Integral of sine of x^2

[jaumzaum]

Hello guys. I was trying to evaluate the integral of sine of x^2 from - infinity to + infinity and ran into some inconsistencies. I know this integral converges to sqrt(pi/2). Can someone help me to figure out why I am getting a divergent answer?

$$ I = \int_{-\infty}^{+\infty} sin(x^2) dx = Im(\int_{-\infty}^{+\infty} e^{ix^2} dx) $$

Now let's call:

$$ A=\int_{-\infty}^{+\infty} e^{ix^2} dx $$
$$ A^2=\int_{-\infty}^{+\infty} e^{ix^2} dx \int_{-\infty}^{+\infty} e^{iy^2} dy $$
$$ A^2=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{i(x^2+y^2)} dx dy $$
$$ A^2=\int_{0}^{2\pi} \int_{0}^{+\infty} e^{ir^2}r dr d\theta $$
$$ A^2= -i\pi \left. e^{ir^2} \right|_0^{\infty} $$
$$ A^2= -i\pi (e^{i\infty}-1) $$

But we know $lim_{M->\infty} e^{iM}$ does not converge, so A^2 would not converge either!

However, if we consider $e^{i\infty}$ as being zero, we get:

$$A^2= i\pi$$
$$A=1/\sqrt{2}\pi(1+i)$$
$$I=\sqrt{\pi/2}$$

But why is $e^{i\infty}$ considered zero if $lim_{M->\infty} e^{iM}$ does not exist?

 

[anuttarasammyak]

I delete my message.
[EDIT]
Going forward your way of
$$A^2=i\pi(e^{iR^2}-1), R\rightarrow \infty$$
$$=-\pi \sin R^2 - i\pi(1-\cos R^2)=\pi \sqrt{2} \sqrt{1-\cos R^2}e^{i\theta}$$
where
$$\theta = \arctan \frac{1-\cos R^2}{\sin R^2}+\pi$$
$$A=\sqrt{\pi \sqrt{2} \sqrt{1-\cos R^2}}e^{i\theta/2}$$
$$I=Im\ A= \sqrt{\pi \sqrt{2} \sqrt{1-\cos R^2}} sin (\theta/2)$$
I let Wolfram plot the graph. I draw red line y=$1/\sqrt{2}$ of the expected limit.

 

[PeroK]

This might be how to do it properly, using line integrals in the complex plane.

 

[PeroK]

An informal way to do it would be to look at the integral: $$\int_0^\infty \sin (x^2) dx = \lim_{n \rightarrow \infty}\int_0^{\sqrt{n\pi}}\sin(x^2)dx$$Then follow your technique to get
$$A^2 = \lim_{n \rightarrow \infty}\int_0^{2\pi} \int_{-2n\pi}^{2n\pi}e^{ir^2}r dr d\theta$$Etc.

 

[Infrared]

I think there is an error in $\int_{-\infty}^{+\infty} e^{ix^2} dx \int_{-\infty}^{+\infty} e^{iy^2} dy =\int_{0}^{2\pi} \int_{0}^{\infty} e^{ir^2} r dr d\theta.$

It looks like you're trying to apply Fubini's theorem, but that requires the integrand to be absolutely integrable, which $e^{i(x^2+y^2)}$ is not. If you try to write out the argument formally with limits, you should see what goes wrong- your integral for $A^2$ would be over a large rectangle, but you would want it to be over a disk for polar coordinates to be useful. This distinction wouldn't matter in the case that your function is absolutely integrable, because then the difference integrals over the two domains would tend to zero.