TITLE: Is the arclength of $x\sin(\pi/x)$ in the interval $(0, 1)$ finite?
QUESTION [8 upvotes]: I'm having trouble determining the arclength of $x\sin(\pi/x)$ in the interval $(0, 1)$ (that is, determining if it converges or not). My first attempt was to write
$$
\begin{aligned}
\int_0^1\sqrt{1+(f'(x))^2}\text{d}x & = \int_0^1\sqrt{1 + \left(\sin\frac \pi x - \frac \pi x\cos\frac \pi x\right)^2}\text{d}x \\
& = \pi\int_1^\infty\sqrt{1 + (\sin x - x\cos x)^2}\frac{\text{d}x}{x^2}
\end{aligned}
$$
Then I wanted to consider the asymptotics function $\sqrt{1 + (\sin x - x\cos x)^2}$, but I can't find an easy way in. One thought was to write
$$
\int_1^\infty\sqrt{1+(\sin x - x\cos x)^2}\frac{\text{d}x}{x^2} > \int_1^\infty\left|\frac{\sin x}{x^2} - \frac{\cos x}x\right|\text{d}x
$$
Next, we notice that $x\cos x + \sin x = 0$ occurs approximately when $x = (k+1/2)\pi$ ($k\in\mathbb{N}$), since these points are the points where $x = \tan x$, and as $x$ increases (so that the function $x$ is higher along the $y$-axis), the function $\tan x$ gets closer and closer to its vertical asymptote, which occurs at $(k+1/2)\pi$, so the intersection of the two functions also occurs close to that asymptote. So we consider 
$$
\int_{(k-1/2)\pi}^{(k+1/2)\pi}\frac{|\sin x - x\cos x|}{x^2}\text{d} x \ge \frac1{[(k+1/2)\pi]^2}\int_{(k-1/2)\pi}^{(k+1/2)\pi}|\sin x - x\cos x|\text{d} x
$$
This is where I get stuck, as I'm not sure how to estimate that integrand. I assume it will be approximately linear in $k$, but I'm not sure how to prove that. If that's true, then the integral is greater than a term that behaves like $\frac 1k$, meaning the original integral is greater than a harmonic sum, and so diverges. So first question: how can I evaluate the integral on the RHS of the inequality above?
Second question: I also feel like there's an easier way of thinking about this problem that I'm missing. Any ideas?
Edit: I just realized that using $\int |f| \ge |\int f|$, we get
$$
\int_{(k-1/2)\pi}^{(k+1/2)\pi}|\sin x - x\cos x|\text{d} x \ge \left|\int_{(k-1/2)\pi}^{(k+1/2)\pi}(\sin x - x\cos x)\text{d}x\right| = |2\pi k\cos(\pi k)| = 2\pi k
$$
However, this doesn't completely answer my question. Yes, it's a solution to the integral, but the argument that we should only look at the intervals $[(k-1/2)\pi, (k+1/2)\pi]$ is kind of handwavy, and it still feels like there should be an easier solution.

REPLY [2 votes]: You can simplify things by noticing
$$\sqrt {1+f'(x)^2} > |f'(x)| = | \cos(\pi/x)/(\pi/x)-\sin(\pi/x)| \ge | \cos(\pi/x)/(\pi/x)|-|\sin(\pi/x)|.$$
Since $\sin(\pi/x)$ is bounded, it suffices to show
$$\int_0^1| \cos(\pi/x)/(\pi/x)|\, dx = \infty.$$