# Asymptotics of a strange oscillatory function

Consider the function $$f:\mathbb{R}\to \mathbb{R}$$ defined by $$f(x)=\sum_{n\geq 1}\sin(x/n^2)$$. It is easy to see that $$f(x) = O(\sqrt{x})$$ for large real $$x$$. Is it true that

1. $$f(x)>0$$ for $$x>0$$, and
2. $$\liminf_{x \to \infty} \frac{f(x)}{\sqrt{x}}>0$$?

Do $$c=\liminf_{x \to \infty} \frac{f(x)}{\sqrt{x}}$$ and $$C=\limsup_{x \to \infty} \frac{f(x)}{\sqrt{x}}$$ have "nice" explicit values?

Here is a graph of the function for $$0 \leq x \leq 600$$.

• By the way, even though the plot might suggest otherwise, thanks to the different scales, $f$ can't really oscillate very heavily since $|f'|\le \sum 1/n^2=\pi^2/6$ (a trivial point, but it fooled me for a while). Feb 21 at 21:02
• If I am not mistaken, $$f(x) = \sqrt {\frac{{\pi x}}{2}} + \frac{1}{{2\pi {\rm i}}}\int_{ + \infty }^{( - 1/4^+)} {\frac{\pi }{{\sin (\pi s)}}\frac{{\zeta (4s + 2)}}{{\Gamma (2s + 2)}}x^{2s + 1} {\rm d}s}$$ for $x>0$, where the contour of interation is a Hankel-type contour encircling the point $-1/4$ in the positive sense.
– Gary
Feb 26 at 7:38

I will show (unless I made a mistake) $$f(x) = a\sqrt{x} + O(x^{1/3}),$$ for some constant $$a > 0$$.

For each $$x > 0$$, let $$g(t) = \sin(x/t^2)$$. Let $$\varepsilon > 0$$ to be chosen later depending on $$x$$. We break the sum up into two pieces by $$f(x) = \sum_{n \leq \varepsilon \sqrt{x}} g(n) + \sum_{n > \varepsilon \sqrt{x}} g(n).$$ Since $$|g(n)| \leq 1$$ for all $$n$$, the first sum is $$O(\varepsilon \sqrt{x})$$. For the second sum, we use Euler-Maclaurin to get $$\sum_{n > \varepsilon \sqrt{x}} g(n) = \int_{\varepsilon \sqrt{x}}^{\infty} g(t) dt + O(\int_{\varepsilon \sqrt{x}} |g'(t)| dt) + O(1).$$ Note that $$g'(t) = -2 \cos(x/t^2) \cdot x/t^3$$, so that $$\int_{\varepsilon \sqrt{x}} |g'(t)| dt \ll \int_{\varepsilon \sqrt{x}}^{\infty} \frac{x}{t^3} dt \ll \varepsilon^{-2}.$$

Finally, we have that the integral with $$g(t)$$ satisfies $$\int_{\varepsilon \sqrt{x}}^{\infty} g(t) dt = \int_0^{\infty} g(t) dt + O(\varepsilon \sqrt{x}),$$ and this error term was already accounted for. In total, we get $$f(x) = \int_0^{\infty} \sin(x/t^2) dt + O(\varepsilon \sqrt{x} + \varepsilon^{-2}).$$ The integral is $$a \sqrt{x}$$ for some $$a > 0$$, which is $$\sqrt{\pi/2}$$ if we can trust Wolfram Alpha. The error term is optimized with $$\varepsilon = x^{-1/6}$$.

Probably this estimate can be improved with more work, which could potentially identify some lower-order terms visible in the graph.

• Nice! We can also get $a$ by hand (in terms of a Fresnel integral, that is) as $a=\int_0^{\infty} \sin (1/t^2)\, dt =\int_0^{\infty} \sin u^2 du/u^2 = 2 \int_0^{\infty} \cos u^2\, du = 2\cdot \sqrt{\pi/8}$, as claimed, by an integration by parts in the last step. Feb 22 at 1:48
• Beautiful! Thanks for this! Feb 22 at 6:05

Yes, to (1) and (2), but the argument below is much too crude to identify $$c$$.

The obvious attempt is to use $$\sin t\ge 2t/\pi$$, $$0\le t\le \pi/2$$ (or something similar), for the terms with large $$n$$. So we can set $$N=N(x)=\lceil \sqrt{2x/\pi}\rceil$$, and we obtain $$f(x)\ge \frac{2x}{\pi} \sum_{n=N}^{\infty} \frac{1}{n^2} - N+1\ge \frac{2x}{\pi (N+1)}-N+1\ge \frac{2x}{\sqrt{2\pi x}+2\pi}-\sqrt{\frac{2x}{\pi}} .$$ This is asymptotically $$(\sqrt{2/\pi}-\sqrt{2/\pi})\sqrt{x}$$, so doesn't quite do it, but even the tiniest improvement will produce answers.

This is easy to do. For example, when $$n\ge 2N$$, then $$x/n^2\le \pi/8$$, and we obtain an improved bound $$\sin x/n^2\ge cx/n^2$$, with $$c>2/\pi$$ for these $$n$$. Since the part $$\sum_{N\le n\le 2N} 1/n^2 \simeq 1/(2N)$$ contributes only a fraction (one half, to be precise) of the full sum, the improvement of the constant on the other part will make itself felt in an improved overall constant.

This shows that $$\liminf f(x)/\sqrt{x}>0$$ (and gives an explicit but crude lower bound), and also that $$f>0$$ since one can keep track of the error terms and $$x$$ of moderate size has already been discussed by the OP (numerically).