# Jacobi Theta Functions?

For the Jacobi theta function $\vartheta_3(z|\tau)$ there exists an equality (by Whittaker & Watson)

$$\vartheta_3(z|\tau) = \sum_{n=-\infty}^{\infty} e^{n^2 \pi i \tau + 2 n i z} = \sqrt{\frac{1}{-i\tau}} \sum_{n=-\infty}^{\infty} e^{\frac{(z-n\pi)^2}{\pi i \tau}}$$

Is there something similar for this "modified version" of the Jacobi theta function?

$$\sum_{n=1}^{\infty} n\,e^{n^2 \pi i \tau}$$

I tried to use the relation shown above, but it results nothing.

Well, I got a relation for the $n^2$ case

$$\sum_{n=1}^{\infty} n^2\,e^{n^2 \pi i \tau}$$

I just 'broke' the sum into parts, $$\sum_{n=-\infty}^{-1} + \sum_{n=1}^{\infty}$$ remembering that $n=0$ contributes with nothing, than I took a derivative with respect to $\tau$ and set $z=0$. I got:

$$\sum_{n=1}^\infty n^2\,e^{n^2 \pi i \tau} = \frac{(-i\tau)^{-3/2}}{2\pi} \left(\frac{1}{2}+\sum_{n=1}^\infty e^{\frac{n^2\pi}{i\tau}}\right) - \pi (-i\tau)^{-5/2}\sum_{n=1}^\infty n^2\,e^{\frac{n^2\pi}{i\tau}}$$

It looks like equalities for $$\sum_{n=1}^\infty n^{2k}\,e^{n^2 \pi i \tau}, \quad k\in\mathbb{N}$$ are allways possible to get by 'recurrence'. The problem is with odd powers...

• I don't think so. After all this identity is a consequence of Poisson summation formula, for which it is essential to sum over all, and not just positive $n$. Sep 10, 2015 at 19:23

The Fourier transform of $1_{y > 0}$ is $PV.(\frac{1}{2i \pi y})+\frac{1}{2}\delta(y)$, thus the Fourier transform of $y \ 1_{y > 0}$ is $\frac{d}{dy}PV.(\frac{-1}{4\pi^2 y})+\frac{1}{4i \pi}\delta'(y)$

and of $T(y) = \sum_{n=1}^\infty n \delta(y-n)$ is $\hat{T}(y)=\sum_n \frac{d}{dy} PV.(\frac{-1}{4 \pi^2 (y-n)})+\frac{1}{4i\pi}\delta'(y-n)=\sum_n \frac{d^2}{d^2y} \frac{\log |y-n|}{-4\pi^2}+\frac{1}{4i\pi}\delta'(y-n)$.

With $\hat{h}(y) = e^{-\pi y^2 x},h(y) = \frac{1}{\sqrt{x}}e^{-\pi y^2 /x}$

$$\sum_{n=1}^\infty n e^{- \pi n^2 x} = \langle T, \hat{h} \rangle=\langle \hat{T},h \rangle = \sum_{n=-\infty}^\infty \frac{i n}{2 x^{3/2}}e^{-\pi n^2 /x}-\frac{1}{4\pi^2} \int_{-\infty}^\infty \log |y-n| h''(y)dy$$

• $$\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} e^{-i k^2 x}=\sum _{n=1}^{\infty } \frac{\pi ^{3/2} e^{\frac{i \pi ^2 n^2}{2 x}} \sqrt[4]{\frac{n^2}{x}} J_{-\frac{1}{4}}\left(\frac{n^2 \pi ^2}{2 x}\right)}{\sqrt[4]{i x}}+\frac{\Gamma \left(\frac{1}{4}\right)}{2 \sqrt[4]{i x}}$$ sorry but i do not see necesary using here fourier transform if you needed i could calculate sum as typo above Jul 14, 2017 at 10:08
• @capea ??? Why do you want to mention $\sum_{n \ge 1} n^{-1/2} e^{- n^2 x}$ ? And where do your formulas come from ? Jul 14, 2017 at 10:27

$$\sum _{k=1}^{\infty } k e^{\pi i k^2 (-x)}=\sum _{n=1}^{\infty } -\frac{i \left(x+\sqrt[4]{-1} \pi n \sqrt{x} e^{\frac{i \pi n^2}{x}} \text{erf}\left(\frac{\sqrt[4]{-1} \sqrt{\pi } n}{\sqrt{x}}\right)\right)}{\pi x^2}-\frac{i}{2 \pi x}$$

• Where is it supposed to come from ? What is the Fourier transform of the distribution $\sum_{n=1}^\infty n \delta(x-n)$ ? Jul 14, 2017 at 8:10
• Why you need a Fourier Transform? Jul 14, 2017 at 8:14
• Because $\langle T, \hat{h} \rangle = \langle \hat{T}, h \rangle$. Here we have $h(y) = e^{-\pi y^2 x}, \hat{h}(y) = \frac{1}{\sqrt{y}} e^{-\pi y^2/x}$, the Poisson summation formula is the case $T = \hat{T} = \sum_n \delta(y-n)$, here we need to apply it with $T = \sum_{n=1}^\infty n \delta(y-n)$. Jul 14, 2017 at 8:17
• See what I get in my answer Jul 14, 2017 at 8:49