Compute the value of Fresnel's Integral So, my teacher wants us to compute the value of the Fresnel integral:
$$\int_0^\infty\cos(x^2)dx=\sqrt{\frac{\pi}{8}}$$
The problem is that we cannot use complex analysis to prove that and we should do that using the Euler identity:
$$\int_0^\infty\cos(x^2)dx=\frac{1}{2}\int_0^\infty e^{iw^2}dw+\frac{1}{2}\int_0^\infty e^{-iw^2}dw$$
But I have that integral of $e^{-iw^2}$ and I cannot solve that :(
I am an engineering student, so basically, I only have the "basic" calculus, just simple/double/triple integrals, some notions about series, ODE's and PDE's, but nothing as deep as in the regular Math degree, so probably there's no need to use hard stuff to figure this out.
 A: 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
$$V = \frac{1}{2}\int_{0}^{\infty}\frac{t^2}{t^4+1}dt$$ 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.
A: 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
$$\int_{0}^{\infty} \cos(z^2)  \ dz =\frac{\sqrt{\pi}}{\sqrt{8}}.$$
A: Hint: note that $$I=\int_{0}^{\infty}\cos\left(x^{2}\right)dx\stackrel{x^{2}=u}{=}\frac{1}{2}\int_{0}^{\infty}u^{-1/2}\cos\left(u\right)du
 $$ $$=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\cos\left(u\right)\int_{0}^{\infty}v^{-1/2}e^{-uv}dvdu
 $$ and now using the Fubini theorem we can exchange the integrals and get $$I=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}\cos\left(u\right)e^{-uv}dudv=\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}v^{-1/2}\int_{0}^{\infty}e^{u\left(i-v\right)}dudv\right)
 $$ $$\textrm{Re}\left(\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{-1/2}}{v-i}dv\right)=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\frac{v^{1/2}}{v^{2}+1}dv\stackrel{z=\sqrt{v}}{=}\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz
 $$ and the last integral is simple to evaluate using partial fractions. Note that $$\int_{0}^{\infty}\frac{z^{2}}{z^{4}+1}dz=\frac{1}{2\sqrt{2}}\left(\int_{0}^{\infty}\frac{z}{z^{2}-\sqrt{2}z+1}-\frac{z}{z^{2}+\sqrt{2}z+1}dz\right)
 $$ $$=\frac{1}{2\sqrt{2}}\left(\lim_{a\rightarrow\infty}\frac{1}{2}\int_{0}^{a}\frac{2z-\sqrt{2}+\sqrt{2}}{z^{2}-\sqrt{2}z+1}-\frac{2z+\sqrt{2}-\sqrt{2}}{z^{2}+\sqrt{2}z+1}dz\right)
 $$ $$=\frac{1}{2\sqrt{2}}\left(\lim_{a\rightarrow\infty}\frac{1}{2}\int_{0}^{a}\frac{2z-\sqrt{2}}{z^{2}-\sqrt{2}z+1}-\frac{2z+\sqrt{2}-\sqrt{2}}{z^{2}+\sqrt{2}z+1}dz+\int_{0}^{a}\frac{\sqrt{2}}{z^{2}-\sqrt{2}z+1}dz\right)$$ and I think you can conclude by yourself from here.
A: After using Euler's, use Gaussian Integral:
$$\int_{-\infty}^{\infty}e^{-a(x+b)^2}dx=\sqrt{
\frac{\pi}{a}}$$
or Integral of a Gaussian Function:
$$\int_{-\infty}^{\infty}ae^{-(x-b)^2/2c^2}dx=\sqrt{2}a\,|c|\sqrt{\pi}$$
A: 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
$$\int_{0}^{\infty}\cos(ax^2) dx =\sqrt{\frac{\pi}{8a}}$$
A: $$
\begin{aligned}
\int_{\infty}^{\infty}\left[\cos \left(t^2\right)-i \sin \left(t^2\right)\right] d t 
= & \int_{-\infty}^{\infty} e^{-t^2 i} d t \\
= & \int_{-\infty}^{\infty} e^{-\left[\frac{(1+i) t}{\sqrt{2}}\right]^2} d t \\
= & \frac{\sqrt{2}}{1+i} \sqrt{\pi} \\
= & \sqrt{\frac{\pi}{2}}(1-i)
\end{aligned}
$$
By comparing the real parts on both sides, we get
$$
\int_{-\infty}^{\infty} \cos \left(t^2\right) d t=\sqrt{\frac{\pi}{2}}
$$
Hence $$
\int_0^{\infty} \cos \left(t^2\right) d t=\frac{1}{2} \sqrt{\frac{\pi}{2}}
$$
