Curve $\gamma(t)=e^{2\pi i t\sin (1/t)}$ is not rectifiable Let $\gamma(t)$ be curve in the complex plane, defined on $[0,2\pi]$ by $$\gamma(t)=e^{2\pi i t\sin (1/t)} \qquad \text{with}\quad\gamma(0)=1.$$
Prove that $\gamma(t)$ is not rectifiable.
My attempt: I'll use the following theorem from PMA Rudin.
In our case: 
$$\gamma'(t)=-2\pi e^{2\pi i t\sin (1/t)}\left(\sin \dfrac{1}{t}-\dfrac{1}{t}\cos \dfrac{1}{t}\right).$$
$$|\gamma'(t)|=2\pi\left|\sin \dfrac{1}{t}-\dfrac{1}{t}\cos \dfrac{1}{t}\right|.$$
So we see that $\gamma'$ is continuous in $(0,1]$ but not at $[0,1]$. So we can't apply theorem $6.27$.
Can anyone explain how to proceed ?
The points of this curve are located on distance 1 from origin. But why Wolfram shows this: 
 A: If $\gamma(t)=e^{2\pi i t\sin \frac{1}{t}}$ then $\gamma'(t)=i\gamma(t)2\pi \left[\sin \frac{1}{t}-\frac{1}{t}\cos \frac{1}{t} \right]$ then $$|\gamma'(t)|=2\pi \left|\sin \frac{1}{t}-\frac{1}{t}\cos \frac{1}{t} \right|$$ Taking $\epsilon\in (0, \frac{3\pi}{4}]$ and $\gamma'(t)\in C[\epsilon,2\pi]$ and $$\Lambda_{[\epsilon,2\pi]}(\gamma)=\int\limits_{\epsilon}^{2\pi}|\gamma'(t)|dt=2\pi\int\limits_{\epsilon}^{2\pi}\left|\sin \frac{1}{t}-\frac{1}{t}\cos \frac{1}{t}\right|dt\geqslant $$ $$\geqslant 2\pi\int\limits_{\epsilon}^{2\pi}\left|\frac{1}{t}\cos \frac{1}{t}\right|dt-4\pi^2.$$ Making substituion $t\mapsto \frac{1}{t}$ integral in last inequality is equal to $$2\pi\int\limits_{1/{2\pi}}^{1/{\epsilon}}\frac{|\cos t|}{t}dt\geqslant 2\pi\sum \limits_{k=1}^{M}\int \limits_{\pi_1(k)}^{\pi_1(k)}\frac{|\cos t|}{t}dt=I$$ where $M=[\frac{1}{\pi \epsilon}-\frac{1}{3}]\ge 1 (\text{since} \quad\epsilon\le \frac{2\pi}{4}),$  $\pi_1(k)=\frac{\pi(2k-1)}{2}+\frac{\pi}{6},$ $\pi_2(k)=\frac{\pi(2k+1)}{2}-\frac{\pi}{6}.$ Since $|\cos t|\geqslant 1/2$ on $[\pi_1(k),\pi_2(k)]$ we get: $$I\geqslant 2\pi\sum \limits_{k=1}^{M}\frac{1}{2}\int \limits_{\pi_1(k)}^{\pi_1(k)}\frac{1}{t}dt\geqslant \pi \sum \limits_{k=1}^{M}(\pi_2(k)-\pi_1(k))f(\pi_2(k))=$$$$=\pi \sum \limits_{k=1}^{M}\frac{2\pi}{3}\dfrac{1}{\frac{\pi(2k+1)}{2}-\frac{\pi}{6}}=2\pi \sum \limits_{k=1}^{M}\frac{1}{3k+1}$$ Letting $\epsilon\to 0$ then $M\to \infty$ and the last series is diverges and $\lim \limits_{\epsilon \to 0}\Lambda_{[\epsilon,2\pi]}(\gamma)=+\infty$ . Thus our curve is non rectifiable.
A: Lemma: Let $U, V \subset \mathbb C$ be open and $F : U \to V$ be globally Lipschitz. If $\gamma : I \to U$ is rectifiable, then $F\circ \gamma$ is also rectifiable. (Try to show this).
In our situation, let $U = \{ z\in \mathbb C : 1/2 < |z| <3/2\}\setminus \{x<0\}$ and $F = \text{Log}$ be the principal $\log$. Thus if $\gamma$ is rectifiable, then so is 
$$\eta(t) = \operatorname{Log}(\gamma(t)) = 2\pi i \ t\sin \left(\frac 1t\right).$$
However, this is not rectifiable: we have 
$$\eta\left(\frac{1}{2n\pi}\right) = 0, \ \ \eta\left(\frac{1}{2n\pi + \pi/2}\right) = \frac{2\pi i}{2n\pi + \pi/2}.$$
This implies that the total variation of $\eta$ is at least
$$\sum_{k=1}^n \frac{2\pi}{2k\pi + \pi/2} \ge \sum_{k=1}^n \frac{2\pi}{4k\pi} =\frac{1}{2}\sum_{k=1}^n \frac 1k.$$
But the harmonic series is unbounded. So $\eta$ is not rectifiable and so $\gamma$ is not rectifiable. 
A: I think this is an intuitive way to see that $\gamma$ is not rectifiable:
Note that the arc length of $f(\varphi) = e^{i\varphi}$ on $[0,t]$ is $t$ by Euler's formula, it is the path of the unit circle. Let $g(t) = 2\pi t \sin(1/t)$, $g(0)=0$ then $\gamma(t) = f(g(t))$. 
But $g$ is not rectifiable (the arc length is infinite) on $[0,1]$ as shown here:
https://math.stackexchange.com/a/90739/286006
Illustration:
The function displayed is $t \mapsto t \sin(1/t)$. The red arcs are longer than their corresponding  green lines but the sum of the lengths of the green lines is $\sum_{k=1}^\infty \frac{1}{4k\pi+1} > \frac{1}{4\pi}\sum_{k=2}^\infty \frac{1}{k} = \infty$.
So the length of $f\circ g$ has to be infinite too.

