Contour integration of $\int_{-\infty}^{\infty}e^{iax^2}dx$ Consider the following integral:
$$\int_{-\infty}^{\infty}e^{iax^2}dx$$
Here I believe we have to consider the two cases when $a<0$ and $a>0$, as they need different contours. For $a>0$ I've been using the upper half circle, radius $R$. I shall call the whole semi circle $\Gamma$ and the curved section $\Gamma_1$
As the function is holomorphic we have:
$$\int_\Gamma e^{iaz^2} dz = \int_{-R}^R e^{iax^2} dx + \int_{\Gamma_1} e^{iaz^2} dz =0$$
$$\Rightarrow \int_{-R}^R e^{iax^2} dx = - \int_{\Gamma_1} e^{iaz^2} dz = -\int_{0}^\pi iRe^{i\theta} e^{iaR^2e^{2i\theta}} d\theta$$
It's this point i'm stuck on, any help would be greatly appreciated!
EDIT: So from the comments, I've seen my method was incorrect, if anyone could outline this other contour it would be great, I have not seen it before.
 A: If $a<0$ we have $b:=-a>0$, and
$$
J(a):=\int_{-\infty}^\infty e^{iax^2}\,dx=\int_{-\infty}^\infty e^{-ibx^2}\,dx=\overline{J(b)}=\overline{J(-a)}.
$$
Therefore, it is enough to evaluate $J(a)$ for $a>0$. So from now on we suppose that $a>0$.
Notice that
$$
J(a)=\int_{-\infty}^\infty e^{iax^2}\,dx=\frac{1}{\sqrt{a}}\int_{-\infty}^\infty e^{ix^2}\,dx=\frac{1}{\sqrt{a}}J(1).
$$
Hence, we only have to compute $J(1)$.
Given $r>0$ we denote by $\Gamma_R$ the boundary of
$$
\Omega_r=\left\{z \in \mathbb{C}:\ |z|\le r,\ 0\le \arg z\le \frac{\pi}{4}\right\}.
$$
Consider the parametrization $\gamma_r:[0,1] \to \mathbb{C}$ of $\Gamma_r$ given by
$$
\gamma_r(t)=\begin{cases}
\gamma_1(3t) & \text{ for } 0 \le t \le \frac13\\
\gamma_2(3t-1) & \text{ for } \frac13 < t \le \frac23\\
\gamma_3(3t-2) & \text{ for } \frac23 < t \le 1
\end{cases},
$$
where
$$
\gamma_1(t)= rt,\
\gamma_2(t)= r\exp\left(i\frac{\pi t}{4}\right),\
\gamma_3(t)= r(1-t)\exp\left(i\frac{\pi}{4}\right).
$$
Since 
$$
f: \mathbb{C} \to \mathbb{C},\ f(z):=e^{iz^2}
$$
is holomorphic and $\Omega_r$ is simply connected, we have thanks to the Cauchy Integral formula:
$$
0=:\int_{\Gamma_r}f(z)\,dz=\sum_{k=1}^3I_k(r),
$$
with
$$
I_k(r):=\int_{\frac{k-1}{3}}^{\frac{k}{3}}\gamma_R'(t)f(\gamma_R(t))\,dt, \ k=1,2,3.
$$
I claim that
$$
\lim_{r\to \infty}I_2(r)=0.
$$
In fact, since $\sin \theta \ge \frac{\theta}{2}$ for $\theta \le \theta \le \frac{\pi}{2}$, we have:
\begin{eqnarray}
|I_2(r)|&=&\left|\int_0^13\gamma_2'(3t-1)f(\gamma_2(3t-1))\,dt\right|=\left|\int_0^1\gamma_2'(t)f(\gamma_2(t))\,dt\right|\\
&\le& \int_0^1\frac{\pi r}{4}\left|\exp\left\{ir^2\cos\left(\frac{\pi t}{2}\right)-r^2\sin\left(\frac{\pi t}{2}\right)\right\}\right|\,dt\\
&=&\int_0^1\frac{r\pi}{4}\exp\left[-r^2\sin\left(\frac{\pi t}{2}\right)\right]\,dt\\
&\le&\int_0^1\frac{r\pi}{4}\exp\left(-\frac{r^2\pi t}{4}\right)\,dt
=\frac{1}{r}\left[1-\exp\left(-\frac{\pi r^2}{4}\right)\right].
\end{eqnarray}
Thus $\lim_{r \to \infty}I_2(r)=0$ as claimed.
Clearly
\begin{eqnarray}
I_3(r)&=&\int_{\frac23}^13\gamma_3'(3t-2)f(\gamma_3(3t-2))\,dt=\int_0^1\gamma_3'(t)f(\gamma_3(t))\,dt\\
&=&-r\exp\left(i\frac{\pi}{4}\right)\int_0^1\exp\left(-r^2t^2\right)\,dt=-\exp\left(i\frac{\pi}{4}\right)\int_0^{r}\exp\left(-x^2\right)\,dx\\
I_1(r)&=&\int_0^{\frac13}3\gamma_1'(3t)f(\gamma_1(3t))\,dt=r\int_0^1\exp\left(ir^2t^2\right)\,dt=\int_0^{r}\exp\left(ix^2\right)\,dx.
\end{eqnarray}
So we have
$$
0=\sum_{k=1}^3I_k(r)=\int_0^{r} e^{ix^2}\,dx-\exp\left(i\frac{\pi}{4}\right)\int_0^{r} e^{-x^2}\,dx+I_2(r).
$$
This implies that
\begin{eqnarray}
J(1)&=&2\int_0^\infty e^{ix^2}\,dx=2\exp\left(i\frac{\pi}{4}\right)\int_0^\infty e^{-x^2}\,dx=\exp\left(i\frac{\pi}{4}\right)\int_{-\infty}^\infty e^{-x^2}\,dx\\
&=&\sqrt{\pi}\exp\left(i\frac{\pi}{4}\right)=(1+i)\sqrt{\frac{\pi}{2}}.
\end{eqnarray}
Thus we have
$$
J(a)=\begin{cases}
(1+i)\sqrt{\frac{\pi}{2a}} & \text{ for } a>0\\
(1-i)\sqrt{\frac{\pi}{-2a}} & \text{ for } a<0
\end{cases}.
$$
