I am trying to prove that $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \sin(x^2+y^2)\, dx\,dy$ diverges and I did it like this:
$x=r\cos \theta$, $y=r\sin\theta$, $\theta \in [0,2\pi]$, $r\in [0,\infty]$, and the jacobian is $r$:
$$\int_{-\infty}^\infty \int_{-\infty}^\infty \sin(x^2+y^2) \, dx \, dy=\int_0^{2\pi}\int_{0}^\infty r\sin(r^2) \, dr \, d\theta = 2\pi \int_0^\infty r\sin(r^2) \, dr$$
Use the variable change $v=r^2$, $dv=2r \, dr$ and so:
$$2\pi \int_0^\infty r\sin(r^2)\,dr = 2\pi \int_{0}^{\infty}r\sin(v)\,\frac{dv}{2r}=\pi \int_0^\infty \sin(v) \, dv $$
I checked wolfram, and it says that $\int_0^\infty \sin(x) \, dx$ diverges, so I must conclude that the original integral diverges as well.
But how do I actually prove that $\int_0^\infty \sin(x)\,dx$ diverges?
We can compute the limiting integral by writing $$\int_0^\infty \sin(x) dx = \lim_{b\to\infty}\int_0^b \sin(x) dx = \lim_{b\to\infty} ( \cos(b)-1 )$$ We know that the limit $\lim_{b\to\infty} \cos(b)$ diverges. This means the integral diverges as well.