This is a problem from Ted Shifrin's book on Differential Geometry; consider the spiral $\vec{\alpha} (t) = {r}(t)(\cos t, \sin t)$ where $r(t)$ is a $C^1$ function bounded between $[0,1]$. Characterize (in terms of the existence of improper integral(s)) the functions ${r}(t)$ for which $\vec{\alpha}(t)$ hasfinite length on $[0,∞)$ and use this to prove that if $\vec{\alpha}(t)$ has finite length on $[0, ∞)$, then: $$\lim_{t \to ∞} r(t)=0$$
Firstly, $\dfrac{d\vec{\alpha}}{dt}=(r'(t)\cos t - r(t) \sin t)\hat{i} + (r'(t)\sin t + r(t) \cos t ) \hat{j}$ so the expression for arc length is
$$L=\int_0^∞ \sqrt{(r'(t)\cos t - r(t) \sin t)^2+(r'(t)\sin t + r(t) \cos t )^2}dt=\int_0^∞ \sqrt{r(t)^2+r'(t)^2}dt$$.
How do I deduce the second part from this? I cannot claim with certainty that the argument of the radical is monotonically decreasing on $[0, ∞)$ or deduce the asymptote from any convergence test. I looked at finite length of a spiral, which is based off the same problem but I didn't understand how to convert the reasoning into a formal proof; why can't $r(t)$ be an oscillatory function therein?