Problem: Prove that $$ \lim_{L \to \infty} \int_{L}^{L-i t} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z =0, $$ where $z$ is a complex number, $L$ is a non-negative real number.
P/S: I am solving a problem that I thing I need to have this result to attain the final result of the problem. So, I wonder that this limitation is equal to 0 or not?
Let $0\le u\le t$. Then we have
$$ \Re[(L-iu)^2]=\Re[L^2-2iuL-u^2]=L^2-u^2 $$
Consequently, we have $$ \left|\int_L^{L-it}{1\over\sqrt{2\pi}}e^{-{z^2\over2}}\mathrm dz\right|\le{t\over\sqrt{2\pi}}e^{{t^2-L^2\over2}}=O(e^{-L^2/2})\to0 $$ as $L\to+\infty$.