Is the arclength of $x\sin(\pi/x)$ in the interval $(0, 1)$ finite?

Question

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.

Answer 1

Expanding the integrand, we have the arclength to be $$ \int_1^\infty\sqrt{\frac2{x^2}-\frac1x\sin\left(2\pi{x}\right)+\left(1-\frac1{x^2}\right)\cos^2\left(\pi{x}\right)}\,\,\frac{\mathrm{d}x}{x} $$ For $x\ge2$, the integrand is greater than $$ \sqrt{\frac34\cos^2(\pi x)-\frac12} $$ For any $k\in\mathbb{Z}$ and $k-\frac16\le x\le k+\frac16$, $$ \sqrt{\frac34\cos^2(\pi x)-\frac12}\ge\frac14 $$ Thus, the arclength is at least $$ \begin{align} \sum_{k=3}^\infty\frac14\cdot\frac13\cdot\frac1{k+\frac16} &=\sum_{k=3}^\infty\frac1{12k+2}\\ &\ge\sum_{k=4}^\infty\frac1{12k} \end{align} $$ which diverges by comparison to the tail of $\frac1{12}$ the Harmonic Series.

Answer 2

My intuitive approach would be to say the convergence of your integral from $1$ to $\infty$ is determined when $x$ is very large. We can make the lower limit large without changing the convergence When $x$ is very large the $1$ and $\sin x$ under the square root don't matter, so for $m$ some convenient large even whole number your integral becomes $$ \pi\int_1^\infty\sqrt{1 + (\sin x - x\cos x)^2}\frac{\text{d}x}{x^2} \approx \pi\int_1^{m \pi}\sqrt{1 + (\sin x - x\cos x)^2}\frac{\text{d}x}{x^2}+\pi \int_{m \pi}^\infty \frac {|\cos x|}x\text{d}x$$ The first integral is some constant, so we can ignore it when we are asking about convergence. Now when $x$ is large the change in $\frac 1x$ is small over the range $x=k\pi$ to $x=(k+1)\pi$, so we can bring the $\frac 1x$ out of the integral. The integral of $|\cos x|$ over one of those intervals is $2$, so we get $$\pi \int_{m \pi}^\infty \frac {|\cos x|}x\text{d}x \gt \pi\sum_{k=m}^\infty\frac 1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\cos x| \text{d}x=\sum_{k=m}^\infty\frac {2}{(k+1)}$$ which diverges as the harmonic series. You may feel the approximation in the first line needs some justification. One way to make it rigorous is to make the integrand $\frac {|x\cos x|-1}{x^2}$ and cut the intervals in the sum down so that $x\cos x \gt 2$. That makes the approximation sign a greater than. The final divergence goes through nicely.

Answer 3

Consider the intervalss $\{(\frac 1 n , \frac 1 {n+1/2})$, $1 \leq n \leq N$. Denoting the given function by f we get $|f(\frac 1 {n+1/2})-f(1/n)|= \frac 1 {n+1/2}$. Summing over N it follows immediately that f is of unbounded variation in $(0,1)$. Hence its length is $\infty$.

Answer 4

You were doing fine up to $\int_0^{\infty}\left|\frac {\sin x}{x^2}=\frac {\cos x}{x}\right|dx.$ This integral is infinite: First observe that $\int_0^{\infty}|f(x)|dx\geq \sum_{n=1}^{\infty}\int_{\pi n-1}^{\pi n+1}|f(x)|dx,$ which is applied in the 2nd line after this one.

For $n\in \Bbb N$ and $x\in [\pi n-1,\pi n+1]$ we have $\left |\frac {\sin x}{x^2}-\frac {\cos x}{x}\right|\geq$ $ \left|\frac {\cos x}{x}\right|-\left|\frac {\sin x}{x^2}\right|\geq$ $ \frac {\cos 1}{\pi n+1}-\frac {\sin 1}{(\pi n -1)^2}.$

Therefore $\int_0^{\infty}\left|\frac {\sin x}{x^2}-\frac {\cos x}{x}\right|dx\geq$ $ 2\sum_{n=1}^{\infty}\left(\frac {\cos 1}{\pi n+1}-\frac {\sin 1}{(\pi n-1)^2}\right).$

Now $\sum_{n=1}^{\infty} \frac {\sin 1}{(\pi n-1)^2}$ converges, by term-by-term comparison, for $n>1,$ with $\sum_{n=2}^{\infty}\frac {1}{\pi^2 (n-1)^2}.$ But $\sum_{n=1}^{\infty}\frac {\cos 1}{\pi n +1}\geq \sum_{n=1}^{\infty}\frac {\cos 1}{\pi n+\pi}=\frac {\cos 1}{\pi}\sum_{n=1}^{\infty}\frac {1}{n+1}=\infty.$

Answer 5

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

$$\int_0^1| \cos(\pi/x)/(\pi/x)|\, dx = \infty.$$