Evaluating integral involving Bessel function $J_{1/2}$ How to evaluate the integral
$$
 \int_0^{\infty}\cos bx\frac{\sin t\sqrt{x^2-a^2}}{\sqrt{x^2-a^2}}\,dx
\quad (t>|b|),
$$
knowing that
$$
\frac{\sin ut}{u}=\sqrt{\frac{\pi t}{2u}}J_{1/2}(ut)
 =\frac{\sqrt{\pi t}}{2}\frac{1}{2\pi i}\int_{C_1}
 \frac{e^{z-u^2t^2/(4z)}}{z^{3/2}}\,dz,
$$
where $C_1=\{z:\Re z=\alpha\}  \; (\alpha>0)$ ? The hint being to write down the consecutive integrals, and then change the order of integration.
The latter equality is easily established using the expansion
$$
J_{\nu}(z)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\Gamma(k+\nu+1)}
 \left(\frac{z}{2}\right)^{\nu+2k},
$$
and the sin series.
One apparent thing is to evaluate the integral with $\cos bx$ replaced through $e^{ibx}$ (or $e^{-ibx}$), but I don't see what to do next.
 A: The integral representation should be written as
\begin{equation}
 \frac{\sin ut}{u}=\sqrt{\frac{\pi t}{2u}}J_{1/2}(ut)
 =\frac{\sqrt{\pi }t}{2}\frac{1}{2\pi i}\int_{C_1}
 \frac{e^{z-u^2t^2/(4z)}}{z^{3/2}}\,dz
\end{equation} 
(prefactor of the integral contains $t$ and not $\sqrt{t}$).
Introducing this integral representation, it comes
\begin{equation}
I=\frac{\sqrt{\pi} t}{2}\frac{1}{2i\pi}\int_{C_1} \frac{e^{z+\frac{a^2t^2}{4z}}}{z^{3/2}}\,dz\int_0^\infty\cos bx \,e^{-\frac{t^2x^2}{4z}}\,dx
\end{equation} 
Using the Fourier transform of the Gaussian (with $\Re z>0$),
\begin{equation}
\int_0^\infty\cos bx e^{-\frac{t^2x^2}{4z}}\,dz=\frac{\sqrt{\pi z}}{t}e^{-\frac{b^2z}{t^2}}
\end{equation} 
it comes
\begin{equation}
I=\frac{\pi}{2}\frac{1}{2i\pi}\int_{C_1}e^{z(1-\frac{b^2}{t^2})+\frac{a^2t^2}{4z}}\frac{dz}{z}
\end{equation} 
As $1-\frac{b^2}{t^2}>0$, it can be written as
\begin{equation}
I=\frac{\pi}{2}\frac{1}{2i\pi}\int_{C_1}e^{s+\frac{a^2\left( t^2-b^2 \right)}{4s}}\frac{ds}{s}
\end{equation} 
We recognize the Schläfli representation for $J_0\left( ia\sqrt{t^2-b^2} \right)$:
\begin{align}
I&=\frac{\pi}{2}J_0\left( ia\sqrt{t^2-b^2} \right)\\
&=\frac{\pi}{2}I_0\left( a\sqrt{t^2-b^2} \right)
\end{align}
