Integral of Bessel function multiplied with sine $\int_0^\infty J_0(bx) \sin(ax) dx$. I need advice on how to solve the following integral:

$$\int_0^\infty J_0(bx) \sin(ax) dx$$

I've seen it referenced, e.g. here on MathSE, so I know the solution is $(a^2-b^2)^{-1/2}$ for $a>b$ and $0$ for $b>a$, but I don't know how to get there.
I have tried to solve it by using the integral representation of the Bessel function and switching the integrals, resulting in
$$
\frac{1}{\pi}\int_0^\pi \int_0^\infty \sin(ax)\cos(bx\sin(\theta))dx d\theta.
$$
Doing the dx-integration, I get
$$
=\frac{1}{\pi}\int_0^\pi \frac{2a}{a^2-b^2\sin^2(\theta)}\left(1-\lim_{x\to\infty}\cos(ax)\cos(bx\sin(\theta)\right)d\theta
$$
and have no idea how to proceed from there.
Is there anything wrong with my calculations? Should I use a totally different approach? Any help appreciated.
 A: We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \,  b >0 )\tag{1}. $$
(See this question  for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that  $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $  converges uniformly for  all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &=  \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $  a < b$,  $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases} 
      \frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \,  & \quad 0 < a < b \\
      \frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0
   \end{cases} $$
A: 
I need advice on how to solve the following integral

Then you shall receive it ! ;-$)$
Use Euler's formula in conjunction with the series expansion of the Bessel function. This will 
require switching the order of summation and integration, and recognizing the binomial series 
of $~\dfrac1{\sqrt{b^2\color{red}+c^2}}~=~\displaystyle\int_0^\infty J_0(bx)~e^{cx}~dx,~$ where the latter converges for $~c<0.~$ Now let 
$c=\epsilon+ia,~$ where $\epsilon\to0.\quad$ :-$)$
