# Integral of cos(x^2)

Problem originally posted in a technical math section, so missing the template.
So i am trying to find the ##\int_{0}^{\infty} cos(x^2) dx##. I used Eulers identity to get ##\int_{0}^{\infty} cos(x^2) - isin(x^2) dx = \int_{0}^{\infty} e^{-i(x^2)} dx##. I squared this integral, changed to polar and evaluated and at the end of this process i got the result of ##\frac{1}{2} \sqrt{\frac{\pi}{i}}##. Where do i go from here? Am i even on the right track? Thanks for your help.

Mark44
Mentor
So i am trying to find the ##\int_{0}^{\infty} cos(x^2) dx##. I used Eulers identity to get ##\int_{0}^{\infty} cos(x^2) - isin(x^2) dx = \int_{0}^{\infty} e^{-i(x^2)} dx##.
I see where your second equation comes from, but how do you get back to ##\int_0^{\infty}\cos^2(x)dx##?
IOW, how is ##\int_0^{\infty}\cos^2(x)dx = \int_{0}^{\infty} \cos(x^2) - i\sin(x^2) dx##?
occh said:
I squared this integral, changed to polar and evaluated and at the end of this process i got the result of ##\frac{1}{2} \sqrt{\frac{\pi}{i}}##. Where do i go from here? Am i even on the right track? Thanks for your help.

well this problem gives you the integrals: ##\int_0^{\infty} cos(x^2) dx ## and ##\int_0^{\infty} sin(x^2) dx## and suggested proceeding in this manner to find the values of both of them by extracting the real and imaginary parts at the end after you solve the gaussian integral. This is where I am lost. I've viewed other threads on this topic this result appears there however they go through other steps to get the final result that i wasn't able to follow.

Mark44
Mentor
What do you mean, "gives you the integrals"? It might be helpful to see the problem as stated.

At any rate, this problem belongs in the homework section, so I'm moving it there.

stevendaryl
Staff Emeritus
This is an unusual integral, because usually, convergent integrals from $0$ to $\infty$ involve a function that goes to zero as $x \rightarrow \infty$. In the case of $cos(x^2)$, it doesn't go to zero, but instead oscillates faster and faster, so the contribution from large values of $x$ tend to cancel out.

The way I would go about it is to use:

$cos(x^2) = Re(e^{i x^2})$

where $Re$ means the real part. So if the integral converges, then we can write:

$\int_0^\infty cos(x^2) dx = Re(\int_0^\infty e^{-i x^2} dx)$

How do you evaluate the right-hand side? I'm a little shaky about how to do it rigorously, but we know that for any $\lambda$ with a positive real part,

$\int_0^\infty e^{-\lambda x^2} dx = \frac{\sqrt{\pi}}{2 \sqrt{\lambda}}$

If we assume (and this is where it gets a little hand-wavy) that this holds even when $\lambda = i$, then we would have:
$\int_0^\infty e^{-i x^2} dx = \frac{\sqrt{\pi}}{2 \sqrt{i}}$

$\frac{1}{\sqrt{i}} =\pm \frac{1-i}{\sqrt{2}}$

So plug that in, take the real part, and (hopefully) you have your answer. You have to choose the $+$ sign so that your integral is positive.

Okay this yields the correct answer, for both the integrals, with the real part being for ##cos(x^2)## and (the same thing) for ##sin(x^2)## in the imaginary. Thank you very much! I guess i just needed to brush up on my manipulation of complex numbers.

Ray Vickson