Proof that the roots of $\mathrm e^{-πx}=\sin πx$ approach integers as $x\to \infty$

Question

This question is inspired by @gt6989b’s comment here.

Numerical analysis suggests that the roots of the equation $\newcommand{\e}{\mathrm{e}} \e^{-πx} = \sin πx$ rapidly and closely approach integers as $x\to\infty$. Here’s a quick list of the first nine solutions:

$$\begin{array}{l} 0.18733579075230\dots \\ 0.98560325090923\dots \\ 2.00059331886993\dots \\ 2.99997431047250\dots \\ 4.00000111005168\dots \\ 4.99999995203014\dots \\ 6.00000000207297\dots \\ 6.99999999991042\dots \\ 8.00000000000387\dots \\ \end{array}$$

How can I prove (or disprove) that these values will get closer and closer to integers?

Wolfie notes that the system has the alternate form $$\newcommand{\i}{\mathrm{i}} \e^{-πx} = \frac{\i\e^{-\i πx} - \i\e^{\i πx}}2$$

Answer 1

Since for a large positive integer $n$, $e^{-\pi n} \approx 0$ and $\sin(\pi x)$ has roots at the integers, we expect that the equation has roots close to positive integers. To gain a better approximation, we choose the approximate value $n \in \mathbb{N}$ for the $n$th root $x_n$ and apply Newton's method once. This yields the better approximation: $$ x_n \approx n + \frac{1}{\pi }\frac{1}{{( - 1)^n e^{\pi n} + 1}}. $$ This shows that $x_n$ converges to $n$ exponentially fast.

Answer 2

Consider the function $$y= e^{\pi x}\,\sin(\pi x)$$ and expand it around $x=n$ $$y=(-1)^n e^{n \pi}\left(\pi (x-n)+\pi ^2 (x-n)^2+\frac{1}{3} \pi ^3 (x-n)^3+O\left((x-n)^5\right)\right)$$ Now, use series reversion to get $$x=n+\frac{e^{-n\pi} }{\pi }y\left(e^{-i \pi n}-e^{(-1-2 i) \pi n} y+\frac{5}{3} e^{(-2-3 i) \pi n} y^2-\frac{10}{3} e^{(-3-4 i) \pi n} y^3+O\left(y^4\right)\right)$$ and now, make $y=1$ to get $$x_n=n+(-1)^n\frac{ e^{-n\pi }}{\pi }-\frac{e^{-2n \pi }}{\pi }+(-1)^n\frac{5 e^{-3 n\pi }}{3 \pi }-\frac{10 e^{-4 \pi n}}{3 \pi }+\cdots$$

Using this truncated formula $$\left( \begin{array}{cc} n & x_n \\ 1 & 0.985603644503601858348373 \\ 2 & 2.000593318869874358850040 \\ 3 & 2.999974310472503086561926 \\ 4 & 4.000001110051677367565600 \\ 5 & 4.999999952030143030217180 \\ 6 & 6.000000002072965152317527 \\ 7 & 6.999999999910419052735504 \\ 8 & 8.000000000003871143731943 \\ 9 & 8.999999999999832712711178 \\ 10 & 10.00000000000000722913923 \\ 11 & 10.99999999999999968760057 \\ 12 & 12.00000000000000001350000 \\ 13 & 12.99999999999999999941661 \\ 14 & 14.00000000000000000002521 \\ 15 & 14.99999999999999999999891 \end{array} \right)$$

$$|x_n-n|\sim \frac 1 \pi e^{- n \pi}$$

Answer 3

I don't know if this answer is satisfactory but $|\sin (\pi x)| >\epsilon$ implies $|e^{\pi x} \sin (\pi x)| >1$ whenever $x >\frac 1 {\pi} \ln (\frac 1 {\epsilon})$ which implies that $x$ is not a root of the equation $e^{-\pi x}= \sin (\pi x)$. Of course, $|\sin (\pi x)| \leq \epsilon$ implies that $x$ is close to an integer.