Integral of $e^{ix^2}$ How does one evaluate 
$$\int_{-\infty}^{\infty} e^{ix^2} dx$$
I know the trick how to evaluate $\int_{-\infty}^{\infty} e^{-x^2}dx$ but trying to apply it here I get a limit which does not converge:
$I = \int_{-\infty}^{\infty} e^{ix^2}dx = \int_{-\infty}^{\infty} e^{iy^2}dy \\\implies I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{i(x^2+y^2)}dxdy = \int_{0}^{2\pi}\int_{0}^{\infty} re^{ir^2} = -\pi i (e^{i\infty}-e^0) $
and $e^{i\infty} $ is not defined. 
Are there any other methods? I am not interested in the result (WolframAlpha can do this for me), but rather the method.
 A: Here is another solution with a different contour.
Let $$I=\int^{\infty}_{-\infty}e^{ix^2}\text{ d}x$$ and let $$f(z)=e^{iz^2}$$ Note that our function is even.
In user279043's answer, the contour they chose was (what I would presume) based on the fact that the complex integrand can be rewritten like this $$\exp\left(z^2e^{\frac{i\pi}2}\right)=\exp\left(\left(ze^{\frac{i\pi}4}\right)^2\right)$$ which implies that the integrand is well behaved along the ray $e^{\frac{i\pi}4}$.
However, we can also note that along the imaginary axis, our function is similarly well behaved. Consider the following contour shown below

consisting of the paths $$\mathcal{C}=B+\Gamma+L$$
Thus, our contour integral about said contour would be $$\oint_{\mathcal{C}}f(z)\text{ d}z=\int_B+\int_{\Gamma}+\int_Lf(z)\text{ d}z=0$$
Each path can be parameterized as follows
\begin{alignat*}{5}
    B&:\text{ }z=x,\qquad &\text{d}z&=\text{d}x,\qquad &x&\in[0, R\,]\\
    \Gamma &:\text{ }z=Re^{i\theta},\qquad &\text{d}z&=iRe^{i\theta}\text{ d}\theta,\qquad &\theta&\in\left[0, \frac{\pi}{2}\right]\\
    L&:\text{ }z=iy,\qquad &\text{d}z&=i\text{d}y,\qquad &y&\in[\,R, 0]
\end{alignat*}
Now let's evaluate each integral. The integral about $B$ is just half of $I$
$$\lim_{R\to+\infty}\int_Bf(z)\text{ d}z=\lim_{R\to+\infty}\int^{R}_0e^{ix^2}\text{ d}x=\frac{1}{2}I$$
The integral about $\Gamma$ vanishes along our integration interval using typical inequalities. Note that I put in red whatever goes to $1$.
\begin{align}
    \left|\int_{\Gamma}f(z)\text{ d}z\right|&\le\int^{\frac{\pi}2}_{0}\left|e^{iR^2e^{2i\theta}}\right|\cdot\color{red}{|i|}|R|\color{red}{\left|e^{i\theta}\right|}\text{ d}\theta\\
    &\le\int^{\frac{\pi}2}_{0}R\color{red}{\left|e^{iR^2\cos(2\theta)}\right|} \left|e^{-R^2\sin(2\theta)}\right| \text{ d}\theta\\
    &\le\int^{\frac{\pi}4}_{0}R e^{-R^2\cdot \frac{4\theta}{\pi}} \text{ d}\theta+\int^{\frac{\pi}2}_{\frac{\pi}4}R e^{R^2\cdot \frac{4\left(\theta-\frac{\pi}{2}\right)}{\pi}} \text{ d}\theta
\end{align}
Which goes to $0$ when we take the limit as $R$ goes to infinity. The third line comes from Jordan's inequality.
Lastly, the integral along $L$ gives
$$\lim_{R\to+\infty}\int_Lf(z)\text{ d}z=\lim_{R\to+\infty}\int^{0}_{R}e^{i(iy)^2}\cdot i\text{ d}y=-i\int^{\infty}_0e^{-iy^2}\text{ d}y$$
So we have
$$\oint_{\mathcal{C}}f(z)\text{ d}z=\int_B+\int_{\Gamma}+\int_Lf(z)\text{ d}z=\frac12 I-i\int^{\infty}_0e^{-iy^2}\text{ d}y=0$$
Lastly, we can rearrange our equation and solve as follows
\begin{align}
    \frac12 I&=i\int^{\infty}_0e^{-iy^2}\text{ d}y\\
    \implies\left(\frac{I}{2i}\right)^2&=\int^{\infty}_0e^{-iy^2}\text{ d}y\cdot\int^{\infty}_0e^{-it^2}\text{ d}t\\
    &=\int^{\infty}_0\int^{\infty}_0e^{-i(y^2+t^2)}\text{ d}y\text{ d}t\\
    &=\int^{\frac{\pi}2}_0\int_0^{\infty}e^{-ir^2}\cdot r\text{ d}r\text{ d}\theta=\frac{\pi}{2}\int_0^{\infty}re^{-ir^2}\text{ d}r\\
    &=\frac{\pi}{4i}\int^{\infty}_0e^{-u}\text{ d}u=\frac{\pi}{4i}\cdot 1\\
    \implies \frac{I}{2i}&=\sqrt{\frac{\pi}{4i}}\\
    \implies I&=2i\cdot\frac{\sqrt{\pi}}{2}\cdot e^{-\frac{i\pi}{4}}=\sqrt{\pi}e^{\frac{i\pi}4}=(1+i)\sqrt{\frac{\pi}{2}}
\end{align}
we can see that line 4 follows from the polar coordinate change of variables
$$y=r\cos(\theta),\,\,t=r\sin(\theta),\qquad J_f=\left[\begin{array}{cc}
     \cos(\theta)& -r\sin(\theta) \\
     \sin(\theta)& r\cos(\theta) 
\end{array}\right],\qquad \left|\det\left(J_f\right)\right|=r$$
and line 5 follows from this simple u-sub
$$u= ir^2,\qquad\frac{\text{d}u}{\text{d}r}=2ir,\qquad\text{d}r=\frac{\text{d}u}{2ir}$$
Hence $$I=\boxed{(1+i)\sqrt{\frac{\pi}{2}}}$$
A: This answer is dedicated to give an alternative proof of
$$
\lim_{R\to+\infty}\int_\Gamma e^{iz^2}\mathrm dz=0
$$
where $\Gamma$ denotes a circular arc connecting $R$ and $e^{i\pi/4}R$. Using the parametrization that $z=Re^{i\theta}$, we have
\begin{aligned}
\left|\int_\Gamma e^{iz^2}\mathrm dz\right|
&\le R\int_0^{\pi/4}e^{-R^2\sin(2\theta)}\mathrm d\theta \\
&=R\left(\int_0^\delta+\int_\delta^{\pi/4}\right)e^{-R^2\sin(2\theta)}\mathrm d\theta \\
&<R\delta+{R\pi\over4}e^{-R^2\sin(2\delta)}
\end{aligned}
Since $\sin u\sim u$ when $u\to0$, we see that for small $\delta>0$ there is $\sin(2\delta)>\delta$, which means when $\delta=R^{-3/2}$ and $R$ is large there is
$$
\left|\int_\Gamma e^{iz^2}\mathrm dz\right|<{1\over\sqrt R}+{\pi R\over2}e^{-\sqrt R}=O\left(1\over\sqrt R\right)
$$
A: All right, let's do some complex analysis!  Let's integrate $e^{iz^2}$ over the closed contour defined in three pieces (the arrows indicating the direction of contour integration)
$$
\begin{cases}
\Gamma_1: & |z|:0\rightarrow R, & \theta=0 \\
\Gamma_2: & |z| = R, & \theta: 0\rightarrow \pi/4\\
\Gamma_3: & |z|:R\rightarrow0, & \theta=\pi/4
\end{cases}
$$
which we will eventually want to take the limit $R\rightarrow\infty$.
It can be seen that
$$
\int_0^\infty e^{ix^2}dx=\lim\limits_{R\rightarrow\infty}\int_{\Gamma_1}e^{iz^2}dz
$$
and since $e^{ix^2}$ is an even function
$$
\int_{-\infty}^{\infty}e^{ix^2}dx=2\int_0^\infty e^{ix^2}dx
$$
so we're heading in the right direction.
Now Cauchy's Theorem states that
$$
\oint_D f(z)dz =0
$$
for $f(z)$ analytic in $D$.  Our function, $e^{iz^2}$, has no singularities and is defined on the entire complex plane, so it is considered an entire function, and Cauchy's Theorem holds for our closed contour:
$$
\int_0^R e^{ix^2}dx+\int_{\Gamma_2}e^{iz^2}dz+\int_{\Gamma_3}e^{iz^2}dz=0
$$
For our second integral above, we show that it vanishes as $R\rightarrow\infty$ using the ML test given by
$$
\left|\int_\Gamma f(z)dz\right|\leq ML
$$
where $M$ is a finite upper bound of $f(z)$ and $L$ is the length of the contour $\Gamma$.  Of course, we need to assume that $f(z)$ is bounded and analytic on $\Gamma$ for this.
In order to apply the ML test, we substitute into our integrand $z=re^{i\theta}$ so that
$$
z^2 = r^2e^{2i\theta} = r^2\cos(2\theta)+ir^2\sin(2\theta)
$$
$$
|e^{iz^2}|=|e^{ir^2\cos(2\theta)-r^2\sin(2\theta)}|\leq e^{-R^2}=M
$$
because $r=R$ on this contour and $\sin(2\theta)\leq1$.  While,
$$
L=\frac{\pi R}{4}
$$
since we are looking at $1/8$th of the perimeter of the circle with radius $R$.  By the ML test
$$
\left|\int_{\Gamma_2} e^{iz^2}dz\right| \leq e^{-R^2}\frac{\pi R}{4}
$$
which goes to $0$ as $R\rightarrow\infty$.
Now we want to deal with the 3rd contour integral $\Gamma_3$.  Fortunately, the contour we picked allows us to easily parameterize this integral, as $y=x$.  We will also need $z^2=(x+iy)^2=x^2-y^2+2ixy$.  Recalling that $dz=dx+idy$ the integral becomes
$$
\int_{\Gamma_3} e^{i(x^2-y^2)-2xy}(dx+idy)
=\int_{R}^{0} e^{-2x^2}dx+i\int_{R}^{0} e^{-2y^2}dy
\rightarrow-\sqrt{\frac{\pi}{8}}(1+i)\ \text{as}\ R\rightarrow0
$$
from our real Gaussian integral identities.
Taking $R\rightarrow\infty$, our results for the contour integrals in our Cauchy's Theorem equation imply that
$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{\pi}{8}}(1+i)
$$
The integral from $-\infty$ to $\infty$ is just twice this.  So boom.
If you want, you can rewrite $e^{ix^2}=\cos(x^2)+i\sin(x^2)$ and equate the real and imaginary parts in the last equation and you will get the limiting values of the Fresnel Integrals.
Boom.
Also, since
$$
(1+i)=\sqrt{2}e^{i\pi/4}=\sqrt{2e^{i\pi/2}}=\sqrt{2i}
$$
we have
$$
\int_0^\infty e^{ix^2}dx = \sqrt{\frac{i\pi}{4}} = \frac{1}{2}\sqrt{-\frac{\pi}{i}}
$$
which exactly matches the well-known Gaussian integral identity
$$
\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{\alpha}}
$$
with $\alpha=-i$.  Boom.  Thus, this suggests that this identity can work for imaginary $\alpha$, and possibly certain complex $\alpha$ with the right combination of real and imaginary parts as well as choice of contours that do not make our integrals blow up.
