Contour Integral 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!
 A: 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.
