Proof of Fresnel's Integral

We have to prove: $$\int_{0}^{\infty} \cos(ax^2) dx =\sqrt{\frac{\pi}{8a}}$$ Now, we consider the LHS. $$ LHS = \int_{0}^{\infty} \cos(ax^2) dx $$

Now, we make the substitution: $$x \rightarrow x^{\frac{1}{4}}$$

Therefore, we get: $$ LHS =\frac{1}{4}\int_{0}^{\infty} x^{-\frac{3}{4}}\cos(a\sqrt{x}) dx $$ Now by Maclaurin series, $$\cos (a\sqrt{x}) = \sum\limits_{n=0}^{\infty} \frac{(-1)^n(a\sqrt x)^{2n}}{2n!}$$ This can also be written as: $$\cos (a\sqrt x) = \sum\limits_{n=0}^{\infty} \frac{(-x)^n(a)^{2n}n!}{2n!n!}$$ On plugging the value into LHS, we get: $$ LHS =\frac{1}{4} \int_{0}^{\infty} x^{-\frac{3}{4}} \sum\limits_{n=0}^{\infty} \frac{(-x)^n(a)^{2n}n!}{2n!n!} dx$$ Now, by Ramanujan's Master Theorem, we get $$ LHS = \frac{1}{4} \int_{0}^{\infty} x^{-\frac{3}{4}}\sum\limits_{n=0}^{\infty}\frac{(-x)^n(a)^{2n}n!}{2n!n!} dx = \frac{1}{4}\frac{\Gamma(\frac{1}{4})\Gamma(\frac{3}{4})}{\sqrt{a}\Gamma(\frac{1}{2})}$$ Therefore, by Euler's Reflection Formula, we get $$\int_{0}^{\infty}\cos(ax^2) dx =\sqrt{\frac{\pi}{8a}}$$

You can use double integrals and change of variables. Consider the surface $e^{-y^2}\cos(x^2)$ in the positive octant, and determine the volume bounded by it. Compute the integrals both in polar coordinates and in Cartesian coordinates, and equate the values. Then you can solve for the value of your integral.

To do it in Cartesian coordinates, you have a product of the Fresnel integral, $I$ and the Gaussian integral with value $\frac{\sqrt{\pi}}{2}$. Then

$$V = \frac{\sqrt{\pi}}{2}I$$

By expressing $V$ with polar cooridinates, and using the substitution $u = p^2\cos^2(\theta)$ then the substitution $\tan(\theta)=t$, you get the integral

$$V = \frac{1}{2}\int_{0}^{\infty}\frac{t^4}{t^4+1}$$ Using partial fractions you can get that this is $\frac{\pi\sqrt{2}}{8}$. Then by equating your two values of $V$ you can solve for I and get $$I = \sqrt{\frac{\pi}{8}}$$ as desired.

Here is a (non-rigorous) proof with Fourier Transforms I discovered. It involves double integrals mostly.

Define the Fourier Cosine Transform of $f(x)$ to be:

$$\mathcal{F}_{c}(f(x))=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(y) \cos(xy) \ dy.$$ One can see that $\mathcal{F}_c$ is self-adjoint by checking the definition using the inner product: $$\langle f(x), g(x) \rangle= \int_{\mathbb{R}} f(x)g(x) \ dx. $$ The definition more specifically is $$\left \langle \mathcal{F}_c(f), g \right \rangle=\left \langle f, \mathcal{F}_c (g) \right \rangle.$$ To prove it is true, expand the left hand side into a double integral, and use Fubini's theorem to get the right hand side.

We compute the inner product of $$\left \langle \mathcal{F}_c \left(\cos(x) \right), \frac{1}{\sqrt{|x|}} \right \rangle$$ in two ways.

Expanding the inner product, we get the inner product equal to $$\left \langle \sqrt{2 \pi} \delta(1-x) , \frac{1}{\sqrt{|x|}} \right \rangle= \sqrt{2 \pi},$$ from the integral property of Dirac Delta function $\delta(x).$

On the other hand, exploiting the self-adjoint property, $$\sqrt{2 \pi}=\left \langle \cos(x), \mathcal{F}_c \left( \frac{1}{\sqrt{|x|}} \right ) \right \rangle =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\cos(x) \cos(xy)}{\sqrt{|y|}} \ dy \ dx.$$

Apply the change of variables $$x=u, y=\frac{v}{u}$$ which has Jacobian $$\frac{\partial(x,y)}{\partial(u,v)}= \frac{1}{u}$$ to get

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\cos(u) \cos(v)}{\sqrt{|uv|}} \ dv \ du.$$

Since the first half of the proof showed that this integral is $\sqrt{2 \pi},$ we see $$2\pi= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{\cos(u) \cos(v)}{\sqrt{|uv|}} \ dv \ du= \left (\int_{-\infty}^{\infty} \frac{\cos(u)}{\sqrt{|u|}} \ du \right)^2.$$

So $$\sqrt{2 \pi}=\int_{-\infty}^{\infty} \frac{\cos(u)}{\sqrt{|u|}} \ du=2 \int_{0}^{\infty} \frac{\cos(u)}{\sqrt{u}} \ du. $$

As a result,

$$\frac{\sqrt{\pi}} {\sqrt{2}}= \int_{0}^{\infty} \frac{\cos(u)}{\sqrt{u}} \ du.$$ Lastly, let $u=z^2, du = 2z \ dz$ to get that

$$\frac{\sqrt{\pi}} {\sqrt{2}}= \int_{0}^{\infty} 2 \cos(z^2) \ dz,$$ so

$$\int_{0}^{\infty} \cos(z^2) \ dz =\frac{\sqrt{\pi}}{\sqrt{8}}.$$