Calculate the following improper integral, and determine
Calculate the following improper integral, and determine for which $\alpha\in\mathbb R$ it converges: $$ \iint_D \frac{dx\,dy}{(1-x^2-y^2)^{\alpha}}, $$ where $D$ is the region $x^2+y^2<1$.
Explanation of Method
The only possible singularity is at the boundary circle $r=1$, where $1-r^2\to0$. Polar coordinates isolate this into a single radial integral of the form $\displaystyle\int_0^1 u^{-\alpha}\,du$ (after substituting $u=1-r^2$), whose convergence at $u=0$ depends purely on whether the exponent $\alpha$ is less than $1$ — exactly analogous to the elementary one-dimensional integral $\int_0^1(1-x)^{-\alpha}dx$.
- Convert to polar coordinates.
- Substitute $u=1-r^2$ to isolate the boundary singularity.
- Determine convergence of $\int_0^1u^{-\alpha}\,du$ by cases on $\alpha$.
Step 1 · Convert to polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, $dx\,dy=r\,dr\,d\theta$, and $0\leqslant r<1$, $0\leqslant\theta\leqslant2\pi$:
$$ I = \int_0^{2\pi}\int_0^1\frac{r}{(1-r^2)^\alpha}\,dr\,d\theta = 2\pi\int_0^1\frac{r}{(1-r^2)^\alpha}\,dr. $$
Step 2 · Substitute $u=1-r^2$
With $du=-2r\,dr$, and $r:0\to1$ corresponding to $u:1\to0$:
$$ \int_0^1\frac{r}{(1-r^2)^\alpha}\,dr = \frac12\int_0^1 u^{-\alpha}\,du. $$
Step 3 · Determine convergence by cases
If $\alpha<1$:
$$ \int_0^1 u^{-\alpha}\,du = \left[\frac{u^{1-\alpha}}{1-\alpha}\right]_0^1 = \frac1{1-\alpha}, $$
since $1-\alpha>0$ makes $u^{1-\alpha}\to0$ as $u\to0^+$. The integral converges.
If $\alpha=1$: $\displaystyle\int_0^1\frac{du}u=+\infty$ — diverges (logarithmically).
If $\alpha>1$: $u^{1-\alpha}=u^{-(\alpha-1)}\to+\infty$ as $u\to0^+$, so $\displaystyle\int_0^1u^{-\alpha}\,du=+\infty$ — diverges.
Step 4 · Assemble the result
For $\alpha<1$:
$$ I = 2\pi\cdot\frac12\cdot\frac1{1-\alpha} = \frac{\pi}{1-\alpha}. $$
For $\alpha\geqslant1$, $I=+\infty$.
Additional Notes
- Sanity check at $\alpha=0$: the formula gives $I=\pi/(1-0)=\pi$, matching the trivial fact that $\iint_D 1\,dx\,dy$ is simply the area of the unit disk.
- This is the direct two-dimensional analogue of $\int_0^1(1-x)^{-\alpha}\,dx$, which converges iff $\alpha<1$; the extra factor of $r$ from the polar Jacobian does not change the threshold, since it is bounded (equal to $1$) exactly where the singularity occurs.
Result
The integral converges precisely for $\alpha<1$: