Example that $u\in W^{1,2}$, but $u \notin W^{1,3}$ I'm doing the calculations about the following assertion
Let $\Omega$ be $\{(x,y):0<y<x^2, 0<x<1\}$. The function $u(x,y)=\log (x^2+y^2)$ belongs to $W^{1,2}(\Omega)$, which you can check by integrating $|\nabla u|^2\approx 1/x^2$ within $\Omega$. We have $\Delta u=0$, which  is the nicest equation one could ask for. However, $u$ does not belong to $W^{1,3}(\Omega)$, which you also can check by integration.
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?

*

*It's correct. Maybe you could justify why is $u$ is $L^2$.

*Let $I(x):=\int_0^{x^2}\frac{dy}{(x^2+y^2)^{3/2}}$. Put $tx=y$, ($xdt=dy$) to get 
$$I(x)=\int_0^x\frac{x dt}{(x^2+x^2t^2)^{3/2}}=\frac x{x^3}\int_0^x\frac{dt}{(1+t^2)^{3/2}}\geq \frac 1{x^2}\frac 1{2^{3/2}}x=\frac 1{2^{3/2}}\frac 1x,$$
and the integral over $(0,1)$ of the last term is infinite. 

