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So I was looking through the homepage of Youtube to see if there were any math problems that I thought that I might be able to solve when I came across this video by Maths$505$ evaluating both $\int_0^\infty\sin(x^2)dx$ and $\int_0^\infty\cos(x^2)dx$, however, I decided to only evaluate $\int_0^\infty\cos(x^2)$ to not put too much strain on myself. Here is my attempt at evaluating $\int_0^\infty\cos(x^2)dx$:

Now here's what we want to do: We want to evaluate $$\int_0^\infty e^{-ix^2}dx$$ because we can rewrite this as $$\int_0^\infty\cos(x^2)dx-i\int_0^\infty\sin(x^2)dx$$because we only need to get the real part of $$\int_0^\infty e^{-ix^2}dx$$because that is what is going to be the value of $$\int_0^\infty\cos(x^2)dx$$

$$\int_0^\infty e^{-ix^2}dx$$Now, using $u$-substitution here,$$x^2=u$$$$x=\sqrt u$$$$dx=\dfrac1{2\sqrt u}du$$$$I=\dfrac12\int_0^\infty u^{-\frac12}e^{-iu}du$$$$\implies I=\dfrac12\int_0^\infty x^{-\frac12}e^{-ix}dx$$$$\implies I=\dfrac12\mathcal{M}\{e^{-ix}\}(s=\dfrac12)$$Where $\mathcal{M}\{f(x)\}(s)$ is a Mellin transform. Now to expand $e^{-ix}$ to represent as a sum. Doing this, we get $$e^{-ix}=\sum_{k\geq0}\dfrac{(-x)^ki^k}{k!}$$where $\sum_{k\geq0}$ is a sum over the nonnegative integers including the element $0$. Our "$w(k)$" function that is needed to perform the Mellin transform is $w(k)=i^k,\quad i=\sqrt{-1}$. We can then expand our $w(k)$ to get $w(k)=e^{a\log i}$, which therefore implies $w(k)$ can be written as $$w(k)=e^{\frac\pi2iz}$$Knowing this, this then implies$$I=\dfrac12\Gamma(\dfrac12)i^{-\frac12}$$$$=\dfrac12\sqrt\pi(e^{i\frac\pi2})^{-\frac12}$$$$=\dfrac12\sqrt\pi e^{-i\frac\pi4}$$$$=\dfrac12\sqrt\pi(\cos(\dfrac\pi4)-i\sin(\dfrac\pi4))$$$$=\dfrac12\sqrt{\dfrac\pi2}-\dfrac i2\sqrt{\dfrac\pi2}$$$$\implies\int_0^\infty\cos(x^2)dx=\sqrt{\dfrac\pi8}$$

My question

Is my evaluation of the integral correct, or what could I do to evaluate it more easily?

Mistakes I might have made

  1. The way I evaluated the integral
  2. U-substitution
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  • also here: https://math.stackexchange.com/questions/4347941/int-0-infty-frac-sin-xxp-dx – MathFail Jul 20 '23 at 19:56
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    This is not necessarily a duplicate because the OP is looking more for a solution verification – FShrike Jul 20 '23 at 19:57
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    This looks ok as an evaluation but to make it fully rigorous you'll have to work harder. In particular, I'm not sure what method you used to arrive at $\frac{1}{2}\Gamma(1/2)i^{-1/2}$ but you should be aware that $i^{-1/2}$ is not a well-defined expression without further context. Whatever the method is, you need to make sure that you get the correct value of $i^{-1/2}$ (one of infinitely many alternatives) – FShrike Jul 20 '23 at 20:53
  • Your Mellin Transform takes inputs $k$? How does $k$ play a role in $w(k)=e^{\frac{\pi}{2} iz}$ and $w(k)=e^{a\log i}$? – Accelerator Jul 20 '23 at 22:57
  • My bad... not the Mellin Transform itself. I meant your $w(k)$ you used for the Mellin Transform. – Accelerator Jul 21 '23 at 00:41

1 Answers1

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(Too long for a comment)

Complex gaussian integrals are actually quite common in quantum field theory and, fortunately, the real case formula is also valid in this context, hence $$ \int_0^\infty e^{-iz^2} \mathrm{d}z = \sqrt{\frac{\pi}{4i}} = \sqrt{\frac{\pi}{4}} e^{-i\pi/4}. $$ A standard derivation of this result uses contour integration on an eighth of a circle, with caution to be taken with the complex phase; you can refer to this document for a brief demonstration.

In consequence, the Fresnel integral you want to compute is simply given by $$ \int_0^\infty \cos(x^2) \,\mathrm{d}x = \Re\left(\sqrt{\frac{\pi}{4}} e^{-i\pi/4}\right) = \sqrt{\frac{\pi}{4}} \cdot \frac{1}{\sqrt{2}} = \sqrt{\frac{\pi}{8}} $$

Abezhiko
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