Prove an integral $\int_0^1r\ln^p r^2 dr$ is finite

Question

Consider the function $f(x,y)=\ln(x^2+y^2)$ on $D=\{(x,y)\in\mathbb{R}^2| x^2+y^2\leq 1\}$. I'm trying to show that $f\in W^{1,p}(D)$ for $p\in[1,2)$, but in $H^1(D)$. We have \begin{equation} \frac{\partial}{\partial x} \ln(x^2+y^2)=\frac{2x}{x^2+y^2} \end{equation} \begin{equation} \frac{\partial}{\partial y} \ln(x^2+y^2)=\frac{2y}{x^2+y^2}. \end{equation} By introducing polar coordinates $x=r\cos\varphi, y=r\sin\varphi$, we get \begin{equation} \int_{D}|\ln(x^2+y^2)|^2dxdy=\int_0^{2\pi}\int_0^1r^2\ln^2 r drd\varphi<\infty. \end{equation} How can I compute the following is finite with $p\in[1,2)$? \begin{equation} \int_{D}|\ln(x^2+y^2)|^pdxdy=\int_0^{2\pi}\int_0^1r\ln^p r^2 drd\varphi<\infty. \end{equation}

Answer 1

hint: The only point where there may be a singularity is $r=0$ Notince that $\lim_{r\rightarrow0}r\log^p(r^2)=0$. Hence, the function $\phi(r)=r\log^p(r^2)$ for $0<r\leq 1$ and $\phi(0)=0$ is continuous. The result follows immediately from this.