Closed form analytical expression for $\int_0^\infty e^{-\lambda a} J_0(\lambda r) \sin(\lambda t) \, \mathrm{d}\lambda$ Consider the following function that is defined by an infinite integral
$$
I(r,t,a) = \int_0^\infty e^{-\lambda a} J_0(\lambda r) \sin(\lambda t) \, \mathrm{d}\lambda \, ,  
$$
where $r, t,$ and $a$ are positive real numbers.
Here, $J_0$ denotes the zeroth order Bessel function of the first kind.
Clearly, the integral above in convergent.
i was wondering whether an explicit analytical expression for $I$ can be obtained analytically. 
Any help or suggestion is highly appreciated.
Thank you
 A: It's easier to start with Bessel's Integral and apply residue theorem. For simplicity make a change of variable to ensure $r = 1$. Then, we have
\begin{align*}I = \int_0^\infty e^{-(a-it)x} J_0(x)\,dx &= \frac{1}{2\pi}\int_0^{\infty} \int_{-\pi}^{\pi} e^{-(a-it)x} e^{ix\sin \tau}\,d\tau\,dx \\&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \int_0^{\infty} e^{-(a-i(t+\sin \tau))x} \,dx \,d\tau \\&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1}{a-i(t+\sin \tau)} \,d\tau \\&= \frac{1}{2\pi} \int_{|z| = 1} \frac{1}{a-i\left(t+\frac{1}{2i}\left(z - \frac{1}{z}\right)\right)} \,\frac{dz}{iz}\tag{$\ast$}\\&= \frac{i}{\pi} \int_{|z| = 1} \frac{1}{z^2 - 2(a-it)z - 1} \,dz\end{align*}
where, in step $(\ast)$ we made the change of variable $z = e^{i\tau}$.
The function has two simple poles at $\alpha^{\pm} = a-it \pm \sqrt{(a-it)^2 + 1}$ (where we are using the principle branch of square root since $a > 0$) and only $\alpha^{-}$ is inside the unit disk. Therefore, the integral is $$I = \frac{i}{\pi}\frac{2\pi i}{\alpha^- - \alpha^+} = \frac{1}{\sqrt{(a-it)^2 + 1}} = \frac{1}{\sqrt{a^2 - t^2 + 1 - 2iat}}.$$
Now, comparing imaginary parts on both sides should give the result.
