I have to prove that $$f(x)=\frac1{(1+x^2)^{\frac{\alpha}{2}}\ln(2+x^2)}, \ \ \ 0<\alpha<1$$ belongs to $W^{1,p}(\mathbf{R})$ for $p\in[\frac1{\alpha},\infty]$ and that it does not belong to $L^p$ for $p\in[1,\frac1{\alpha})$.
Since $f$ is even and positive I can only discute the convergence of this integral $$\int_2^\infty f(x)^p dx.$$ $f$ is defined on the whole real line, so I need only to discute its behaviour at $\infty$.
But I have that $$f(x)\sim \frac1{(x^2)^{\frac{\alpha}{2}}\ln(x^2)}=\frac1{2x^\alpha\ln(x)}$$ therefore $$\int_2^\infty f(x)^p dx\sim\frac12\int_2^\infty\frac1{x^{\alpha p}\ln^p(x)} dx.$$
Since the last integral converges iff $\alpha p\ge1$, I can assert that $$f\in L^p(\mathbf{R})\Leftrightarrow p\in[\frac1{\alpha},\infty].$$
Am I wrong? And, now, I could I prove that $f\in W^{1,p}(\mathbf{R})$ for such $p$?