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.