$\sum \exp(-x^2)$ vs $\sum x^2 \exp(-x^2)$

Question

I am curious about the following sum, for $\alpha \in (0,1)$:

$$\sum_{k = -\infty}^{\infty} (1-(2k - 1 + \alpha)^2) \exp(-\frac{1}{2} (2k - 1 + \alpha)^2)$$

I have reasons to believe sum should be zero when $\alpha = 1/2$, but I don't know how to prove it. And according to Mathematica, the value at $\alpha = 1/2$ is

$-2.6474039 \times 10^{-7}$,

which is notably not zero (though I'm not convinced that this isn't a rounding error). So: prove me wrong! (Is it possible to show that the value of this sum is non-zero?)

For some 'intuition' about why the sum is so close to zero, note that

$$\int_{-\infty}^{\infty} (1-(2x - \alpha + 1)^2)\exp(-\frac{1}{2} (2x - \alpha + 1)^2) \, dx = 0$$

for all $\alpha \in \mathbb{R}$. This is because the value of the integral doesn't depend on $\alpha$, and computing the value at $\alpha = 0$ is an easy exercise. The sum above is a Riemann sum for this integral. Of course, there is always error in changing from a Riemann sum to an integral -- I am just suspicious of how small the error is!

I would also like to know if there is a 'nicer' expression for the sum, or some relationship to well-known special functions. I suspect there are some clever Fourier-analysis type ideas to help evaluate this kind of sum.

Answer 1

Let $f(k) = (1-(2k - 1 + \frac12)^2) \exp(-\frac{1}{2} (2k - 1 + \frac12)^2)$, so that you are asking about the sum $S = \sum_{k=-\infty}^\infty f(k)$.

We can calculate that $$ \sum_{k=-2}^3 f(k) \approx -2.37104\times10^{-7} $$ (and presumably you are comfortable trusting this finite calculation). This proves that $S<0$, since the omitted summands $f(k)$ are negative for all $k\le-3$ and all $k\ge4$.

Indeed, we can also calculate $$ \sum_{k=-4}^5 f(k) \approx -2.647403940047578\times10^{-7} \tag{$*$} $$ and, since $|f(t)|$ is increasing for $t<-2$ and decreasing for $t>2$, $$ \sum_{k=6}^\infty |f(k)| < \int_5^\infty |f(t)|\,dt = |F(5)| < 1.2\times10^{-19} $$ and $$ \sum_{k=-\infty}^{-5} |f(k)| < \int_{-\infty}^{-4} |f(t)|\,dt = |F(-4)| < 8.7\times10^{-16}, $$ where $$ F(t) = \int f(t)\,dt = \frac{1}{4} e^{-2t^2+t-\frac{1}{8}} (4t-1); $$ these calculations show that the right-hand side of $(*)$ is the correct value of the infinite sum $S$ to 7 or 8 significant digits.

Answer 2

As suggested in comments, Poisson summation formula yields $$\sum_{k=-\infty}^\infty\big(1-(2k-1+\alpha)^2\big)e^{-(2k-1+\alpha)^2/2}=2^{1/2}\pi^{5/2}\sum_{n=1}^\infty n^2 e^{-(n\pi)^2/2}\cos\alpha n\pi,$$ so that the exact value at $\alpha=1/2$ equals $(2\pi)^{5/2}\sum_{n=1}^\infty(-1)^n n^2 e^{-2(n\pi)^2}$.

The term at $n=1$ is $\approx-2.64740394\times10^{-7}$, the term at $n=2$ is $\approx2.02764454\times10^{-32}$, etc.