Analysis of the function $\prod_{n=-\infty}^{ \infty }(1-e^{-a{(x+n)^2} })$

It's not an answer, but there are a lot of graphs which can't be put into a comment.

The function $\prod\limits_{n=-\infty}^{ \infty }(1-e^{-a{(x+n)^2}})$ is very close to $C \sin^2(\pi x)$ for $a \leq 1$, as can be seen by numerical experiments:

1) For $a=1$ the scaling constant $C=0.0390070548953618...$ and the functions are practically indistinguishable.

2) For $a<<1$ the functions are also very close with appropriate scaling.

3) However, for $a>>1$ the functions are different, since $f(x)$ approaches a constant around its maxima.

The overall $a$ dependence can be seen in the following 3D graph and the $f(1/2)$ graphs for the dependence around a maximum. The latter has the character of a logistic function.

I didn't have much luck with analytics so far, but I will expand my post into a proper answer if something useful comes up.

Update. As asked by the OP, I tested their approximation, it's not very good:

I post this answer separately, so that the analytics are not lost among the illustrative material from my first post.

This answer deals with Fourier series, not for the function $f(x)$ itself, but for its logarithm.

It's often easier to deal with the logarithms of infinite products, since they are represented as infinite series.

$$\ln f(x)=\sum_{n=-\infty}^\infty \ln(1-e^{-a{(x+n)^2}})=$$

Since $a(x+n)^2>0$ we can always expand the logarithm into its Taylor series:

$$=-\sum_{n=-\infty}^\infty \sum_{k=1}^\infty \frac{1}{k} e^{-ka(x+n)^2}=-\sum_{k=1}^\infty \frac{1}{k} e^{-kax^2} \sum_{n=-\infty}^\infty e^{-kan(n+2x)}$$

The latter sum fits the definition of a Jacobi Theta Function $\vartheta_3$, thus:

$$\sum_{n=-\infty}^\infty e^{-kan(n+2x)}=\sqrt{\frac{\pi}{ka}} e^{kax^2} \vartheta_3 \left(\pi x, e^{-\frac{\pi^2}{ka}}\right)$$

This result is confirmed by Wolfram Alpha.

This looks complicated, but there is another series representation for $\vartheta_3$ which might help:

$$\vartheta_3 \left(\pi x, e^{-\frac{\pi^2}{ka}}\right)=1+2\sum_{l=1}^\infty e^{-\frac{\pi^2l^2}{ka}} \cos(2 \pi l x)$$

Now we obtain:

$$\ln f(x)=-\sqrt{\frac{\pi}{a}} \sum_{k=1}^\infty \frac{1}{k^{3/2}} \left(1+2\sum_{l=1}^\infty e^{-\frac{\pi^2l^2}{ka}} \cos(2 \pi l x) \right)$$

And finally:

$$\ln f(x)=-\sqrt{\frac{\pi}{a}} \zeta \left( \frac32 \right) -2\sqrt{\frac{\pi}{a}} \sum_{l=1}^\infty \sum_{k=1}^\infty \frac{1}{k^{3/2}} e^{-\pi^2l^2/(ka)} \cos(2 \pi l x)$$

Or, introducing:

$$C_l (a)=\sum_{k=1}^\infty \frac{1}{k^{3/2}} e^{-\pi^2l^2/(ka)}$$

The familiar form of a Fourier series emerges (with the first term being the constant term $l=0$):

$$\ln f(x)=-\sqrt{\frac{\pi}{a}} \zeta \left( \frac32 \right) -2\sqrt{\frac{\pi}{a}} \sum_{l=1}^\infty C_l (a) \cos(2 \pi l x)$$

The convergence is not very good, however the $l$ sum converges much faster than the $k$ sum (which should be obvious by looking at the exponential term in $C_l$). Thus, we don't need a lot of Fourier terms, but the coefficients have to be computed precisely.

Here is an illustration for $a=1$:

This Fourier series allows us excellent insight into similarities with the $C \sin^2 (\pi x)$ function.

$$\frac{(\sin^2 (\pi x))'}{\sin^2 (\pi x)}=2 \pi \cot (\pi x)$$

$$\frac{f'(x)}{f(x)}=(\ln f(x))'=4 \pi \sqrt{\frac{\pi}{a}} \sum_{l=1}^\infty l~C_l \sin(2 \pi l x)$$

It turns out, these new functions are very close for $a=1$ (if we compute $C_l$ accurately enough):

However, there is a correlation - the larger $a$, the faster $k$-sum converges, but the slower $l$-sum converges. Thus, if $a$ is small you need to take more terms in $k$-sum, but if $a$ is large you ned to take more terms in the $l$-sum.

Here are plots for the same large number of terms for $a=0.2,1,5$:

Disclaimer

The Fourier series has been introduced in this answer, I have not seen it before writing my own, even though it's linked above in the comments to the OP. I acknowledge that Semiclassical got there first.