Show that $g'$ is integrable iff $\alpha>\beta$ (Lebesgue Integral) 
Problem. Let
  $$g(x) = \begin{cases}x^{\alpha}\sin(x^{-\beta}),& 0<x\leq 1 \\ 0,& x=0. \end{cases}$$
  Show that $g$ is continuous and $g'$ exists and is continuous in $x \neq 0$. Moreover, show that $g'$ is integrable on $|x|<1$ iff $\alpha>\beta$.


I proved the first part. For the second:
\begin{eqnarray*}
\int_{|x|<1}|g'(t)|dt & = & \int_{|x|<1}|\alpha t^{\alpha-1}\sin(t^{-\beta}) - \beta t^{\alpha-\beta-1}\cos(t^{-\beta})|dt\\
                      & \leq & \alpha \int_{|x|<1}|t^{\alpha-1}|dt - \beta \int_{|x|<1}|t^{\alpha-\beta-1}|dt
\end{eqnarray*}
So, only $\beta\int_{|x|<1}|t^{\alpha-\beta-1}|dt$ dependes on $\alpha$ and $\beta$ and the problem becomes:

  
*
  
*$|x^{\alpha-\beta-1}$| is integrable on $|x|<1$ iff $\alpha > \beta$.
  

My thought without rigor is: if $\alpha \leq \beta$, the result of integral is $\frac{1}{t^{\gamma}}$ with $\gamma > 0$ and this is unbounded on $|x|<1$.
I do not know how to prove it formally. Can someone help me?

If $\alpha -\beta - 1\ge 0$, the integrand is bounded and obviously the function is integrable. So we'll assume $\alpha - \beta - 1< 0$. 
Consider the sequence of intervals $I_n = (2^{-n-1},2^{-n}]$. Clearly $\cup_n I_n = (0,1]$ and $m(I_n) = 2^{-n-1}$. Also, if $t\in I_n$, then $2^{n\gamma}\le t^{-\gamma} \le 2^{(n+1)\gamma}$. So we must have
\begin{multline}
\sum_{n=0}^\infty 2^{-n-1}2^{n\gamma}\le \int_{(0,1]} t^{-\gamma}dt\le \sum_{n=0}^\infty 2^{-n-1}2^{(n+1)\gamma}
\\ \Longrightarrow\frac{1}{2}\sum_{n=0}^\infty \left[2^{\gamma-1}\right]^n\le \int_{(0,1]} t^{-\gamma}dt\le 2^{\gamma-1}\sum_{n=0}^\infty \left[2^{\gamma - 1}\right]^n.
\end{multline}
The geometric series converges if and only if $\gamma < 1$, so the same condition clearly applies to the integral. Plugging in $\gamma = 1 + \beta - \alpha$, we see $\gamma < 1$ is equivalent to $\alpha > \beta$.
