The limit $\lim_{x \to 0-} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4} \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt$ A while back I derived the following expression valid for $x>0$:
$$\sum_{n=1}^\infty e^{-(n+x)^2}= \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt$$
While the integral doesn't converge for $x=0$, it has a right limit:
$$\lim_{x \to 0+} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt=\sum_{n=1}^\infty e^{-n^2}=\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)-1\right)$$
But, despite the fact that the integral converges for $x<0$, it converges to a different limit from the left side, as can be seen from the numerical plot by Mathematica:

Surprisingly enough, by numerical integration with Mathematica, we seem to have:

$$\lim_{x \to 0-} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt \approx -\frac{1}{2} \left(\vartheta _3\left(0,\frac{1}{e}\right)+1\right)$$
In other words, the limits are related as $L^- = -1-L^+$. Is this correct? And why?

Here's the derivation of the first equality https://math.stackexchange.com/a/2751575/269624.

The problematic part is $\frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } $.    
Rewrite $$\frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 }=\frac{\left(e^{2x} \cos t-1\right)-e^{2x}\left(e^{2x}-\cos t\right)+e^{2x}\left(e^{2x}-\cos t\right)}{e^{2x}\left(e^{2x}-\cos t\right)-\left(e^{2x}\cos t-1 \right)},$$
which gives $$\frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 }= -1-\frac{e^{2x}\left( \cos t-e^{2x}\right)}{1-2e^{2x} \cos t+e^{4x}}.\tag 1$$
Set $h=-x$ for $x<0$ and compute $$\frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 }=\frac{e^{-2h} \cos t-1}{e^{-4h}-2e^{-2h} \cos t+1 }= \frac{e^{2h}\left( \cos t-e^{2h}\right)}{1-2e^{2h} \cos t+e^{4h}}.$$ Compare with $(1).$
