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?

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The integral is a Fresnel one. Usually, it's evaluated in a
  'pizza-slice' closed contour wich subtends a $\ds{\pi/4}$-angle in the complex plane first quadrant ( details can be seen elsewhere ). Since there are not any poles inside such contour, the integral along $\ds{\pars{0,\infty}}$
  satisfies:

\begin{align}
\int_{0}^{\infty}\expo{\ic x^{2}}\dd x & =
\lim_{r \to \infty}\bracks{-\int_{0}^{\pi/2}
\exp\pars{\ic\bracks{r\expo{\ic\theta}}^{2}}r\expo{\ic\theta}\ic\,\dd\theta -
\int_{r}^{0}\exp\pars{\ic\bracks{\xi\expo{\pi\ic/4}}^{2}}\expo{\pi\ic/4}\dd\xi}
\\[5mm] & =
-\ic\,\overbrace{\lim_{r \to \infty}\int_{0}^{\pi/2}\exp\pars{\ic r^{2}\cos\pars{2\theta}}
\exp\pars{-r^{2}\sin\pars{2\theta}}r\expo{\ic\theta}\dd\theta}^{\ds{=\ 0}}
\\[5mm] &
\mbox{} + \expo{\pi\ic/4}\
\underbrace{\int_{0}^{\infty}\expo{-\xi^{2}}\,\dd\xi}
_{\ds{=\ {\root{\pi} \over 2}}}\ =\
\bbox[8px,border:0.1em groove navy]{{\root{\pi} \over 2}\,\expo{\pi\ic/4}}
\end{align}

Note that

\begin{align}
0 & < \verts{\int_{0}^{\pi/2}\exp\pars{\ic r^{2}\cos\pars{2\theta}}
\exp\pars{-r^{2}\sin\pars{2\theta}}r\expo{\ic\theta}\dd\theta} <
\int_{0}^{\pi/2}\exp\pars{-r^{2}\sin\pars{2\theta}}r\,\dd\theta
\\[5mm] & =
{1 \over 2}\,r\int_{0}^{\pi}\exp\pars{-r^{2}\sin\pars{\theta}}\dd\theta =
r\int_{0}^{\pi/2}\exp\pars{-r^{2}\sin\pars{\theta}}\dd\theta <
r\int_{0}^{\pi/2}\exp\pars{-\,{2r^{2} \over \pi}\,\theta}\dd\theta
\\[5mm] & =
{\pi \over 2}\,{1 - \expo{-r^{2}} \over r}\,\,\,\stackrel{r\ \to\ \infty}{\to} 0
\end{align}
