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

\begin{equation}
\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}}
\end{equation}

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...
The Fourier transform of $1_{y &gt; 0}$ is $PV.(\frac{1}{2i \pi y})+\frac{1}{2}\delta(y)$, thus the Fourier transform of $y \ 1_{y &gt; 0}$  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 $\hat{h}(y) = e^{-\pi y^2 x},h(y) = \frac{1}{\sqrt{x}}e^{-\pi y^2 /x}$ 

$$\sum_{n=1}^\infty n e^{- \pi n^2 x} = \langle T, \hat{h} \rangle=\langle \hat{T},h \rangle 
 = \sum_{n=-\infty}^\infty \frac{i  n}{2 x^{3/2}}e^{-\pi n^2 /x}-\frac{1}{4\pi^2} \int_{-\infty}^\infty \log |y-n| h''(y)dy$$
$$\sum _{k=1}^{\infty } k e^{\pi  i k^2 (-x)}=\sum _{n=1}^{\infty } -\frac{i \left(x+\sqrt[4]{-1} \pi  n \sqrt{x} e^{\frac{i \pi  n^2}{x}} \text{erf}\left(\frac{\sqrt[4]{-1} \sqrt{\pi } n}{\sqrt{x}}\right)\right)}{\pi  x^2}-\frac{i}{2 \pi  x}$$