Verify $ \limsup_{\epsilon \rightarrow 0^+} \int_{D}\frac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}}\frac{1}{\sqrt{1-\sqrt{x^2+y^2}}} \, dx \, dy <+\infty$

Question

I want to know whether or not $$ \limsup_{\epsilon \rightarrow 0^+} \int_{D}\dfrac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}} \frac{1}{\sqrt{1-\sqrt{x^2+y^2}}} \, dx \, dy <+\infty.$$

Here $D $ denotes the disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}$.

Mathematica indicates that the limit is finite

In:= NIntegrate[(1/Sqrt[(x - 0.99999999999)^2 + 
         y^2])*(1/Sqrt[1 - Sqrt[x^2 + y^2]]), {x, -1, 1}, {y, -Sqrt[1 - x^2], Sqrt[1 - x^2]}]


Out:= 13.3727774796526

but along with the warning:

Numerical integration converging too slowly; suspect one of the 
following: singularity, value of the integration is 0, highly 
oscillatory integrand.

The $L^1$-norm of $\dfrac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}}$ is uniformly bounded, and $\dfrac{1}{\sqrt{1-\sqrt{x^2+y^2}}}$ is integrable. However I don't know how to handle the singularity when the two are combined.

Answer 1

Transforming to polar coordinates, $$ \int_{D}\dfrac{1}{\sqrt{(x-(1-\epsilon))^2 +y^2}} \frac{1}{\sqrt{1-\sqrt{x^2+y^2}}} \, dx \, dy = $$ $$ \int_{0}^{1} dr \frac{1}{\sqrt{1-r} } \int_{0}^{2\pi } d\phi \frac{1}{\sqrt{1-2 ((1-\epsilon)/r) \cos \phi + (1-\epsilon)^2 /r^2 } } $$ The inner, $\phi $, integral can be evaluated in closed form, $$ I = \int_{0}^{2\pi } d\phi \frac{1}{\sqrt{1-2 ((1-\epsilon)/r) \cos \phi + (1-\epsilon)^2 /r^2 } } =\frac{4}{1+ ((1-\epsilon)/r)} K\left( \frac{4(1-\epsilon)/r }{(1+((1-\epsilon)/r))^2 }\right) $$ where $K$ is the complete elliptic integral of the first kind. $K$ diverges logarithmically when its argument approaches unity, i.e., when $r\rightarrow 1-\epsilon $, such that $$ I= -2 \log |(1-\epsilon )-r| \ \ + \ \ \mbox{terms regular at} \ \ r=1-\epsilon $$ This singularity is, of course, integrable. Moreover, when $\epsilon \rightarrow 0$, the $r$-integrand has the singularity $\log (1-r)/\sqrt{1-r} $, which remains integrable.