In a complex integral related question, I have difficulty in trying to find the estimation of the integral along the vertical side.
The integral is \begin{equation} \int_{L_{R}} \frac{e^{-\frac{z^2}{2}}}{z} \mathrm{d}z \end{equation} where the line segment $L_{R}$ is $z=R+iy, R>0, 0\leq y\leq R$. I want to find the limit of it when $R$ goes to infinity. My attempt is as follows \begin{equation} \left\vert\int_{L_{R}} \frac{e^{-\frac{z^2}{2}}}{z} \mathrm{d}z\right\vert=\left\vert\int_{0}^{R} \frac{e^{-\frac{1}{2}(R^2-y^2+2iRy)}}{R+iy}i\mathrm{d}y\right\vert\leq\frac{1}{R}\int_{0}^{R}e^{-\frac{R^2}{2}}e^{\frac{y^2}{2}}\mathrm{d}y\leq\frac{1}{R}\times R=1 \end{equation} which is weird.
How to find the estimation order of $R$ when it goes to infinity?
$\frac{1}{R}\int_{0}^{R}e^{-\frac{R^2}{2}}e^{\frac{y^2}{2}}\mathrm{d}y\le \frac{1}{R}\int_{0}^{R}e^{-\frac{R^2}{2}}e^{\frac{Ry}{2}}\mathrm{d}y$. Now compute the integral to finish the proof.