Let $\Omega = \{(x,y)\in\mathbb R^2: \sqrt{x^2+y^2}<\frac{1}{2}\}$ be a bounded domain and $v(x,y) = \ln\left|\ln\sqrt{x^2+y^2}\right|$. I'm trying to show that $v\in H^1(\Omega)$ where $H^1(\Omega)$ is the Sobolev-space $W^{1,2}(\Omega)$.
I therefore calculated the partial derivatives $$D_x v = \frac{x}{(x^2+y^2)\ln\sqrt{x^2+y^2}}$$
$$D_y v = \frac{y}{(x^2+y^2)\ln\sqrt{x^2+y^2}}$$
The only thing left to proof now, I think, is that the derivatives are in $L^2(\Omega)$. I tried to show this by proofing, that
$$\int_\Omega (D_x v)^2 < \infty$$
but I don't know how to do that. By introducing polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ we get
$$\int_0^{1/2}\int_0^{2\pi} \frac{\cos^2\theta}{r^2\ln^2(r)}\mathrm{d}\theta\mathrm{d}r = \int_0^{1/2} \frac{\pi}{r^2\ln^2(r)}\mathrm{d}r$$ But how do I solve the integral over $r$? Or is there a majorant I could use to show that the integral is finite?
The integral you have to prove convergence is actually $\int_0^{1/2}\frac 1{r\log^2(r)}dr$ (don't forget the determinant of Jacobian).
First look at $\int_\varepsilon^{1/2}\frac 1{r\log^2(r)}dr$ for $\varepsilon>0$, then make the substitution $s:=\log r$.
It's a good example of an element of $H^1(\Omega)$ which is not essentially bounded.