Integral of the form $e^{i K \sqrt{x^2+a^2}}/\sqrt{x^2+a^2}$ I am trying to solve integrals of the following form $$\int\frac{ {\rm e}^{\imath K \sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} {\rm d} x,$$
where $K$ and $a$ are real positive constants. Mathematica did not provide any solution so I have serious doubts whether it can be solved analytically. And yet, it looks so elementary?
Substituting the square root doesn't seem to help since for $p = \sqrt{x^2+a^2}$ we have $\frac{ {\rm d}p}{{\rm d}x} = \frac x p = \frac{\sqrt{p^2-a^2}}{p}$, leading to $$\int\frac{ {\rm e}^{\imath K p}}{\sqrt{p^2-a^2}} {\rm d} p.$$ We've linearlizted the exponent but the demoninator looks no better. Mathematica still gives up.
Somehow these square-root-of-sum-of-squares terms seem to call for a trigonometric substitution but I just cannot put my finger on what it might look like.
Is this a standard integral? Would someone have a clue whether it has an analytical solution or not?
edit: after Chappers' comment I had a look at the infinite indefinite integral and it seems this can be solved in closed form. Mathematica claims $$\int_{-\infty}^\infty\frac{ {\rm e}^{\imath K \sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}} {\rm d} x = -\pi(Y_0(a K) + \jmath J_0(a K)),$$ where $J_0(z)$ and $Y_0(z)$ are the Bessel functions of the first and second kind, respectively. Unfortunately, I need the integral in a finite interval $[b,b+c]$, where both $b$ and $c$ can be arbitrary. I guess this means I need to integrate numerically.
 A: Changing $x=a\sinh t$ in the integral,
\begin{equation}
 I=\int_\beta^{\gamma}e^{ika\cosh t}\,dt
\end{equation} 
where $\beta=\sinh^{-1}b/a$ and $\gamma=\sinh^{-1}(b+c)/a$. These kinds of integrals are related to incomplete Bessel functions. Their properties are described by Jones in Incomplete Bessel functions and its companion paper. In particular, for $0<\sigma<\pi$
\begin{equation}
 H_0^{(1)}(ka,w)=\frac{2}{i\pi}\int_w^{\infty+i\sigma}e^{ika\cosh t}\,dt
\end{equation} 
Then
\begin{equation}
 I=\frac{2}{i\pi}\left[ H_0^{(1)}(ka,\beta)- H_0^{(1)}(ka,\gamma) \right]
\end{equation} 
It can also be expressed as
\begin{equation}
 I= K_0(-ika,\beta)- K_0(-ika,\gamma)
\end{equation}
where many properties of the incomplete Bessel function $K_0(z,w)$ are given in the cited reference (series expansion, asymptotics...).
A: Consider function:
$$y=\sqrt{x^2+a^2} \ln e^{\sqrt {x^2 +a^2}}$$
$$y'=\frac{x}{\sqrt {x^2+a^2}}.(e^ {\sqrt{x^2+a^2}})(1+\sqrt{x^2+a^2})$$
Now compare this with the relation in question we get:
$$k=-[\frac {\ln x(1+\sqrt{x^2+a^2})}{\sqrt {x^2+a^2}}+1]i$$
This is condition for initial relation to be integrable.
