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.
Solution 1:
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 $$ z^2 = r^2e^{2i\theta} = r^2cos(2\theta)+ir^2sin(2\theta) $$ $$ |e^{iz^2}|=|e^{ir^2cos(2\theta)-r^2sin(2\theta)}|\leq e^{-R^2}=M $$ 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 $e^{ix^2}=cos(x^2)+isin(x^2)$ 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.