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

Solution 1:

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 $\sin(\alpha +\beta)$.

Solution 2:

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.