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I learned about a example which does not have finite arc length:

The curve $\alpha(t) =\left( \begin{array}{c} t\\ t^2 \sin(\frac{1}{t^2})\\ \end{array} \right)$ has a infinite arc length between $t=0$ and $t=1$

An estimate for the arc length is to take the length of the paths between the points $\alpha(\frac{1}{\sqrt{2 \pi n + \pi/2}})$ and $\alpha(\frac{1}{\sqrt{2 \pi n}})$ and sum them up.

Now comes the calculation in the example:

$$\sum_{n=1}^{\infty} \|\alpha(\frac{1}{\sqrt{2 \pi n + \pi/2}}) -\alpha(\frac{1}{\sqrt{2 \pi n}})\|=\sum_{n=1}^{\infty}\|(\frac{1}{\sqrt{2 \pi n + \pi/2}} - \frac{1}{\sqrt{2 \pi n}})^2+(\frac{1}{2 \pi n + \pi/2})^2 \|$$

I am not really understanding how to get from the left side to the right side. Here are my calculations

$$\sum_{n=1}^{\infty} \|\left( \begin{array}{c} \frac{1}{\sqrt{2 \pi n + \pi/2}}\\ \frac{1}{2 \pi n + \pi/2} \sin(\frac{1}{2 \pi n + \pi/2})\\ \end{array} \right) - \left( \begin{array}{c} \frac{1}{\sqrt{2 \pi n}}\\ \frac{1}{2 \pi n} \sin({\frac{1}{2 \pi n}})\\ \end{array} \right)\| \\=\sum_{n=1}^{\infty} \sqrt{ (\frac{1}{\sqrt{2 \pi n + \pi/2}}-\frac{1}{\sqrt{2 \pi n}})^2 + (\frac{1}{2 \pi n + \pi/2} \sin(\frac{1}{2 \pi n + \pi /2})-\frac{1}{2 \pi n } \sin(\frac{1}{2 \pi n }))^2} $$

Now the part where I struggle:

How does $(\frac{1}{2 \pi n + \pi/2} \sin(\frac{1}{2 \pi n + \pi /2})-\frac{1}{2 \pi n }\sin(\frac{1}{2 \pi n }))^2$ equal $(\frac{1}{2 \pi n + \pi/2})^2$ ?

I tried to continue calculation but I am stuck at:

$$(\frac{1}{2 \pi n + \pi/2}\sin(\frac{1}{2 \pi n + \pi /2})-\frac{1}{2 \pi n }\sin(\frac{1}{2 \pi n }))^2\\=(\frac{1}{2 \pi n + \pi/2})^2 \sin^2 (\frac{1}{2 \pi n + \pi /2})-2\frac{1}{2 \pi n + \pi/2}\sin(\frac{1}{2 \pi n + \pi /2})\frac{1}{2 \pi n }\sin(\frac{1}{2 \pi n })+(\frac{1}{2 \pi n })^2 \sin^2 (\frac{1}{2 \pi n})$$

Robert Z
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Walli
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    The second coordinate of your map is $t^2\sin\left(\frac{1}{t^2}\right)$ not $t^2\sin t^2$. – mathcounterexamples.net Feb 11 '22 at 14:37
  • @mathcounterexamples.net thank you, this makes much more sense now – Walli Feb 11 '22 at 14:47
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    For what it's worth, calculations such as this are much more complicated than they need to be. For oscillating sine type functions, if you suspect the length is infinite, then you can usually show this by direct comparison with an appropriate $p$-series of the sum of the vertical distances to the $x$-axis of the relative extrema (example), and if you suspect convergence, then you can usually show this by direct comparison with a $p$-series of an appropriate "circumscribed square wave". – Dave L. Renfro Feb 11 '22 at 18:16

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