Determine whether the following integral converges

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

Step 1 · Convert to polar coordinates

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

$$ I = \int_0^{2\pi}\int_0^a \frac{e^{-r^2}}{r^2}\cdot r\,dr\,d\theta = 2\pi\int_0^a\frac{e^{-r^2}}r\,dr. $$

Step 2 · Compare near $r=0$

For $0<r<1\leqslant a$ (or simply for $r$ in a neighborhood of $0$), $e^{-r^2}\geqslant e^{-1}$ is bounded below by a positive constant, so:

$$ \frac{e^{-r^2}}r \geqslant \frac{e^{-1}}r \qquad\text{for } 0<r<1. $$

Step 3 · Apply the comparison test

Since $\displaystyle\int_0^{1}\frac{dr}r=+\infty$ (the elementary logarithmic divergence), the comparison $\dfrac{e^{-r^2}}r\geqslant\dfrac{e^{-1}}r>0$ forces:

$$ \int_0^{a}\frac{e^{-r^2}}r\,dr \geqslant e^{-1}\int_0^{1}\frac{dr}r = +\infty. $$

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

Additional Notes

• This contrasts sharply with Problem 104: there, the singularity ($\ln r$, multiplied by the Jacobian factor $r$) was weaker than $1/r$ and hence integrable; here the singularity ($1/r^2$ in the original integrand, multiplied by the same Jacobian factor $r$) is exactly $1/r$, the borderline non-integrable case.
• General principle: in polar coordinates, a radial factor $g(r)\sim r^{-k}$ combined with the Jacobian $r$ gives an integrand $\sim r^{1-k}$ near the origin, integrable iff $1-k>-1$, i.e. $k<2$. Here $k=2$ sits exactly on the (non-convergent) boundary.