Updated: Prove completely $\int^\infty_0 \cos(x^2)dx=\frac{\sqrt{2\pi}}{4}$ using Fresnel Integrals

Question

Prove completely $\int^\infty_0 \cos(x^2)dx=\frac{\sqrt{2\pi}}{4}$

I've tried substituting $ x^2 = t $ but could not proceed at all thereafter in integration. Any help would be appreciated.

I should mentioned at the start that I am trying to use Fresnel Integrals.

That's why I was trying to substitute t=x^2 since I'm nearly positive that is the first step. However, thereafter I am lost.

Answer 1

As is common, use $f(z)=e^{-iz^2}=\cos(z^2)-i\sin(z^2)$

Now $$\int_{-\infty}^{\infty}e^{-iz^2}{\rm d}z=\int_{-\infty}^{\infty}e^{-\left(e^{i\pi/4}z\right)^2}{\rm d}z=\frac1{e^{i\pi/4}}\int_{-\infty}^{\infty}e^{-x^2}{\rm d}x=e^{-i\pi/4}\sqrt{\pi}=\sqrt{\frac{\pi}2}-i\sqrt{\frac{\pi}2}$$ Now, since $f(z)$ is even: $$\int_0^{\infty}\cos(x^2){\rm d}x=\Re\left(\frac12\int_{-\infty}^{\infty}e^{-iz^2}{\rm d}z\right)=\frac12\sqrt{\frac{\pi}2}=\frac{\sqrt{2\pi}}4$$

Answer 2

Define $I(a)=\int_0^{\infty}\cos ax^2 dx$ for $a>0$. Note that $I(-a)=I(a)$. Now take a Laplace transform with respect to $a$ to obtain \begin{align} \mathcal{L}_{a \rightarrow s}\{I(a)\}&=\int_0^{\infty}\frac{s}{s^2+x^4} dx\\ &=\frac{\pi}{2\sqrt{2s}}. \end{align} Now take inverse Laplace transform to obtain \begin{align} I(a)&=\mathcal{L}^{-1}_{s \rightarrow a}\{\frac{\pi}{2\sqrt{2s}}\}\\ &=\frac14\sqrt{\frac{2\pi}{a}}. \end{align} For your problem we then have $I(1)=\frac{\sqrt{2\pi}}{4}$.