Prove that length of a curve is finite / infinite

Question

I have problems proving the following:
Let $\alpha > 0 $. Consider the curve $\gamma : [0,1] \to \mathbb{R}^2$ given by
$\gamma (0)=(0,0) , \gamma(t) = (t^\alpha \cos ( \frac {1}{t}), t^\alpha \sin(\frac {1}{t})$
Prove that the length $L(\gamma)=\infty $ for $ \alpha \leq 1$, and that $L(\gamma)<\infty$ for $\alpha >1$
To do this, I computed $|\dot\gamma|=$ $\sqrt{\alpha^2t^{2\alpha-2}+t^{2\alpha-4}}$
And now one has to estimate the integral
$\int_0^1 |\dot\gamma|dt $
But i cant seem to find a smaller estimate, that approaches $\infty$, for $\alpha \leq 1$ or a bigger estimate that doesnt approach $\infty$ for $\alpha >1$.

Answer 1

hint :$\alpha-1=m$ so $m>0$ and $$\\\sqrt{\alpha^2t^{2\alpha-2}+t^{2\alpha-4}}>\sqrt{\alpha^2t^{2\alpha-2}}=\alpha t^{\alpha-1}=(m+1)t^{m}\\$$ now $$\int_1^{+\infty} (m+1)t^mdt \rightarrow \infty$$