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.

We have
\begin{align}
L^-
&=\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\\
&=\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\\
&=\lim_{x \to 0^+} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{e^{2x} \cos t-e^{4x}}{1-2e^{2x} \cos t+e^{4x} } dt
\end{align}
hence
\begin{align}
L^++L^-
&=\lim_{x \to 0^+} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4}  \frac{-1+2e^{2x} \cos t-e^{4x}}{1-2e^{2x} \cos t+e^{4x} }dt\\
&=-\lim_{x \to 0^+} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4} dt\\
&=-1
\end{align}
by the wellknow $\int_0^\infty e^{-t^2/4}dt=\sqrt\pi$.
