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How can I show that the length of the curve

$$f(x)=\begin{cases} \sqrt{x}\cos(\pi/x) & \text{ if } x\neq0\\ 0 & \text{ if } x=0 \end{cases}$$

is infinite on $[0,1]$?

I tried using the arc length formula, but ended up with the very nasty integral: $$L=\int_0^1\sqrt{1+\left(\frac{x\cos(\pi/x)+2\pi\sin(\pi/x)}{2x^{3/2}}\right)^2}dx.$$

Perhaps the way to go here is to use the definition of arc length I defined in this thread, but I am not certain.

Do you guys have any ideas?

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Why not show that the integral you obtained diverges? –  Sasha Apr 27 '12 at 17:25
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1 Answer

up vote 4 down vote accepted

Choosing $x$ so that

$$\pi/x \;\; = \;\; \frac{3 \pi}{2}, \; \frac{4 \pi}{2},\; \frac{5 \pi}{2},\; \frac{6 \pi}{2},\; \frac{7 \pi}{2},\; \frac{8 \pi}{2},\; \frac{9 \pi}{2},\; \frac{10 \pi}{2}, \; \frac{11 \pi}{2},\; \frac{12 \pi}{2},\; \frac{13 \pi}{2},\; ...$$

so that

$$x \;\; =\;\; \frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; \frac{2}{6}, \; \frac{2}{7}, \; \frac{2}{8}, \; \frac{2}{9}, \; \frac{2}{10}, \; \frac{2}{11}, \; \frac{2}{12}, \; \frac{2}{13}, \; ... $$

we get

$$\sqrt{x} \cos\left(\pi/x\right) \; \; = \;\; 0, \;\; \sqrt{\frac{1}{2}}, \;\; 0, \;\; -\sqrt{\frac{1}{3}}, \;\; 0, \;\; \sqrt{\frac{1}{4}}, \;\; 0, \;\; -\sqrt{\frac{1}{5}}, \;\; 0, \;\; \sqrt{\frac{1}{6}}, \;\; 0, \; ...$$

Now, if you plot these points (not necessarily to scale, just make a rough hand sketch), it will be clear that the sum of the lengths of the inscribed segments corresponding to the partition

$$\cal{P}_{n} \;\; =\;\; \left\{1, \;\frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; ..., \; \frac{2}{2n}, \; 0\right\}$$

is greater than

$$\sqrt{\frac{1}{2}} \;\; + \;\; \sqrt{\frac{1}{3}} \;\; + \;\; ... \;\; + \;\; \sqrt{\frac{1}{n}}$$

Note that this last sum diverges to $+\infty$ as $n \rightarrow \infty$, as it corresponds to the $p$-series from elementary calculus for $p = \frac{1}{2}.$

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