Square of integral of sum: What can I do? Let $a>0$ and consider the following expression
$$\Big( \int_{-a}^a \sum_{n=-\infty}^{\infty} e^{-(x+4na)^2}-e^{-(x+2a-4na)^2}dx\Big)^2.$$
Is there a way to calculate this expression or at least get a good upper bound?
I have started with Cauchy-Schwarz (which actually isn't a "good" upper bound) and estimated
$$\leq  2a\int_{-a}^a \Big(\sum_{n=-\infty}^{\infty} e^{-(x+4na)^2}-e^{-(x+2a-4na)^2}\Big)^2dx$$
and now I am having trouble with the squared sum. Does anybody have an idea on how to proceed or maybe even know a better way?
 A: Here is a way to obtain an estimate: 
Step 1. Change of variable $x = ay-a$. Then $$\int_{-a}^a \sum_n e^{-(x+4na)^2}-e^{-(x+2a-4na)^2}\, dx = a\int_0^2 \sum_n e^{-a^2[(4n-1)+y]^2}-e^{-a^2[(4n-1)-y]^2}\, dy.$$
Step 2. Term-wise estimate. Fix any $n$, then for all $y\in (0,2)$, we have \begin{align}
\left|e^{-a^2[(4n-1)+y]^2}-e^{-a^2[(4n-1)-y]^2}\right| & = 
\left| e^{-a^2[(4n-1)^2+y^2]}\cdot \left\{ e^{-2a^2(4n-1)y}-e^{2a^2(4n-1)y} \right\}\right| \\
& \leq e^{-a^2[(4n-1)^2+y^2]}\cdot \left( \left| e^{-2a^2(4n-1)y}\right| + \left|e^{2a^2(4n-1)y} \right|\right)\\
& \leq 2e^{-a^2[(4n-1)^2+y^2]}\cdot e^{2a^2\cdot|4n-1|y}\quad \text{since } e^\alpha + e^{-\alpha} \leq 2e^{|\alpha|}\  \forall \alpha \in \mathbb{R}\\
& = 2e^{-a^2(|4n-1|-y)^2}. 
\end{align}
Step 3. Substitute back to the original integral. For simplicity, call the original integral $A$. So we have $$
A \leq 2a\int_0^2 \sum_n e^{-a^2(|4n-1|-y)^2}\, dy. $$
To calculate the explicit value, let's turn the integrand into something that's easier to compute. 
Step 4. Note that $y\in [0,2]$, so $y$ is bounded. For large values of $n$, $|4n-1| \gg y$, so we have $$(|4n-1|-y)^2 \geq (|4n-1|/2)^2 =\frac14 |4n-1|^2\geq \frac18 (|4n-1|^2 + y^2).$$ Therefore, $$
A \lesssim 2a\int_0^2 \sum_n e^{-\frac18 a^2(|4n-1|^2+y^2)}\, dy. $$
Step 5. Note that for an eventually decreasing function $f$, we have $\sum_n f(n) \lesssim \int_0^\infty f(x)\, dx$ (this can be proven by drawing the graph of $f$ and estimating it using Riemann sum.) Thus, \begin{align}
A & \lesssim 2a\int_0^2 \int_0^\infty  e^{-\frac18 a^2(x^2+y^2)}\, dxdy.
\end{align}
Now the estimate is easy to obtain using polar coordinates. 
