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: $\frac{\sin(n+a)x}{(n+a)x}$?

To go beyond the limitation $-2\pi<x<2\pi$ for the sine series, we can use the representation G&R (6.677.6) (or Ederlyi TI p.57 1.13.47)
\begin{equation}
\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}
\end{equation} 
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):
\begin{equation}
\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}}
\end{equation} 
for $2p\pi<z<2(p+1)\pi$, 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
\begin{equation}
\frac{\pi}{2}J_0(s)=\int_0^1\frac{\cos sx}{\sqrt{1-x^2}}\,dx
\end{equation} 
the result can be written as
\begin{equation}
\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)
\end{equation} 
which seems to be numerically correct. Unfortunately, no corresponding form of (1) for the cosines seems to exist.
