Consider the function $ f(x) = e^{-\alpha x^2} $ with $\alpha>0$. Define the infinite sum \begin{align} g(x) = \sum_{n=-\infty}^\infty f(x-2n) \end{align}
I am trying to see if the following relationship holds \begin{align} \int_0^1 {\rm d} x_0 \, \bigg( g(x-x_0) + g(x+x_0) \bigg) \bigg( g(x'-x_0) + g(x'+x_0) \bigg) \stackrel{?}{=} \int_{-\infty}^\infty {\rm d}x_0 \, f(x-x_0) \, f(x'-x_0) \end{align} From a naive examination, it appears to me this holds, since the combination of the sum and integration on the l.h.s seems to cover the whole real axis. This is based on the following analysis: $ \int_0^1 dx_0\, [f(x-x_0-2n) + f(x+x_0-2n)] = \int_{2n-1}^{2n+1} dx_0 f(x-x_0) \,$ and then $\sum_{n} \int_{2n-1}^{2n+1} \to \int_{-\infty}^\infty$. But since in the actual equation I'm trying to show, there are two separate summations multiplied by each other, I am not sure how this could be constructed.
Does this relationship hold? Is there a more rigorous way to 'show' it?
Update
I think this relationship does not hold, as the l.h.s includes some terms that are not present on the r.h.s (basically from combinations like $g(x-x_0) g(x+x_0)$ ). I now want to see if the difference between l.h.s and r.h.s could be estimated (or in the unlikely case, be represented in a closed form).
Update
I now realized that the sum can be expressed in terms of the Jacobi theta function (through Poisson summation as in this post) \begin{equation} g(x) = \frac{ \sqrt{\pi} }{2 \sqrt{\alpha} } \vartheta_3 ( \frac{\pi x}{2} , e^{-\frac{\pi^2}{4\alpha}} ) \end{equation} So I am really looking for an identity and/or an approximation (for $\alpha\ll 1$ and $\alpha \gg 1$) for \begin{align} I(x,x',\alpha)=\int_0^1 {\rm d}x_0 \, &\Big( \vartheta_3 ( \frac{\pi (x-x_0)}{2} , e^{-\frac{\pi^2}{4\alpha}} ) + \vartheta_3 ( \frac{\pi (x+x_0)}{2} , e^{-\frac{\pi^2}{4\alpha}} ) \Big) \\ \times&\Big( \vartheta_3 ( \frac{\pi (x'-x_0)}{2} , e^{-\frac{\pi^2}{4\alpha}} ) + \vartheta_3 ( \frac{\pi (x'+x_0)}{2} , e^{-\frac{\pi^2}{4\alpha}} )\Big) \end{align} when $x,x' \in (0,1)$ as well.
