Determine whether the following integral converges
Determine whether the following integral converges: $$ \iint_D \frac{\sin(x^2+y^2)}{(x^2+y^2)^{3/2}}\,dx\,dy, $$ where $D$ is the region $x^2+y^2<a^2$.
Explanation of Method
The only candidate singularity is at the origin. In polar coordinates, the Jacobian factor $r$ combines with the $r^{-3}$ singularity to leave $\sin(r^2)/r^2$ — and since $\sin(r^2)\sim r^2$ for small $r$, this ratio tends to the finite limit $1$ rather than blowing up. The singularity is therefore removable, and the integral is not actually improper at all.
- Convert to polar coordinates.
- Examine the behavior of $\sin(r^2)/r^2$ as $r\to0^+$ using the Taylor expansion of $\sin$.
- Conclude convergence, since the radial integrand extends to a bounded, continuous function on $[0,a]$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$:
$$ I = \int_0^{2\pi}\int_0^a \frac{\sin(r^2)}{r^3}\cdot r\,dr\,d\theta = 2\pi\int_0^a\frac{\sin(r^2)}{r^2}\,dr. $$
Step 2 · Examine the singularity at $r=0$
Using $\sin t=t-\dfrac{t^3}6+O(t^5)$ with $t=r^2$:
$$ \sin(r^2) = r^2-\frac{r^6}6+O(r^{10}), \qquad \frac{\sin(r^2)}{r^2} = 1-\frac{r^4}6+O(r^8). $$
So the radial integrand extends continuously to $r=0$ with value $1$:
$$ f(r)=\begin{cases}\dfrac{\sin(r^2)}{r^2}, & r>0,\\[4pt] 1, & r=0,\end{cases} $$
is continuous (indeed smooth) on all of $[0,a]$.
Step 3 · Conclude convergence
A continuous function on the closed bounded interval $[0,a]$ is automatically Riemann-integrable there, with no impropriety to check. Hence:
$$ \int_0^a \frac{\sin(r^2)}{r^2}\,dr $$
is a genuine (proper) integral, and therefore $I=2\pi\displaystyle\int_0^a f(r)\,dr$ converges.
Additional Notes
- This is structurally identical to Problem 104's removable-singularity mechanism ($r\ln r\to0$), but here the cancellation comes from a Taylor expansion rather than a standard limit formula.
- Contrast with the closely related Problem 107, $\displaystyle\iint_D\frac{\cos(x^2+y^2)}{x^2+y^2}\,dx\,dy$: there $\cos(r^2)\to1$ (it does not vanish at $r=0$), so no cancellation occurs and that integral diverges. The presence of $\sin$ rather than $\cos$ — and the matching power of $r$ in the denominator — is exactly what makes this integral convergent.
- Because $D$ is a bounded region, only the behavior at the origin needs to be checked; there is no issue at infinity to consider, unlike the improper integrals over unbounded regions elsewhere in this collection.
Result
The integral converges: