Evaluating $\int_C e^{-z^2} dz$ as radius goes to infinity

Question

I was trying to calculate the integral $$\lim_{R \rightarrow \infty}\int_{C_R} e^{-z^2}dz$$ where $C_R$ is parameterized by $C(\theta) = Re^{i\theta}$ for $\theta \in [-\frac{\pi}{4}, 0]$. I tried using the inequality $|\int_C f(z) dz| \leq length(C)\max_{z \in C}|f(z)|$ to see if the integral goes to $0$ which yields $$ |\int_{C_R} e^{-z^2}dz| \leq |e^{-R\cos2\theta}||kR|$$ but on each $C_R$ there are $z$ with $|z|$ arbitrarily close to $1$, so this doesn't work and also the integral probably isn't zero. How do you evaluate this integral?

Answer 1

Let $\phi=2\theta+\pi/2$. Then \begin{align} \left|\int_{C_R} e^{-z^2} dz\right|&=\left|\int_{-\pi/4}^0e^{-R^2\cos{2\theta}}e^{-iR^2\sin{2\theta}}iRe^{i\theta}d\theta\right| \\ &=\frac1{2}\left|\int_{0}^{\pi/2}e^{-R^2\sin{\phi}}e^{iR^2\cos{\phi}}iRe^{\frac1{2}i(\phi-\pi/2)}d\phi\right| \\ &\leqslant\frac{R}{2}\int_{0}^{\pi/2}e^{-R^2\sin{\phi}}d\phi \\ &\leqslant\frac{R}{2}\int_{0}^{\pi/2}e^{-R^2\frac{2}{\pi}\phi}d\phi\tag1 \\ &=-\frac{\pi}{4R}e^{-R^2\frac{2}{\pi}\phi}\bigg|_{0}^{\pi/2} \\ &=\frac{\pi}{4R}(1-e^{-R^2}) \\ &\to0 \quad\text{as }\:R\to\infty \end{align} $(1)$: by Jordan Lemma, for $x\in[0,\pi/2]$, $\:\sin x>\frac{2}{\pi}x$.

Hence $$ \lim_{R \rightarrow \infty}\int_{C_R} e^{-z^2}dz=0 $$