When is this function integrable? 
I need to find values of $\alpha$ and $\beta$ such that $$|x|^{\alpha}|\log |x||^{\beta}$$ is integrable over $\{x\in \mathbb{R} : |x|<\frac{1}{2}\}$ and over $\{x\in\mathbb{R} : |x|>2\}$

I could only see the easy cases i.e. for the first region if we just take $\beta=0$ then for any $\alpha$ the function would be integrable. Please guide as to how I can cover all possible cases and what are the valid cases.
 A: Without loss of generality, we assume that $x\in[0,1/2]$.  Enforcing the substitution $x\to e^{-x}$ reveals, 
$$\int_0^{1/2}x^\alpha |\log(x)|^{\beta}\,dx=\int_{\log(2)}^\infty e^{-(1+\alpha) x}x^\beta\,dx \tag 1$$

Now, for any $\gamma >0$, $x<\frac{1}{\gamma}e^{\gamma x}$.  
Therefore, we assert that for any $\gamma >0$, $e^{-(1+\alpha)x}x^\beta<\frac1\gamma e^{(\gamma \beta -(1+\alpha))x}$.
If $\alpha >-1$, then we for any fixed $\beta$ we can choose $\gamma>0$ such that $\gamma\beta-(1+\alpha)<0$ and the integral on the right-hand side of $(1)$ converges by comparison.

If $\alpha =-1$, then $\int_0^{1/2}x^\alpha |\log(x)|^\beta \,dx$ converges for $\beta <-1$ and diverges elsewhere since
$$\int_\epsilon^{1/2}\frac{|\log(x)|^\beta}{x} \,dx=\frac{(-\log(\epsilon))^{\beta+1}-(\log(2))^{\beta+1}}{\beta +1}$$.

If $\alpha <-1$, then $\int_0^{1/2}x^\alpha |\log(x)|^\beta\,dx=\int_{\log(2)}^\infty e^{|1+\alpha|x}x^\beta\,dx$ diverges for all $\beta$ due to the exponential growth of the integrand.

NOTE:  For $x>2$ repeat the analysis by substituting $x\to 1/x$.
