Determine whether the following integral converges

Determine whether the following integral converges: $$ \iint_D \frac{\cos(x^2+y^2)}{x^2+y^2}\,dx\,dy, $$ where $D$ is the region $x^2+y^2<a^2$.

Step 1 · Convert to polar coordinates

With $x=r\cos\theta$, $y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$:

$$ I = \int_0^{2\pi}\int_0^a \frac{\cos(r^2)}{r^2}\cdot r\,dr\,d\theta = 2\pi\int_0^a\frac{\cos(r^2)}r\,dr. $$

Step 2 · Examine the behavior near $r=0$

Using $\cos t=1-\dfrac{t^2}2+O(t^4)$ with $t=r^2$:

$$ \cos(r^2) = 1-\frac{r^4}2+O(r^8) \longrightarrow 1 \quad\text{as } r\to0. $$

So, for $0<r<1\leqslant a$ say, $\cos(r^2)\geqslant\cos(1)>0$, giving:

$$ \frac{\cos(r^2)}r \geqslant \frac{\cos(1)}r \qquad\text{for } 0<r<1. $$

Step 3 · Apply the comparison test

Since $\displaystyle\int_0^1\frac{dr}r=+\infty$:

$$ \int_0^a\frac{\cos(r^2)}r\,dr \geqslant \cos(1)\int_0^1\frac{dr}r = +\infty. $$

So the radial integral diverges, and hence so does $I$.

Additional Notes

• Comparison with Problem 106: $\displaystyle\iint_D\frac{\sin(x^2+y^2)}{(x^2+y^2)^{3/2}}\,dx\,dy$ converges precisely because $\sin(r^2)\sim r^2$ vanishes at the origin at exactly the rate needed to cancel the singularity's extra power of $r$; here $\cos(r^2)\to1\neq0$, so the same cancellation mechanism is unavailable, and this integral diverges. The distinguishing factor is whether the oscillatory numerator vanishes at the origin.
• This can also be seen via the substitution $u=r^2$, reducing the radial integral (up to a constant) to $\int_0^{\epsilon}\frac{\cos u}u\,du$, the classical (divergent) cosine-integral kernel near $u=0$.