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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?

Integrand
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convxy
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  • You are not wrong. The improper integral converges. – Mark Viola Jun 18 '20 at 17:08
  • Looking at your last four posts, I feel obliged to tell you that this site is not a homework-checking service, even though that’s what people are doing for you. – gen-ℤ ready to perish Jun 18 '20 at 18:15
  • Careful! The integrand changes sign, so you need to check the positive and negative parts separately. https://en.wikipedia.org/wiki/Improper_integral#Functions_with_both_positive_and_negative_values – Hans Lundmark Jun 18 '20 at 18:49

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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 ${\pi^2}/{2}$.

Integrand
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    All you did was rewrite what the OP did, using $t$ instead of $u$, and then tell him the value. –1 for sure – gen-ℤ ready to perish Jun 18 '20 at 18:35
  • It's not the most insightful answer, I'll give you that, but it does answer the OP's question. In retrospect, this might be best as a comment, but since it's accepted I can't delete it. – Integrand Jun 18 '20 at 20:54