Computing the integral $\int \exp(ix^2) dx$ I'm trying to compute the following integral: $I_1 =  \int_{-\infty}^\infty \exp(iu^2) du$. 
This is what I did, but it is wrong and I don't know why:
$$I_1^2 = \left (\int_{-\infty}^\infty \exp(iu^2) du \right)^2 = \int_{-\infty}^{\infty} \exp(iu^2) du \int_{-\infty}^{\infty} \exp(iv^2) dv = \iint_{\mathbb{R}^2} \exp{i(v^2+y^2)} dA$$
Using polar coordinates:
$$I_1^2 = \lim_{c\to \infty} \int_0^{2\pi} d\varphi \int_0^{c} \exp(ir^2) r dr$$
But the last integral doesn't converge. What is wrong here?

This and related ones are known as Fresnel Integrals,  see HERE
to find that (improper Riemann integrals)
$$
\int_{-\infty}^\infty e^{ix^2}dx = 
\int_{-\infty}^\infty \cos(x^2)\;dx + i\int_{-\infty}^\infty \sin(x^2)\;dx
= (1+i)\sqrt\frac{\pi}{2}
$$
