Contour Integral

Question

I have this question:

I'm aware that $e^{iz^2}$ is analytic, and hence $I_R = 0$ by Cauchy's Integral theorem. I'm not really sure what to do from there. Thanks!

Answer 1

The closed contour is split into 3 pieces:

The original integral:

$$\int_0^R dx \, e^{i x^2}$$

The integral over the angled line of the contour; parametrize with $z=e^{i \pi/4} t$:

$$e^{i \pi/4} \int_R^0 dt \, e^{- t^2}$$

(because $i e^{i \pi/2} = -1$). The third piece is the integral over the circular arc subtending an angle of $\pi/4$ with respect to the origin. Parametrize using $z=R e^{i \theta}$, $\theta \in [0,\pi/4]$:

$$i R \int_0^{\pi/4} d\theta \, e^{i \theta} \, e^{-R^2 \sin{2 \theta}} e^{i R^2 \cos{2\theta}}$$

which, in absolute value, is

$$\le R \int_0^{\pi/4} d\theta \, e^{-R^2 \sin{2 \theta}} \le R \int_0^{\pi/4} d\theta \, e^{-4 R^2 \theta/\pi} \le \frac{\pi}{4 R}$$

as $R \to \infty$. Note that I used the relation $\sin{2 \theta} \ge \frac{4 \theta}{\pi}$ when $\theta \in [0,\pi/4]$. So when we set $R \to \infty$, this integral vanishes. Thus, in this limit, we are left with

$$\int_0^{\infty} dx \, e^{i x^2} - e^{i \pi/4} \int_0^{\infty} dt \, e^{- t^2} = 0$$

Thus, you may now evaluate the integral on the left in terms of the integral on the right, which you know converges.