2% \omega_{1}}\right)\sin\left(\frac{\pi z}{2\omega_{1}}\right)\*\prod_{n=1}^{% \infty}\frac{1-2q^{2n}\cos\left(\pi z/\omega_{1}\right)+q^{4n}}{(1-q^{2n})^{2}},$
 23.8.7 $\sigma\left(z\right)=\frac{2\omega_{1}}{\pi}\exp\left(\frac{\eta_{1}z^{2}}{2% \omega_{1}}\right)\sin\left(\frac{\pi z}{2\omega_{1}}\right)\prod_{n=1}^{% \infty}\frac{\sin\left(\pi(2n\omega_{3}+z)/(2\omega_{1})\right)\sin\left(\pi(2% n\omega_{3}-z)/(2\omega_{1})\right)}{{\sin}^{2}\left(\pi n\omega_{3}/\omega_{1% }\right)}.$