Problem: Let $\phi : [0,1] \to \mathbb C$ be given by $$\phi(t) := t + it\sin\left(\frac{\pi}{t^p}\right)$$ for $t \in (0,1]$, and define $\phi(0) := 0$. Then $\phi$ is continuous. For what values $p > 0$ is the arc-length finite? Infinite?
Attempt: Recall that arc-length is given by $$AL(\phi) := \int_0^1 |\phi'(t)|\,\mathrm{d}t$$ It turns out that a direct computation of arc-length is absolutely miserable. Instead, I tried various ways of bounding it. For example, using that $$\sqrt{1 + x^2} \leq 1 + |x|$$ This yields, at best, an opportunity to Taylor expand, but doing this led me to believe that this particular approximation is not even convergent. Unfortunately, I'm absolutely stumped!
To determine the values of $p$ for which the arc length is integrate, I was hoping to use the inequality $$\mathrm{AL}(\phi) \geq \frac{\left|\int_{\mathrm{Im}(\phi)} f(z) \,\mathrm{d}z\right|}{\max\limits_{z \in \mathrm{Im}(\phi)} |f(z)|}$$ with a clever choice of $f$. But I just haven't been clever enough yet!
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Let $\theta = \pi t^{-p}$. After some work, we can show that the arclength is given by $$K_p\int_\pi^\infty \sqrt{1 + (\sin \theta - p\theta \cos \theta)^2} \frac{d\theta}{\theta^{1+1/p}}$$ where $K_p$ is some constant depending on $p$ for which all we care about is that it is finite. Dropping the unimportant constant, we can further change it to $$\int_\pi^\infty \sqrt{\frac{1 + \sin^2 \theta}{\theta^2} - \frac{p\sin 2\theta}{\theta} + p^2\cos \theta^2} \frac{d\theta}{\theta^{1/p}}$$
For sufficiently large $\theta$, the first two terms are less than $p^2$, so we have $$\sqrt{\frac{1 + \sin^2 \theta}{\theta^2} - \frac{p\sin 2\theta}{\theta} + p^2\cos \theta^2} \le \sqrt{p^2+ p^2\cos \theta^2} = p|\sin \theta| \le p$$ which is sufficient to handle the case of $p < 1$, and should hopefully suggest ways of looking at $p \ge 1$ as well.