When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length? When does the curve $y = {x^p}\cos \frac{\pi }{x}(0 < x \leqslant 1)$ have finite length?
I cannot figure out how to solve this problem. Thanks for your help.
 A: the problem arises because there are an unbounded number of shorter and shorter "cycles" as $x$ approaches zero.
think of two piecewise linear figures one of which underestimates, and the other of which overestimates, the required length. 
to get the idea first look at the length $L$ of $y=\cos x$ on $[0,2\pi]$. simple geometry shows that:
$$
4\sqrt{1 + (\frac{\pi}2)^2} \lt L \lt 4 + 2\pi
$$
of course the devil is in the detail, and that is your problem, but you can probably already see that the nub of the matter will be in the convergence of a series related to
$$
\sum^{\infty} n^{-p}
$$
A: You need to integrate the arc-length $ds$ through the curve, i.e., $L=\int_0^1\,ds $. The arc-length is found by the formula
$$
ds = \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}\, dx.
$$
Summarizing, you have to consider $y=f(x)$, and then compute the integral
$$
\int_0^1 \sqrt{1+f'(x)^2}\, dx,
$$
and check when is it convergent in terms of $p$.
A: You have to translate the condition
$$ \sqrt{1+y'^2}\in L^1(0,1) \tag{1}$$
in terms of $p$. $(1)$ is equivalent to:
$$ \sqrt{1+x^{2p-4}\left(px\cos\frac{\pi}{x}+\pi\sin\frac{\pi}{x}\right)^2}\in L^1(0,1) \tag{2}$$
and $\left(px\cos\frac{\pi}{x}+\pi\sin\frac{\pi}{x}\right)^2$ is bounded by $(p+\pi)^2$. That gives that the LHS of $(2)$ behaves like $x^{p-2}$. Can you finish from there?
A: Thanks for everyone! With your hints I have figured out a solution.
We need to check if the integral
$\begin{gathered}
  \int_0^1 {\sqrt {1 + y'{{(x)}^2}} dx}  = \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + y'{{(\frac{1}{t})}^2}} dt}  \hfill \\
   = \int_1^{ + \infty } {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt}  \hfill \\ 
\end{gathered} $
is convergent.
If $p > 1$, we have an estimate
$\begin{gathered}
  \int_n^{n + 1} {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt}  \hfill \\
   \leqslant \int_n^{n + 1} {\frac{1}{{{t^2}}}(1 + \frac{{p + \pi t}}{{{t^{p - 1}}}})dt}  \leqslant \frac{1}{{{n^2}}}(1 + \frac{{p + 2\pi n}}{{{n^{p - 1}}}}) \hfill \\ 
\end{gathered} $
from which we know the integral is convergent.
If $p \leqslant 1$, we have an estimate
$\begin{gathered}
  \int_{2n + \frac{1}{6}}^{2n + \frac{1}{3}} {\frac{1}{{{t^2}}}\sqrt {1 + \frac{1}{{{t^{2(p - 1)}}}}{{(p\cos \pi t + \pi t\sin \pi t)}^2}} dt}  \hfill \\
   \geqslant \int_{2n + \frac{1}{6}}^{2n + \frac{1}{3}} {\frac{1}{{{t^2}}}(\frac{{p + \pi t}}{{2{t^{p - 1}}}} - 1)dt}  \geqslant \frac{1}{6}\frac{1}{{{{(3n)}^2}}}(\frac{{p + 2\pi n}}{{2{{(2n)}^{p - 1}}}} - 1) \hfill \\ 
\end{gathered} $
from which we know the integral is not convergent.
