I'm doing the calculations about the following assertion
Let $\Omega$ be $\{(x,y):0
My efforts:
$| \nabla u |^2 = 4/(x^2+y^2)$ and \begin{equation} \int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{x^2+y^2}dy dx = \int_{0}^{1} \dfrac{\arctan}{x}dx < \infty \end{equation} because $\lim_{x\rightarrow 0}\dfrac{\arctan}{x} = 1$ and $\dfrac{\arctan}{x}$ is bounded in(0,1). Hence $\nabla u \in L^{2}(\Omega)$.
Am I right here?
I don't know how $\int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{(x^2+y^2)^{^3/2}}dy dx$ diverges. Thank you.
3. \begin{eqnarray} \int_{0}^{1}\int_{0}^{x^2} \ln(x^2+y^2)dydx &=& \int_{0}^{1}\int_{0}^{x^4} u\ln udu dx\\ & = &\int_{0}^{1} \left \{\dfrac{1}{2} u^2 \ln + \dfrac{1}{4}u^2 \right \}_{0}^{x^4}dx\\ &=& \int_{0}^{1} (x^8 \ln(x^4) -\dfrac{1}{4}x^8) dx. \end{eqnarray} In the second equality used that $\lim_{x\rightarrow 0}x^2 \ln x = 0$. Moreover, we obatain that $x^2 \ln(x^2)$ is finite in $(0,\infty)$ and the integral above is finite. Am I right here too?