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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.

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  • $\begingroup$ Try Google or searching the site ;) $\endgroup$ Commented Dec 9, 2016 at 16:07
  • $\begingroup$ Please clarify your question. What Fresnel integral are you trying to solve? Write it out in Latex. $\endgroup$ Commented Dec 9, 2016 at 16:07
  • $\begingroup$ @SimpleArt I've read tons of articles of Fresnel's integrals and nothin :( $\endgroup$ Commented Dec 9, 2016 at 16:23
  • $\begingroup$ @Rohan I believe so, my teacher talks about it like it is very simple $\endgroup$ Commented Dec 9, 2016 at 16:39
  • $\begingroup$ The integrals of both $\sin$ and $\cos$ are evaluated in this answer. $\endgroup$
    – robjohn
    Commented Dec 12, 2016 at 6:21

6 Answers 6

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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 $$\int_{0}^{\infty}\cos(ax^2) dx =\sqrt{\frac{\pi}{8a}}$$

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  • $\begingroup$ Thank you, but we didn't see those theorems and formulas hahahahah. I aprecciate ur help tho $\endgroup$ Commented Dec 9, 2016 at 16:56
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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.

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  • $\begingroup$ I'm very sorry, but could you be a little more specific? hahaha $\endgroup$ Commented Dec 9, 2016 at 16:32
  • $\begingroup$ Oh I got it now, thanks.. The only problem is that we didn't learn that Gaussian integral, I believe that sadly that is not the way, :( $\endgroup$ Commented Dec 9, 2016 at 16:50
  • $\begingroup$ You could show the Gaussian integral by essentially the same methods... But then your problem would take up several pages of computation. It may be a horrible way to go, but it is a way. $\endgroup$ Commented Dec 9, 2016 at 16:54
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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

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

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    $\begingroup$ OMG, you are a genius!!! This question is about Fourier Transforms basically.. I need to proof that the function $\frac{1}{\sqrt{x}}$ is self reciprocal, and when we do that we get that Fresnel integral and the teacher wants us to proof it! Thank you!!! $\endgroup$ Commented Dec 9, 2016 at 17:53
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    $\begingroup$ By self reciprocal I mean self adjoint I guess...English is not my native language, that is probably a false cognate $\endgroup$ Commented Dec 9, 2016 at 17:55
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    $\begingroup$ this: $$\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)cos(tx)dx=f(t)=\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)sin(tx)dx$$ $$f(x)=\frac{1}{\sqrt{x}}$$ $\endgroup$ Commented Dec 9, 2016 at 18:24
  • $\begingroup$ You are right, I wasn't familiar with the term self-reciprocal at all until I just looked it up. I thought about it as an eigenfunction. But self-reciprocal does not mean self-adjoint. $\endgroup$ Commented Dec 9, 2016 at 18:25
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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.

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  • $\begingroup$ What does that "Re" means? I think that you solved the answer hahaha $\endgroup$ Commented Dec 9, 2016 at 17:27
  • $\begingroup$ @RafaelVignoli It means the real part. $\endgroup$ Commented Dec 9, 2016 at 17:27
  • $\begingroup$ Can I do the same thing to proof the Fresnel Integral for sin(x)? $\endgroup$ Commented Dec 11, 2016 at 20:41
  • $\begingroup$ And how do you get to the answer $\sqrt{\frac{\pi}{8}}$ $\endgroup$ Commented Dec 11, 2016 at 20:43
  • $\begingroup$ @RafaelVignoli About the first question, the answer is yes. You have to consider the imaginary part instead of the real part. About the second question, you have to evaluate the last integral but it is not difficult, you may integrate it using partial fractions. $\endgroup$ Commented Dec 11, 2016 at 21:37
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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:

$$\int_{-\infty}^{\infty}ae^{-(x-b)^2/2c^2}dx=\sqrt{2}a\,|c|\sqrt{\pi}$$

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$$ \begin{aligned} \int_{\infty}^{\infty}\left[\cos \left(t^2\right)-i \sin \left(t^2\right)\right] d t = & \int_{-\infty}^{\infty} e^{-t^2 i} d t \\ = & \int_{-\infty}^{\infty} e^{-\left[\frac{(1+i) t}{\sqrt{2}}\right]^2} d t \\ = & \frac{\sqrt{2}}{1+i} \sqrt{\pi} \\ = & \sqrt{\frac{\pi}{2}}(1-i) \end{aligned} $$ 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 $$ \int_0^{\infty} \cos \left(t^2\right) d t=\frac{1}{2} \sqrt{\frac{\pi}{2}} $$

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