Does $\iint_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dx\,dy$ converge?

Question

Consider $$\iint_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dx\,dy$$
My try: changing to polar coordinates and then calculate the integral $$\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,dr\,d\theta\underset{r^2=u}{=}\pi\int_{0}^{\infty}\frac{\sin(u)}{u}\,du,$$ which converges but I saw in the solutions of a test that this integral diverges. Was I wrong?

Answer 1

As you suggested, let's change it to polar coordinates: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dxdy$$
$$=\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,rdr d\theta$$
$$=\pi\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,2rdr$$ Let $t=r^2$, $dt= 2r\,dr$: $$ \Rightarrow \pi\int_{0}^{\infty}\frac{\sin(t)}{t}\,dt $$

This integral is convergent and its value is $\pi/2$; see, for instance, https://dlmf.nist.gov/5.9.7 or Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?. So the final answer is ${\pi^2}/{2}$.