Integral of weak derivative is finite I want to show that the integral over the unit ball of the weak derivative of $u(x)=|x|^{-\gamma}, 0< \gamma< \frac{n-p}{p}$ is finite, in order to show that the function is in $W^{1,p}$.
How can we find the waek derivative of $u(x)$ ?
I have tried the following:
$\langle \frac{\partial|x|^{-\gamma}}{\partial{x_i}}, \phi \rangle=-\langle |x|^{-\gamma}, \frac{\partial{\phi}}{\partial{x_i}} \rangle=- \int_{B(0,1)} |x|^{-\gamma} \frac{\partial{\phi}}{\partial{x_i}} dx$ 
Is it right so far? How could we continue?
 A: Let $u(x)=|x|^{-\gamma}$ defined for $x \in \Omega=B(0,1) \setminus \lbrace 0 \rbrace$. Assuming temporarily wich $u$ is smooth, and can be differentiated pointwise with $\partial_{x_j} u= -\gamma x_j |x|^{-\gamma -2}$, and hence also $|\nabla u(x)|=|\gamma||x|^{-\gamma - 1}$. Recall a criterion of improper interation (with polar coordinate in $\mathbb{R}^n$), tells us that  
$\displaystyle \int_{\Omega} \frac{1}{|x|^\gamma} dx < \infty$
if $\gamma < n$. In this sense we have $|\nabla u| \in L^1(\Omega)$ for $\gamma +1 < n$, and also $|\nabla u| \in L^p(\Omega)$ for $(\gamma +1)p < n$.
So we must assume $\gamma +1 < n$ and $(\gamma +1)p < n$.
Let $\varphi \in \mathcal{D}(\Omega)$ with $\mathrm{supp}(\varphi) \subset \Omega$, then $\varphi(x)=0$ on $\partial B(0,1)$, and with integration by parts we have
$\displaystyle \int_{\Omega} u \partial_{x_j} \varphi dx =-\int_{\Omega } \partial_{x_j}u \varphi dx + 0$
for the assumptions made. So that $\partial_{x_j} u$ is weak derivative of $u$, and $u \in W^{1,p}(\Omega)$ if $u, |\nabla u| \in L^{p}(\Omega)$ for $(\gamma + 1)p < n$, i.e. for $\gamma < (n-p)/p$.
