On the curve $y=\frac{\sin (\pi x)}{x^p},x>0$, for what values of $p$ does the product of all the arc lengths between neighboring roots exist?

Question

Consider the curve $y=\frac{\sin (\pi x)}{x^p}, x>0$, shown here with $p=0.75$.

It occurred to me that if $p$ is large enough, then the curve flattens quickly, so the arc lengths between neighboring roots approach $1$ quickly, so the product of all those arc lengths should converge.

So my question is:

For what values of $p$ does the product of all the arc lengths between neighboring roots exist?

That is, given $f(x)=\dfrac{\sin (\pi x)}{x^p}$, for what real values of $p$ does the following limit exist:

$$L=\lim_{n\to\infty}\prod_{k=1}^n \int_{k}^{k+1}\sqrt{1+(f'(x))^2}dx$$

Numerical investigation suggests that when $p=1$, $L\approx 4.93$; and when $p=0.5$, $L$ does not exist. If that's true, then there must be some critical value of $p$ between $0.5$ and $1$, and I wonder what it is.

Possibly related: Convergence $I=\int_0^\infty \frac{\sin x}{x^s}dx$

Context: I am interested in limits of geometrical products. Here is another question about a limit of a product of arc lengths. And here is the question that got me interested in such limits.

Answer 1

Using the same approach as the other answer, setting $a_k=-1+\int_k^{k+1}\sqrt{1+(f'(x))^2}dx=\int_k^{k+1}\left(\sqrt{1+(f'(x))^2}-1\right)dx$ to be the arc-length between $k$ and $k+1$, minus 1, we have that $$\prod_{k=1}^\infty(1+a_k)<\infty\iff\sum_{k=1}^\infty a_k<\infty$$

The other answer showed that this sum converges when $p>1/2$, so I'll focus on $p=1/2$. Consider $$\sqrt{1+(f'(x))^2}-1$$ We have that $$f'(x)=\frac{2\pi x\cos(\pi x)-\sin(\pi x)}{2x\sqrt{x}}\to 0$$

and using the series expansion of $\sqrt{1+x^2}-1$, we get that $$\sqrt{1+(f'(x))^2}-1\ge C f'(x)^2$$

If we look at $\int_k^{k+1}f'(x)^2dx$, we see that $$\int_k^{k+1}f'(x)^2dx=\int_k^{k+1}\frac{4\pi^2x^2\cos^2(\pi x)+\sin^2(\pi x)-2\pi x\sin(2\pi x)}{4x^3}dx$$

We can bound $$\int_k^{k+1}\left|\frac{\sin^2(\pi x)-2\pi x\sin(2\pi x)}{4x^3}\right|dx\le \int_k^{k+1}\frac{1+2\pi x}{4x^3}dx=\frac{1}{8}\left(\frac{1}{k^{2}}-\frac{1}{(k+1)^2}\right)+\frac{\pi}{2}\left(\frac{1}{k}-\frac{1}{k+1}\right)\le \frac{1}{4k^3}+\frac{\pi}{2k^2}$$

So the important part is $\int_k^{k+1}\frac{4\pi^2x^2\cos^2(\pi x)}{4x^3}dx=\int_k^{k+1}\frac{\pi^2\cos^2(\pi x)}{x}dx$. We want to show this grows with $1/k$. For this, for $k\le x\le k+1$, we have that $0\le 4(x-(k+1/2))^2\le \cos^2(\pi x)$, which means $$\int_k^{k+1}\frac{\pi^2\cos^2(\pi x)}{x}dx\ge \int_k^{k+1}\frac{\pi^2}{x}4(x-(k+1/2))^2dx=4\pi^2\left(-k-\frac{1}{2}+\left(k+\frac{1}{2}\right)^{2}\ln\left(1+\frac{1}{k}\right)\right)\ge 4\pi^2\left(\frac{1}{12k}-\frac{1}{24k^2}-\frac{1}{6k^3}-\frac{1}{16k^4}\right)$$

The important part is that it is asymptotically $o(1/k)$, which then indicates $\sum_{k=1}^\infty a_k$ diverges and the product of the arc-lengths as well. Then for $p<1/2$, since each arc-length is at least as big as for $p=1/2$, the product of those arc-lengths will diverge as well.

Answer 2

The extrema of the curve are at the half-integers, for which $|\sin(\pi x)| = 1$, and the curve flattens exactly as $x \mapsto 1/x^p$ does when $p$ is increased.

Remember that if $a_k \to 0$, $a_k>0$, then $\prod\limits_{k=1}^{\infty} (1 + a_k)$ and $\sum\limits_{k=1}^{\infty} a_k$ converge or diverge simultaneously. We have $||f'(x)||_{[k,k+1]} = \left|\dfrac{\pi k \cos(\pi k) - p \sin(\pi k)}{k^{p+1}}\right| = \pi/k^p$, and $\sqrt{1 + a} = 1 + a/2 + O(a^2)$, $a \to 0$.

So $\int\limits_{k}^{k+1} \sqrt{1 + (f'(x))^2} dx \le C k^{-2p}$, and for $1/2<p\le1$ the series converges (for larger $p$ the curve is infinitely long on $(0,1]$). If you also want a careful consideration of $p \le 1/2$ you'll probably have to wait for an (amateur or not) analyst...

I'm curious what the motivation for this question is.

Answer 3

Strating from @Varun Vejalla's answer

$$I_k=\int_k^{k+1}f'(x)^2dx=\int_k^{k+1}\frac{4\pi^2x^2\cos^2(\pi x)+\sin^2(\pi x)-2\pi x\sin(2\pi x)}{4x^3}\,dx$$ the antiderivative is rather simple to compute and $$I_k=\frac{ \pi ^2}{4} (\text{Ci}(2 k \pi )-\text{Ci}(2 (k+1) \pi ))+\frac{1}{2} \pi ^2 \log \left(\frac{k+1}{k}\right)$$

Expanded for large values of $k$ $$I_k=\frac{ \pi ^2}{4} \left(\frac{2}{k}-\frac{1}{k^2}+\frac{8 \pi ^2-6}{12 \pi^2 k^3}+\frac{9-6 \pi ^2}{12 \pi ^2 k^4}+O\left(\frac{1}{k^5}\right)\right)$$

Numerically, this gives $I_{10}=\color{red}{0.47022}15$