I am a graduate student.I have been studying complex analysis from Stein Shakarchi's book.In chapter $2$ ,there is an exercise which is as follows:
As given in the hint,I assume that the countour is $C$ and I used Cauchy's theorem,to get $\int_C e^{-z^2}dz=0$.
Now I split the integral into three parts.First one is along the line segment from $0$ to $R$ which will give me the desired integral when $R\to \infty$.But I am unable to manipulate the integral along the arc which turns out to be ,$\int_0^{\pi/4} e^{-R^2e^{2it}}.Rie^{it}dt$.Can someone help me with this?
First, we want to put an uppoer bound on the magnitude of the integral. Moving the absolute value inside is a good start to get rid of annoying phase terms: $$ \left|iR\int_0^{\pi/4}e^{-R^2e^{2it}}e^{it}dt\right|\le R\int_0^{\pi/4}\left|e^{-R^2e^{2it}}e^{it}dt\right| = R\int_0^{\pi/4}e^{-R^2\cos(2t)}dt. $$ Next, note that $\cos(2t) \ge 1 - 4t/\pi$ on this interval, which leads to $$ R\int_0^{\pi/4}e^{-R^2\cos(2t)}dt \le R\int_0^{\pi/4}e^{-R^2(1-4t/\pi)}dt = \frac{\pi}{4R}\left(1-e^{-R^2}\right)\le \frac{\pi}{4R}. $$ Putting these together we have $$ \left|iR\int_0^{\pi/4}e^{-R^2e^{2it}}e^{it}dt\right|\le \frac{\pi}{4R} $$ and thus the integral goes to zero as $R\rightarrow\infty$.