Computation of an iterated integral

Question

I want to prove $$\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty\frac{\sin(x^2+y^2)}{x^2+y^2}dxdy=\frac{\pi^2}{2}.$$ Since the function $(x,y)\mapsto\sin(x^2+y^2)/(x^2+y^2)$ is not integrable, I can't use the Theorem of Change of Variable. So, I'm trying to use residue formulae for some suitable holomorphic function to compute the inner integral, but I can't continue. Can someone suggest me a hint to solve this problem?

Addendum: I may be wrong, but I suspect Theorem of Change of Variable (TCV) is not the answer. The reason is the following: the number $\pi^2/2$ is gotten if we apply polar coordinates, but TCV guarantees that if we apply any other change of variable we can get the same number, $\pi^2/2$. If this function were integrable, this invariance property would be guaranteed, but it is not the case. Thus we may have strange solutions to this integral.

Answer 1

using polar coordinates we have:

$$ \lim_{R\rightarrow\infty}\int_{x^2+y^2 \le R^2} \frac{\sin(x^2+y^2)}{x^2+y^2}dxdy=\lim_{R\rightarrow\infty}\int_0^{R}\int_0^{2\pi}\frac{\sin (r^2)}{r^2}r d\theta dr= $$ $$ =2\pi\int_0^{+\infty} \frac{\sin (r^2)}{r}dr=2\pi \frac{\pi}{4}=\frac{\pi^2}{2} $$

Answer 2

Let $D$ be any Jordan domain in $\mathbb{R}^2$, containing origin in its interior, whose boundary $\partial D$ has the form $r = f(\theta)$ in polar coordinates where $f \in C[0,2\pi]$.

Consider following integral as a functional of $D$: $$\mathcal{I}_D \stackrel{def}{=} \int_D \phi(x,y) dx dy \quad\text{ where }\quad\phi(x,y) = \frac{\sin(x^2+y^2)}{x^2+y^2} $$

Since the origin is a removable singularity for $\phi(x,y)$, as long as $D$ is of finite extent, there isn't any issue about integrability or change of variable. We have $$\mathcal{I}_D = \int_0^{2\pi} \int_0^{f(\theta)}\frac{\sin(r^2)}{r^2} rdr d\theta = \frac12\int_0^{2\pi} \left[\int_0^{f(\theta)^2}\frac{\sin t}{t} dt \right] d\theta $$ For any non-increasing, non-negative function $g$ on $(0,\infty)$. Using integration by part (the RS version), one can show that

$$\left|\int_a^b g(x) \sin(x) dx \right| \le 2 g(a)\quad\text{ for }\quad 0 < a < b < \infty$$

For any $R > 0$ where $B(0,R) \subset D$. By setting $g(x)$ to $1/x$, above inequality leads to following estimate for $\mathcal{I}_D$.

$$\left| \mathcal{I}_D - \mathcal{I}_{B(0,R)} \right| = \frac12 \left| \int_0^{2\pi} \left[\int_{R^2}^{f(\theta)^2}\frac{\sin t}{t} dt \right] d\theta \right| \le \frac12 \int_0^{2\pi} \left|\int_{R^2}^{f(\theta)^2}\frac{\sin t}{t} dt\right| d\theta \le \frac{2\pi}{R^2} $$

For any fixed $Y$, the integrand $\phi(x,y)$ is Lebesgue integrable over $(-\infty,\infty)\times [-Y,Y]$. Double integral of the form below is well defined. With help of DCT, one can evaluate it as a limit

$$\int_{-Y}^Y \int_{-\infty}^{\infty}\phi(x,y) dxdy = \lim_{X\to\infty}\int_{-Y}^Y \int_{-X}^X \phi(x,y) dxdy = \lim_{X\to\infty}\mathcal{I}_{[-X,X]\times[-Y,Y]}$$

We will combine this with above estimation. By setting $R = Y$ and $[-X,X] \times [-Y,Y]$ taking the role of $D$, one get

$$\left|\int_{-Y}^{Y} \int_{-\infty}^{\infty}\phi(x,y) dxdy - \mathcal{I}_{B(0,Y)}\right| \le \limsup_{X\to\infty}\left|\int_{-Y}^{Y} \int_{-X}^{X}\phi(x,y) dxdy - \mathcal{I}_{B(0,Y)}\right| \le \frac{2\pi}{Y^2}$$

Since following two limits exist, $$\lim_{Y\to\infty} \mathcal{I}_{B(0,Y)} = \lim_{Y\to\infty} \pi\int_0^{Y^2}\frac{\sin t}{t}dt = \pi\int_0^\infty \frac{\sin t}{t} dt = \frac{\pi^2}{2} \quad\text{ and }\quad \lim_{Y\to\infty}\frac{2\pi}{Y^2} = 0$$ By squeezing, the double integral at hand exists as an improper integral!

$$\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(x,y) dxdy \stackrel{def}{=} \lim_{Y\to\infty} \int_{-Y}^Y \int_{-\infty}^\infty \phi(x,y) dxdy = \lim_{Y\to\infty} \mathcal{I}_{B(0,Y)} = \frac{\pi^2}{2}$$