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Let $$f(x)=\int_0^\infty\Big|\sin(t)\cdot\sin(x\,t)\cdot e^{-t}\Big|\,dt,$$ where $|\dots|$ denotes the absolute value. We are concerned only with positive values of $x$ (i.e. let the domain of the function be $\mathbb{R}^+$).

The graph of this function looks roughly as follows (modulo possible errors in numeric algorithms used to plot it): The function graph

The function values at rational points can be evaluated in a closed form, e.g. $$f\left(\frac32\right)=\frac2{145}\left(24+\frac{27\,\sqrt3\,\left(e^{\frac\pi3}+e^{-\frac\pi3}\right)-24\,\left(e^{\frac\pi3}-e^{-\frac\pi3}\right)-4}{e^\pi-e^{-\pi}}\right).$$


Questions:

  • Is the function $f(x)$ continuous? Is it smooth? Is it analytic?
  • How many local extrema does it have?
  • What is the $\lim\limits_{x\,\to\,\infty}f(x)$, if it exists?
  • Can we find a closed-form value of $f(x)$ at an explicit irrational point (given by a closed-form expression)?
  • Is there a general formula for values of $f(x)$ at rational points?
Vladimir Reshetnikov
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    I think for $\lim_{x\to\infty}f(x)$ you can replace $|\sin xt|$ by its average value $\frac{2}{\pi}$ and get $\frac1\pi\coth\frac\pi2=0.347$. – Kirill Oct 29 '13 at 18:10
  • I intuitively feel so, but do not know a theorem to that effect. – Vladimir Reshetnikov Oct 29 '13 at 18:24
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    Instead of $\lim_{x\to\infty}f(x)$ compute instead $\lim_{L\to\infty}\frac1L\int_0^L f(x)\,dx$ and interchange the integrals and limits. This doesn't show the limit exists, but computes it anyway. http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation – Kirill Oct 29 '13 at 18:28

2 Answers2

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My calculation shows that

$$f(x) = \frac{1}{\pi}\coth\left(\frac{\pi}{2}\right) + \frac{2}{\pi} \sum_{n=1}^{\infty} \frac{(2n^{2}x^{2} - 1) \sinh\pi + 2nx\sin (2\pi n x)}{(4n^{2} - 1)(4n^{4}x^{4} + 1) (\cosh \pi -\cos (2\pi n x))}. \tag{1} $$

This allows (at least theoretically) calculate the value of $f$. This shows that

  1. The series $\text{(1)}$ defines a holomorphic function, say $\tilde{f}$, on $\Bbb{C} \setminus \Bbb{R}$ with poles $$ \frac{\pm 1 \pm i}{2n} : n = 1, 2, \cdots \quad\text{and}\quad \frac{k}{n} \pm \frac{i}{2n} : k \in \Bbb{Z}, n = 1, 2, \cdots. $$ Because these poles accumulate to $\Bbb{R}$ as $n\to\infty$, I'm not sure if $\tilde{f}$ and $f$ can be amalgamated to yield a nice function.

  2. This sum converges uniformly on $\Bbb{R}$: putting $y = nx$, $$ \left| \frac{(2n^{2}x^{2} - 1) \sinh\pi + 2nx\sin (2\pi n x)}{(4n^{4}x^{4} + 1) (\cosh \pi -\cos (2\pi n x))} \right| \leq \frac{(2y^{2} + 1)\sinh \pi + 2|y|}{(4y^{4}+1)(\cosh \pi - 1)}, $$ which is uniformly bounded by some constant $C > 0$. Thus we can take limit $x \to \infty$ pointwise to $\text{(1)}$, proving that $$ \lim_{x\to\infty} f(x) = \frac{1}{\pi}\coth\left(\frac{\pi}{2}\right). $$

  3. The graph of its derivative is shown below: I'm unable to find any clear pattern that the zeros of $f'(x)$ must satisfy. One exception is that some zeros seem to be very close to integers.

enter image description here


To derive $\text{(1)}$, I just exploited the following formula:

$$ \left| \sin x \right| = \frac{2}{\pi} \sum_{n \in \Bbb{Z}} \frac{1 - \cos(2nx)}{4n^{2} - 1}, \quad x \in \Bbb{R} $$

Since each summand is non-negative, Tonelli says that we can plug this to the integral and interchange the order of limiting operators to obtain

$$ f(x) = \frac{4}{\pi^{2}} \sum_{m \in \Bbb{Z}} \sum_{n \in \Bbb{Z}} \frac{1}{(4m^{2}-1)(4n^{2} - 1)} \int_{0}^{\infty} (1 - \cos 2mt)(1 - \cos 2nxt) e^{-t} \, dt. $$

Calculating the integral and utilizing the fact $\sum_{n\in\Bbb{Z}} (4n^{2}-1)^{-1} = 0$, this reduces to

$$ f(x) = \frac{4}{\pi^{2}} \sum_{m \in \Bbb{Z}} \sum_{n \in \Bbb{Z}} \frac{1}{(4m^{2}-1)(4n^{2} - 1)(4(nx+m)^{2} + 1)}. $$

Further simplification then gives $\text{(1)}$.


EDIT.

  1. Numerical observation suggest that $f$ is $C^{2}(\Bbb{R}^{+})$, and heuristic investigation suggests that $f^{(3)}$ has jump discontinuity at every point in $\frac{1}{2}\Bbb{Z} \cup \frac{1}{3}\Bbb{Z}$.

enter image description here

Sangchul Lee
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  • If $j_k$ is the jump $\lim_{x\to k+}-\lim_{x\to k-}f^{(3)}(x)$, then numerically $k^2 j_k=\text{const}$. Also the root of $f'$ just below integer $k$ is approximately $k-\frac{0.277}{k-0.877}$. – Kirill Oct 30 '13 at 14:30
  • I think it has more jumps than $\frac12\Bbb{Z}\cup\frac13\Bbb{Z}$, but they are hard to study numerically. I find jumps at rational values $\frac pq$ that are exponentially small in $q$ and maybe $p$ but with no obvious pattern. Be careful with your numerical implementation of $f^{(3)}$, and if you use finite differences with step $\epsilon$ to compute derivates, don't use "almost-rational" values of $\epsilon$ when studying a function with possible jumps at rational values. – Kirill Oct 30 '13 at 21:34
  • @Kirill, that was also my first thought (as you can check at the edit history of this answer), but after rewriting the series expansion of $f^{(3)}$ in the form $$f^{(3)}(x) = \sum_{n=1}^{\infty} \frac{c_{1}(n,x)}{n} + \frac{c_{2}(n,x)}{n^{2}} + \frac{c_{3}(n,x)}{n^{3}} + \cdots $$ with each $c_{k}(n,x)$ being a trigonometric rational function of $n$, I realized that only the leading term can contribute to the discontinuity of $f^{(3)}(x)$, which turns out to be precisely $\frac{1}{2}\Bbb{Z} \cup \frac{1}{3}\Bbb{Z}$. That's why I edited my answer. – Sangchul Lee Oct 30 '13 at 21:49
  • @Kirill, I think I made a mistake. While deducing an answer, I inappropriately ignored the denominator of $c_{1}(n, x)$, but it seems to contribute exponentially small jumps (with size $\sim \sech^{n} \pi$) at each $x = k/n$ with $(k, n) = 1$ when $n \geq 4$. Jumps at $x = k/n$ with $n \leq 3$ are quite large compared to others because the numerator of $c_{1}(n, x)$ consists of trigonometric functions with period $1$, $\frac{1}{2}$ and $\frac{1}{3}$. – Sangchul Lee Oct 30 '13 at 22:04
  • Hm, that's interesting. From my data, $\frac pq$ between $2$ and $3$ with $q\leq 11$, it appears that the jump is $e^{-c q}$, but I don't have enough precision to estimate $c = 1.8\pm 0.6$, although I think it's consistent with $\log\cosh\pi$. Very nice work, by the way. – Kirill Oct 30 '13 at 23:39
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The general formula for rational $x$ can be derived as follows. Consider, first, a convergent sum of the form $\sum_k \frac{p(k)}{q(k)}$, where $p$ and $q$ are polynomials, $\deg q\geq \deg p+2$, and $q$ only has simple roots. If $\rho$ is a variable that runs over the roots of $q$, then $$ \frac1{q(x)} = \sum_{q(\rho)=0}\frac{1}{q'(\rho)}\frac1{x-\rho}, $$ so the sum can be written as $$ \sum_k \frac{p(k)}{q(k)} = \sum_{q(\rho)=0} \frac1{q'(\rho)} \sum_k \frac{p(k)\bmod k-\rho}{k-\rho} + \lfloor p(k)/(k-\rho)\rfloor. $$ Now, $p(k)\bmod k-\rho = p(\rho)$ is a constant, $$ \sum_{k\geq0} \frac{z^k}{k-\rho} = \Phi(z,1,-\rho), $$ by definition of Lerch's transcendent function, and the fact that the original sum is convergent assures us we can ignore the terms $\sum_\rho\frac{\lfloor p(k)/(k-\rho)\rfloor}{q'(\rho)}$, as they must necessarily sum to zero.

Therefore, since $\Phi$ has an asymptotic expansion $$ \Phi(z,1,-\rho) = -\log(1-z) - \gamma - \psi(-\rho), $$ ($\gamma$ is Euler's gamma, and $\psi$ is the digamma function), and since the fact that the original sum converges assures us that we can ignore all terms of $\Phi(z,1,-\rho)$ independent of $\rho$ as we take the limit $z\to1-$ to recover the sum, we therefore have $$ \sum_{k\geq0} \frac{p(k)}{q(k)} = -\sum_{q(\rho)=0} \frac{p(\rho)}{q'(\rho)}\psi(-\rho). $$

The function $f$ has the form $$ f(x) = -\frac{\coth\frac\pi2}{\pi} + \frac2\pi\sum_{n\geq0} S(n, \sin(2\pi nx), \cos(2\pi nx)), $$ where $S(n,\gamma,\delta) = \frac{p(n,\gamma)/q(n,\delta)}{\cosh\pi-\delta}$ is given as in sos440's answer by $$ p(n,\gamma) = (2n^2x^2-1)\sinh\pi+2nx\gamma, \qquad q(n)=(4n^2-1)(4n^4x^4+1). $$

When $x=r/s$ is rational, this can be simplified. Rewrite the sum as $$ \sum_{n\geq0}S(n,\sin2\pi nx, \cos2\pi nx) = \sum_{0\leq\alpha<s} \frac{1}{\cosh\pi-\cos2\pi\alpha x} \sum_{m\geq0} \frac{p(ms+\alpha, \sin2\pi\alpha x)}{q(ms+\alpha)}, $$ and now the inner sum over $m$ is a convergent sum of terms rational in $m$, so can be written as $$ \sum_{q(\rho)=0}\frac{-1}{q'(\rho)} \frac1{s} \sum_{0\leq\alpha<s} \frac{p(\rho, \sin2\pi\alpha x)}{\cosh\pi-\cos2\pi\alpha x} \psi\left(\frac{\alpha-\rho}{s}\right), $$ where $\rho$ runs over the roots of $q(n)$: $\{\pm\frac12, \frac{e^{\pi i j/4}}{x\sqrt{2}}\}$, $j$ odd.

This is pretty much a closed form, it is a finite sum of digamma terms, but you might not be able to simplify it much, especially by hand. For example, Mathematica now gives: $$ f({\textstyle\frac52}) = \left( 68-68 \sqrt{5}-500 \sqrt{10-2 \sqrt{5}} \cosh\frac{\pi }{5}+272 \cosh\frac{2 \pi }{5}-\left(125 \sqrt{10-2 \sqrt{5}}+125 \sqrt{5 \left(10-2 \sqrt{5}\right)}\right) \cosh\pi-\left(160+160 \sqrt{5}\right) \sinh\frac{\pi }{5}+160 \sinh\frac{3 \pi }{5}-160 \sinh\frac{7 \pi }{5} \right)/\left( 689(-1+\sqrt{5}-4 \cosh\frac{2 \pi }{5})\sinh\pi\right) $$

Kirill
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