I have a sum of the form:
$$\sum_{n,m=-N}^N e^{-\alpha (n-m)^2}$$
where $\alpha > 0$ is some constant, and $n,m$ take the integer values: $-N,..,N$. I know there is a possibility of exchanging the "variable of summation" by introducing some "relative coordinates" $\tau = n-m$ and $\eta = \frac{m+n}{2}$, by I don't know how to continue from there.
Thank you
On
Since integrals are continuous sums,
$$S\simeq\int_{-N}^N\underbrace{\int_{-N}^Ne^{-\alpha(x-y)^2}dx}_\text{The Gaussian Integral}dy\simeq\int_{-N}^N\sqrt\frac\pi\alpha dy=2N\sqrt\frac\pi\alpha$$
To be more specific,
$$\begin{align}0<\alpha<2.5&\iff S\simeq2N\sqrt{\frac\pi\alpha} \\\\\\\\\\ 3<\alpha<3.5&\iff S\simeq2N\sqrt{\frac\pi{\alpha-\frac12}} \\\\\\\\\\ \alpha>4&\iff S\simeq2N\end{align}$$
On a related note, $\displaystyle\vartheta_3(0,a)=\sum_{n=-\infty}^\infty a^{n^2}$ is the Jacobi theta function. See also.
As @LutzL said, it might be difficult to get a closed form, but you can simplify it at least down to a single sum.
Using the transformation $(\tau,\eta)=(n-m,n+m)$, your grid is modelled by going over the values $$\tau=-2N,-2N+1,...,2N\\ \eta=-K(\tau),-K(\tau)+2,...,K(\tau)$$ where $K(\tau)=2N-|\tau|$. So you have $$f(N)=\sum_{\tau=-2N}^{2N}e^{-\alpha\tau^2}\sum_{\eta=-K(\tau)}^{K(\tau),+2}1$$ where the second sum goes over every second number, denoted by the superscript $(+2)$. Now you can simplify this by using $K(-\tau)=K(\tau)$ to $$\sum_{\tau=-2N}^{2N}e^{-\alpha\tau^2}\sum_{\eta=-K(\tau)}^{K(\tau),+2}1 = 2\sum_{\tau=1}^{2N}e^{-\alpha\tau^2}\sum_{\eta=-K(\tau)}^{K(\tau),+2}1 + e^0\sum_{\eta=-K(0)}^{K(0),+2}1$$
Finally, if you use $$\sum_{\eta=-k}^{k,+2}1=k+1,$$ you get to $$f(N)=(2N+1)+2(2N+1)\sum_{\tau=1}^{2N}e^{-\alpha\tau^2}-2\sum_{\tau=1}^{2N}\tau e^{-\alpha\tau^2}.$$ If you define $$S_N(\alpha):=\sum_{\tau=1}^{2N}e^{-\alpha\tau^2},$$ your result can be written as
$$f(N)=(2N+1)(1+2S_N(\alpha))+\frac{1}{\alpha}S_N'(\alpha).$$
Edit According to the comments on this question, there is no closed form solution for $S_N(\alpha)$, but if you want to do numerics, you might be lucky and find a fast implementation, which makes it applicable for you.