<s>[INST] 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: $\frac{\sin(n+a)x}{(n+a)x}$?
 [/INST] COMMENT to users: Achille-Hui and Random-Variable
How about this ?:
$$\sum _{n=0}^{\infty } \frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}=\frac{1}{2} \pi  J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\sum _{n=0}^{\infty } \mathcal{L}_x\left[\frac{\sin \left(x \sqrt{n^2+a^2}\right)}{\sqrt{n^2+a^2}}\right](s)=\frac{1}{2} \pi  J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\sum _{n=0}^{\infty } \frac{1}{a^2+n^2+s^2}\right](x)=\frac{1}{2} \pi  J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}+\frac{\pi  \sqrt{-a^2-s^2} \cot \left(\pi  \sqrt{-a^2-s^2}\right)}{2 \left(a^2+s^2\right)}\right](x)=\frac{1}{2}
   \pi  J_0(a x)+\frac{\sin (a x)}{2 a}$$
$$\mathcal{L}_s^{-1}\left[\frac{1}{2 \left(a^2+s^2\right)}\right](x)+\mathcal{L}_s^{-1}\left[-\frac{\pi  \cot \left(\pi  \sqrt{-a^2-s^2}\right)}{2
   \sqrt{-a^2-s^2}}\right](x)=\frac{1}{2} \pi  J_0(a x)+\frac{\sin (a x)}{2 a}$$
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:

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