# Does there exist a closed form for the sinc function series $\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}}$?

Here I want to get the closed form solution of the following summation

$$\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}} \qquad(1)$$

Or the more general form ($x$ be an arbitrary real number, and $a\geq0$ is a constant):

$$f_a(x) = \sum_{n=1}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(2)$$

I tried the numeircal simulations before I post the question. I truncated the first $1,000,000$ terms of equation (1) and it turned $0.781233190560320$.

Anyone can help me?

In fact, I made it the reduced case when $a=0$. It can be proved by Fourier series: $$f_0(x)=\sum_{n=1}^\infty \frac{\sin nx}{n} = \left\{ \matrix{\dfrac{\pi-x}{2}, 0<x<2\pi\\0, x=0,2\pi} \right.$$ And the function is periodical: $$f_0(x) = f_0(x+2\pi)$$

Edit: How about this one? $$g_a(x) = \sum_{n=1}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(3)$$

We get the "diverging wave solution" in physics when we combine the equation (2) and (3): $$h_a(x)=g_a(x)+\text{i}f_a(x)=\sum_{n=1}^\infty \frac{\exp\left(\text{i}x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\qquad(4)$$

Edit #2: I tested the solution solved by Random Variable (see the most ranked answer and thousands thanks to it!) compared with the truncating results:

$$\sum_{n=1}^N\frac{\sin \left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}, N = 1,000,000, a=1$$

Here is the solution by @Random Variable: the solution of equation (2) as the following:

$$\sum_{n={1}}^\infty \frac{\sin\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}} = \frac{\pi}{2} J_0(ax) -\frac{\sin(ax)}{2a}, a>0, 0<x<2\pi\qquad(2*)$$ where $J_0(ax)$ is the Bessel function of the first kind of order zero.

Here is the comparison:

It can be found that the both agree well when $0 <x<2\pi$,but differ in other domain. So, how about the solution beyond $(0,2\pi)$?

Edit #3:

Inspired by Random Variable's answer, I found the solution of equation (3) as the following:

$$\sum_{n={1}}^\infty \frac{\cos\left(x\sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}} = -\frac{\pi}{2} Y_0(ax) -\frac{\cos(ax)}{2a}, a>0, 0<x<2\pi\qquad(3*)$$ where $Y_0(ax)$ is the Bessel function of the second kind of order zero.

Here is the comparison: Note that equation (3) is divergent when $x=0$.

Possible relating QUESTIONS:

Does there exist a closed form for the non-integer shifted sinc-function series: $\frac{\sin(n+a)x}{(n+a)x}$?

• It's equal to $$\sum_{n=1}^\infty \prod_{k=1}^\infty \left(1-\frac{n^2+1}{k^2\pi^2}\right)$$ but I don't immediately see how that helps. – orlp Nov 27 '17 at 14:28
• Does $(3*)$ seem to hold numerically? I wasn't sure if the Abel-Plana formula would be applicable due to the fact that $\frac{\cos(x \sqrt{z^{2}+a^{2}})}{\sqrt{z^{2}+a^{2}}}$ has branch points on the imaginary axis. – Random Variable Nov 28 '17 at 5:09
• I just reasoned out the solution by analogy according to the equation 10.9.9 HERE. The numerical results does hold on $(-2\pi,2\pi)$! – Jiaxin Zhong Nov 28 '17 at 5:17
• There was a missing condition about the behavior of $f(z)$ as $\operatorname{Re}(z) \to +\infty$. I'm not quite sure how to justify that the Abel-Plana formula is applicable here. And the justification I used in my answer to that other question won't work here. – Random Variable Nov 28 '17 at 12:26
• Do you have a link to the original question/answer? There's some context missing here – Dylan Nov 28 '17 at 12:50

UPDATE:

To apply the Abel Plana formula, the behavior of $f(z)$ as $\operatorname{Re}(z) \to + \infty$ is also important. This was omitted from my answer.

A sufficient condition, now stated HERE, is $f(z) \sim O(e^{2\pi|\Im z|}/|z|^{1+\epsilon})$ as $\operatorname{Re}(z) \to \infty$.

The function here does not satisfy that condition. But as achille hui explains, this condition is about ensuring that $\lim_{b\to\infty} f(b) = 0$ and $$\lim_{b\to\infty}\int_0^\infty \frac{f(b+it)-f(b-it)}{e^{2\pi t} - 1}dt = 0.$$

And achille hui informed me that that latter is indeed satisfied here.

I will ask achille hui to post a brief answer to explain this.

We can use the version of the Abel-Plana formula stated in achille hui's answer HERE.

Alternatively, we could also use the approach I used HERE to evaluate a series involving Bessel functions. Both approaches are related.

(EDIT: I used the DCT in that other answer, which won't work here.)

First notice that the singularities of $f(z) = \frac{\sin \left( x\sqrt{z^{2}+a^{2}}\right)}{\sqrt{z^{2}+a^{2}}}$ are removable.

Also, for $x>0$, $\left|\sin \left( x\sqrt{z^{2}+a^{2}}\right)\right| \sim \frac{e^{x\left|\operatorname{Im}(z)\right|}}{2}$as $\operatorname{Im}(z) \to \pm \infty$.

So if $0 < x < 2 \pi$, the conditions of the Abel-Plana formula are satisfied, and we get \begin{align} \sum_{{\color{red}{n=0}}}^{\infty} \frac{\sin \left( x\sqrt{n^{2}+a^{2}}\right)}{\sqrt{n^{2}+a^{2}}} &= \int_{0}^{\infty} \frac{\sin \left( x\sqrt{t^{2}+a^{2}}\right)}{\sqrt{t^{2}+a^{2}}} \, dt + \frac{1}{2} f(0) + i (0) \\ &= \int_{0}^{\infty} \frac{\sin \left( x\sqrt{t^{2}+a^{2}}\right)}{\sqrt{t^{2}+a^{2}}} \, dt + \frac{\sin (ax)}{2a}. \end{align}

But from this answer, we know that $$\int_{0}^{\infty} \frac{\sin \left( x\sqrt{t^{2}+a^{2}}\right)}{\sqrt{t^{2}+a^{2}}} \, dt = \frac{\pi}{2} J_{0}(ax), \quad (a>0, \ x>0), \tag{1}$$ where $J_{0}(x)$ is the Bessel function of the first kind of order zero.

(To see that $(1)$ is related to the Mehler–Sonine integral representation of the Bessel function of the first kind, you only need to make the initial substitution in that answer).

Therefore, $$\sum_{{\color{red}{n=0}}}^{\infty} \frac{\sin \left( x\sqrt{n^{2}+a^{2}}\right)}{\sqrt{n^{2}+a^{2}}} = \frac{\pi}{2} J_{0}(ax) + \frac{\sin (ax)}{2a}, \quad (a>0, \ 0<x < 2 \pi).$$

To recover the $a=0$ case, you'll have to pull out the $n=0$ term and take the limit on both sides of the equation as $a \to 0^{+}$.

It should be noted that the series converges by Dirichlet's test since $$\sin \left( x\sqrt{t^{2}+a^{2}}\right) \sim \sin(tx) + \mathcal{O} \left(\frac{1}{t} \right)$$ as $t \to \infty$, which can be shown by expanding $\sqrt{t^{2}+a^{2}} = t \sqrt{1+ \frac{a^{2}}{t^{2}}}$ at $t= \infty$ and using the trig identity for $\sin(\alpha +\beta)$.

• awesome, i wouldn't have bet a penny on a closed form but here we go....(+10) – tired Nov 27 '17 at 22:56
• it is interesting that $\sum...=\int...$ becomes exact for $x=k\pi/a$ for all $k\in \mathbb{Z}$. Any idea why this is the case? – tired Nov 27 '17 at 23:00
• @tired I could have evaluated that Bessel function series using the Abel-Plana formula, but at the time I didn't know that you could weaken the conditions. I think the proof would be similar to what I did in that answer. The contour would just be a bit different. – Random Variable Nov 27 '17 at 23:21
• Impressive ... But it will take me some time to understand your answer... You may consider the other case as seen in my last edit. – Jiaxin Zhong Nov 28 '17 at 2:01
• What does the growth conditions mean? – Jiaxin Zhong Nov 28 '17 at 7:20

Per request, this is a supplement to Random Variable's answer.

In Frank W. J Olver's book: Asymptotics and Special Functions, the Abel-Plana formula on finite sum appears in essentially following form:

Let $S$ be the strip $a \le \Re z \le b$ where $a, b \in \mathbb{N}$. For any function $f(z)$

1. continuous on $S$ and analytic on interior of $S$.
2. $f(z) \sim o(e^{2\pi|\Im z|} )$ as $\Im z \to \pm \infty$, uniformly with respect to $\Re z$.

We have \begin{align}\sum_{n=a}^b f(n) = &\int_a^b f(x) dx + \frac12\left( f(a) + f(b)\right) \\& + i \int_0^\infty \frac{f(a+it) - f(a-it) - f(b+it) + f(b-it)}{e^{2\pi t}-1} dt\end{align}

For $f(z) = \frac{\sin(x\sqrt{z^2+a^2})}{\sqrt{z^2+a^2}}$ with $0 < x < 2\pi$, above conditions is satisfied for $a = 0$ and any $b \in \mathbb{Z}$. To obtain the version of AP formula for infinite sum used in Random Variable's answer, we just need:

$$\lim_{b\to \infty}f(b) = 0\quad\text{ and }\quad \lim_{b\to\infty}\int_0^\infty \frac{f(b+it) - f(b-it)}{e^{2\pi t}-1} dt = 0$$ The first condition is trivial. For the second condition, notice for any $n > 0$, $$\left|\sqrt{(n\pm it)^2+a^2}\right| = \left|\sqrt{(n \pm i(t+a))(n \pm i(t-a))}\right| \ge n$$ We find for large $b$ and $t$, $$\frac{\left|f(b\pm it)\right|}{e^{2\pi t}-1} \le \frac{\left|\sin\left(x(b\pm it) + O\left(\frac{a^2}{b}\right)\right)\right|}{b(e^{2\pi t}-1)} \sim \frac{1}{2b}e^{-(2\pi - x)t}\left( 1 + O\left(\frac{a^2}{b}\right)\right)$$ For large $b$ but small $t$, we have $$\frac{\left|f(b + it) - f(b - it)\right|}{e^{2\pi t}-1} \sim O\left(\frac{1}{b}\right)$$ instead (the pole at $t = 0$ from denominator is cancelled by the differences in numerator).

Combine these, we have following estimate of the integral appears in second condition:

$$\int_0^\infty \frac{f(b+it)-f(b-it)}{e^{2\pi t}-1} dt = O\left(\frac{1}{b(2\pi - x)}\right)$$ The second condition is satisfied and the use of AP formula in answering this question is justified.

• Thanks for your supplement and concern! – Jiaxin Zhong Nov 29 '17 at 5:53

To go beyond the limitation $$-2\pi for the sine series, we can use the representation G&R (6.677.6) (or Ederlyi TI p.57 1.13.47) $$$$\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}=\int_0^x J_0\left( n\sqrt{x^2-t^2} \right)\cos at\,dt\tag{1}$$$$ valid for $$x>0$$. (For $$x<0$$, we will use the fact that the series is an odd function of $$x$$, as remarked by @RandomVariable). The summation can be computed using the Schlömilch series (G&R 8.521.1): $$$$\sum_{n=1}^\infty J_0(nz)=-\frac{1}{2}+\frac{1}{z}+2\sum_{m=1}^p\frac{1}{\sqrt{z^2-4\pi^2m^2\pi^2}}$$$$ for $$2p\pi, which defines $$p=\lfloor \frac{z}{2\pi}\rfloor$$. Choosing $$z=\sqrt{x^2-t^2}$$, \begin{align} S(x)&=\sum_{n=1}^\infty\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}\\ &=\int_0^x\sum_{n=1}^\infty J_0\left( n\sqrt{x^2-t^2} \right)\cos at\,dt\\ &=\int_0^x\left[-\frac{1}{2}+\frac{1}{\sqrt{x^2-t^2}}+2\sum_{m=1}^{\lfloor \frac{\sqrt{x^2-t^2}}{2\pi}\rfloor}\frac{1}{\sqrt{x^2-t^2-4\pi^2m^2\pi^2}}\right] \cos at\,dt\\ &=\int_0^x\left[-\frac{1}{2}+\frac{1}{\sqrt{x^2-t^2}}\right]\cos at\,dt+2\sum_{1\le m< \lfloor \frac{x}{2\pi}\rfloor}\int_0^{\sqrt{x^2-4\pi^2m^2}}\frac{\cos at\,dt}{\sqrt{x^2-4\pi^2m^2-t^2}} \end{align} (The summation does not exist if $$\lfloor \frac{x}{2\pi}\rfloor=0$$). With the classic integral representation $$$$\frac{\pi}{2}J_0(s)=\int_0^1\frac{\cos sx}{\sqrt{1-x^2}}\,dx$$$$ the result can be written as $$$$\sum_{n=1}^\infty\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}=\frac{\pi}{2}J_0(ax)-\frac{\sin ax}{2a}+\pi\sum_{1\le m< \lfloor \frac{x}{2\pi}\rfloor}J_0\left( a\sqrt{x^2-4\pi^2m^2-t^2}\right)$$$$ which seems to be numerically correct. Unfortunately, no corresponding form of (1) for the cosines seems to exist.

• Great. This is more of a complete answer compared to others. – Fengshan Xiong Sep 29 at 5:52

COMMENT to users: Achille-Hui and Random-Variable

$$\sum _{n=0}^{\infty } \frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

$$\sum _{n=0}^{\infty } \mathcal{L}_x\left[\frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\right](s)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

$$\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } \frac{1}{a^2+n^2+s^2}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}+\frac{\pi \sqrt{-a^2-s^2} \cot \left(\pi \sqrt{-a^2-s^2}\right)}{2 \left(a^2+s^2\right)}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}\right](x)+\mathcal{L}_s^{-1}\left[-\frac{\pi \cot \left(\pi \sqrt{-a^2-s^2}\right)}{2 \sqrt{-a^2-s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

for $a>0$ and $s>0$

$$\frac{\sin (a x)}{2 a}+\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$ $$\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)=\frac{1}{2} \pi J_0(a x)$$

$$\mathcal{L}_x\left[\mathcal{L}_s^{-1}\left[\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\right](x)\right](s)=\mathcal{L}_x\left[\frac{1}{2} \pi J_0(a x)\right](s)$$

$$\frac{\pi \coth \left(\pi \sqrt{a^2+s^2}\right)}{2 \sqrt{a^2+s^2}}\neq \frac{\pi }{2 \sqrt{a^2+s^2}}$$

and then:

$$\sum _{n=0}^{\infty } \frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\neq \frac{1}{2} \pi J_0(a x)+\frac{\sin (a x)}{2 a}$$

EDITED:

Comparison a numeric InverseLaplaceTransform and Bessel function:

• OP's simulation already tell us $f_a(x) \ne \frac12\pi J_0(ax) + \frac{\sin(ax)}{2a}$ outside the interval $(0,2\pi)$. If you want to show this also fails for $x \in (0,2\pi)$, what you have done is not enough, you need to compute the Laplace transform with functions clipped within $(0,2\pi)$. – achille hui Nov 29 '17 at 18:43
• @achillehui.Yes you're right is true only in interval 0<x<2*Pi – Mariusz Iwaniuk Nov 29 '17 at 20:13
• Thanks for your concerns! – Jiaxin Zhong Nov 30 '17 at 13:56