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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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  • $\begingroup$ Wrong arclength formula. That is the arclength of the graph of a real-valued function. But $\phi$ is a parametrized curve in $\Bbb C$, and you are calculating the arclength of its image, not its graph. the formula you want is $\int_0^1 |\phi '(t)|dt$ $\endgroup$ Commented Sep 19, 2015 at 4:21
  • $\begingroup$ Thank you for pointing that out. It turns out (because of the question), that this is equivalent to finding the arc-length of the graph of the imaginary part of the function - something I still haven't been able to do. $\endgroup$ Commented Sep 20, 2015 at 7:43

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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.

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See (fix the url): http://i.stack.imgur.(com)/w6bEr.png

Given any of the oscillations of phi you can get an upper and lower bound on the arclength of the oscillation by considering these red and blue curves in this quickly-made image. Picking the most natural points for where the red and blue curves peak gives nice bounds so that once you sum over all oscillations, you manage all p-cases at once.

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