Determine whether the following integral converges
Determine whether the following integral converges, and if so, evaluate it: $$ \iint_D \ln\sqrt{x^2+y^2}\,dx\,dy, $$ where $D$ is the region $x^2+y^2<a^2$.
Explanation of Method
The only candidate singularity is at the origin, where $\ln\sqrt{x^2+y^2}\to-\infty$. In polar coordinates this becomes $\int_0^a r\ln r\,dr$; the extra factor of $r$ from the Jacobian tames the logarithmic blow-up completely ($r\ln r\to0$ as $r\to0^+$), so the integral is not actually improper at all once written this way — it converges, and can be evaluated directly by integration by parts.
- Convert to polar coordinates.
- Verify that $r\ln r\to0$ as $r\to0^+$, so no singularity survives.
- Evaluate $\int_0^a r\ln r\,dr$ by integration by parts.
Step 1 · Convert to polar coordinates
Since $\ln\sqrt{x^2+y^2}=\frac12\ln(x^2+y^2)=\ln r$, and $dx\,dy=r\,dr\,d\theta$:
$$ I = \int_0^{2\pi}\int_0^a (\ln r)\cdot r\,dr\,d\theta = 2\pi\int_0^a r\ln r\,dr. $$
Step 2 · Confirm the singularity is removable
By L'H\^opital's rule (or the standard limit $x\ln x\to0$):
$$ \lim_{r\to0^+} r\ln r = \lim_{r\to0^+}\frac{\ln r}{1/r} = \lim_{r\to0^+}\frac{1/r}{-1/r^2} = \lim_{r\to0^+}(-r) = 0. $$
So $r\ln r$ extends continuously (by the value $0$) to $r=0$, and $\int_0^a r\ln r\,dr$ is a proper (not improper) integral.
Step 3 · Evaluate by integration by parts
Let $u=\ln r$, $dv=r\,dr$, so $du=\frac{dr}r$, $v=\frac{r^2}2$:
$$ \int_0^a r\ln r\,dr = \left[\frac{r^2}2\ln r\right]_0^a - \int_0^a\frac{r}2\,dr. $$
The boundary term at $r=0$ vanishes (by the same limit argument as Step 2, since $r^2\ln r\to0$ even faster):
$$ \int_0^a r\ln r\,dr = \frac{a^2}2\ln a - \frac12\left[\frac{r^2}2\right]_0^a = \frac{a^2}2\ln a - \frac{a^2}4. $$
Step 4 · Assemble the result
$$ I = 2\pi\left(\frac{a^2}2\ln a-\frac{a^2}4\right) = \pi a^2\ln a - \frac{\pi a^2}2 = \pi a^2\left(\ln a-\frac12\right). $$
Additional Notes
- Sanity check at $a=1$: the formula gives $I=\pi\left(0-\frac12\right)=-\frac\pi2$, matching the direct computation $2\pi\int_0^1 r\ln r\,dr=2\pi\left(-\frac14\right)=-\frac\pi2$.
- More generally, $\displaystyle\int_0^{\epsilon}r^\beta\ln r\,dr$ converges for any $\beta>-1$ (here $\beta=1$), since the polynomial factor $r^{\beta+1}$ always beats the logarithm's divergence at the origin — this is the same mechanism that makes $x^\beta\ln x\to0$ as $x\to0^+$ for $\beta>0$.
Result
The integral converges, and its value is: