Proving the identity $\sum_{n=-\infty}^\infty e^{-\pi n^2x}=x^{-1/2}\sum_{n=-\infty}^\infty e^{-\pi n^2/x}.$

Question

Can you help prove the functional equation: $$\sum_{n=-\infty}^\infty e^{-\pi n^2x}=x^{-1/2}\sum_{n=-\infty}^\infty e^{-\pi n^2/x}.$$

Specifically, I am looking for a solution using complex analysis, but I am interested in any solutions.

Thanks!

Answer 1

Define $\theta(t) = \sum_{n \in \mathbb{Z}} e^{-\pi i n^{2} t}$. The identity that you quote is the Jacobi theta functional identity \begin{align} \theta(t) = t^{-1/2} \theta(t^{-1}). \end{align} This identity can be used to prove the Riemann zeta functional identity. To prove it, first recall that the Fourier transform of an integrable function $f \colon \mathbb{R} \to \mathbb{C}$ is simply \begin{align} \tilde{f}(s) = \int_{\mathbb{R}} f(x) e^{2 \pi i x s} dx. \end{align} and (presupposing that $f$ is also uniformly continuous) the Poisson summation formula is the identity \begin{align} \sum_{n \in \mathbb{Z}} \tilde{f}(n) = \sum_{n \in \mathbb{Z}} f(n). \end{align} Now, recall the integral \begin{align} e^{-\pi s^{2}} = \int_{\mathbb{R}} e^{-\pi x^{2}} e^{2 \pi i x s} dx \end{align} Observe that \begin{align} \sum_{n \in \mathbb{Z}} e^{- \pi n^{2} s^{2}} = \sum_{n \in \mathbb{Z}} \int_{\mathbb{R}} e^{-\pi x^{2}} e^{2 \pi i x (ns)} dx = \sum_{n \in \mathbb{Z}} s^{-1} \int_{\mathbb{R}} e^{-\pi (x^{\prime}/s)^{2}} e^{2 \pi i n x^{\prime}} dx^{\prime}, \end{align} where we have changed variables, $x^{\prime} = x s$. Now use the fourier transform, \begin{align} \sum_{n \in \mathbb{Z}} s^{-1} \int_{\mathbb{R}} e^{-\pi (x^{\prime}/s)^{2}} e^{2 \pi i n x^{\prime}} dx^{\prime} = s^{-1} \sum_{n \in \mathbb{Z}} e^{- \pi (n / s)^{2}}. \end{align} Take $s = \sqrt{t}$ and conclude the result.

Answer 2

Hint: Apply the Poisson Summation Formula to a suitable function.

Answer 3

The following is not a proof but it gives some physical intuition why such a formula is true.

If at time $t=0$ a unit of heat is concentrated at $x=0$ and dissipates according to the heat equation $u_t=u_{xx}$ along the $x$-axis then the temperature $x\mapsto u(x,t)$ is a Gaussian becoming flatter and flatter as $t$ increases: $$u(x,t)={1\over\sqrt{4\pi t}}e^{-x^2/(4t)}\qquad(t>0)\ .$$ Now if at time $t=0$ we have such a unit of heat at each integer point $k$ then the resulting temperature will be $$u(x,t)={1\over\sqrt{4\pi t}}\sum_{k\in{\mathbb Z}}e^{-(x-k)^2/(4t)}\ .$$ In particular the temperature at $x=0$ will be $$U(t)={1\over\sqrt{4\pi t}}\sum_{k\in{\mathbb Z}}e^{-k^2/(4t)}\qquad(t>0)\ .\qquad {\rm (a)}$$ On the other hand the process considered here is periodic in $x$ with period $1$. So the temperature $u(x,t)$ must have a description of the form $$u(x,t)=\sum_{k\in{\mathbb Z}}a_k(t)e^{2\pi i k x}$$ for certain functions $a_k(t)$. Plugging this into the heat equation gives $a_k(t)=c_k \exp(-4\pi^2 k^2 t)$ for constants $c_k$, so that we now have $$u(x,t)=\sum_{k\in{\mathbb Z}}c_k \exp(-4\pi^2 k^2 t)e^{2\pi i k x}\ .$$ The $c_k$ have to be determined by the initial condition which is a delta-function at $x=0$. Here we have to cheat a little: We replace the delta-function by a rectangle of width $2\epsilon$ and area $1$. The computation gives $c_0=1$ and $$c_k={\sin(2\pi k\epsilon)\over 2\pi k\epsilon}\qquad(k\ne 0)$$ which tends to $1$ with $\epsilon\to0$. This means that "in the limit" we have $$u(x,t)=\sum_{k\in{\mathbb Z}} \exp(-4\pi^2 k^2 t)e^{2\pi i k x}\qquad(t>0)\ .$$ Putting $x=0$ here gives $$U(t)=\sum_{k\in{\mathbb Z}} \exp(-4\pi^2 k^2 t)\qquad(t>0)\ .\qquad{\rm (b)}$$ If you believe that (a) and (b) are the same thing then you have the stated formula with $4\pi t$ instead of $x$.

Answer 4

Biane, Pitman and Yor describe one classical analytic approach that admits a probabilistic interpretation: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.17.7395

(Not sure if the link is behind a paywall. Just google the title if so).

Answer 5

See the book Elliptic Functions by Chandrasekharan, K. at page 73 the section entitled

$\S \;8.\; The\;transformation\;formula\;connecting\;\theta_{3}(v,\tau)\;and\;\theta_{3}\left(v,-\frac{1}{\tau}\right)$

Answer 6

Answers have already been accepted but you wanted an approach based on complex analysis. You can derive the transformation properties of the theta functions directly from the residue theorem.

Let $C$ be the rectangle whose corners are $\pm(N+\frac12)\pm i$. We have from the residue theorem

$$\oint_C\frac{e^{-\pi tz^2}}{e^{2\pi iz}-1}dz=\sum_{n=-N}^{N}e^{-n^2\pi t}$$

I put $t$ as the variable in the series so as not to confuse it with $z=x+iy.$ The vertical parts of the contour integral decay as $N\to\infty$ so we will have

$$\sum_{n=-\infty}^{\infty}e^{-n^2\pi t}=\int^{\infty-i}_{-\infty-i}\frac{e^{-\pi tz^2}}{e^{2\pi iz}-1}dz+\int^{\infty+i}_{-\infty+i}\frac{e^{-\pi tz^2}}{1-e^{2\pi iz}}dz.$$

After some algebra, expanding the denominators as geometric series, I claim that the sum of these two integrals can be written as

$$e^{\pi t}\int^{\infty}_{-\infty}e^{-\pi tx^2}\sum_{n=-\infty}^{\infty}e^{2n\pi}\cos2\pi(n+t)xdx.$$

This class of integral can also be evaluated by the residue theorem. Since this is not the point of this answer, I will skip this working and just state that it can be shown without much labour that for all $t>0,$

$$\int^\infty_{-\infty}e^{-\pi tx^2}\cos2\pi(n+t)xdx=\frac{e^{-\pi t-2n\pi-n^2\pi/t}}{\sqrt{t}}.$$

After cancellation, it follows immediately that

$$\sum_{n=-\infty}^{\infty}e^{-n^2\pi t}=\frac{1}{\sqrt{t}}\sum_{n=-\infty}^{\infty}e^{-n^2\pi /t}$$

for all $t>0$, and this completes the proof, using only basic complex analysis, as required.

Even though this question was asked 9 years ago I hope this is illustrating to someone.