The limit $\lim_{x \to 0-} \frac{e^{-x^2}}{\sqrt{\pi}} \int_0^\infty e^{-t^2/4} \frac{e^{2x} \cos t-1}{e^{4x}-2e^{2x} \cos t+1 } dt$

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 $\int_0^\infty e^{-t^2/4}dt=\sqrt\pi$.

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 $(1).$