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

  • 3
    $\begingroup$ 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 $n$. $\endgroup$ Sep 10, 2015 at 19:23

2 Answers 2


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

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

  • $\begingroup$ $$\sum _{n=1}^{\infty } \frac{1}{\sqrt{k}} e^{-i k^2 x}=\sum _{n=1}^{\infty } \frac{\pi ^{3/2} e^{\frac{i \pi ^2 n^2}{2 x}} \sqrt[4]{\frac{n^2}{x}} J_{-\frac{1}{4}}\left(\frac{n^2 \pi ^2}{2 x}\right)}{\sqrt[4]{i x}}+\frac{\Gamma \left(\frac{1}{4}\right)}{2 \sqrt[4]{i x}}$$ sorry but i do not see necesary using here fourier transform if you needed i could calculate sum as typo above $\endgroup$
    – capea
    Jul 14, 2017 at 10:08
  • $\begingroup$ @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 ? $\endgroup$
    – reuns
    Jul 14, 2017 at 10:27

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

  • $\begingroup$ Where is it supposed to come from ? What is the Fourier transform of the distribution $\sum_{n=1}^\infty n \delta(x-n)$ ? $\endgroup$
    – reuns
    Jul 14, 2017 at 8:10
  • $\begingroup$ Why you need a Fourier Transform? $\endgroup$
    – capea
    Jul 14, 2017 at 8:14
  • $\begingroup$ 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 $T = \sum_{n=1}^\infty n \delta(y-n)$. $\endgroup$
    – reuns
    Jul 14, 2017 at 8:17
  • $\begingroup$ See what I get in my answer $\endgroup$
    – reuns
    Jul 14, 2017 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.