Laplace transform of the Bessel function of the first kind I want to show that $$ \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx = \frac{(\sqrt{a^{2}+b^{2}}-a)^{n}}{b^{n}\sqrt{a^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(a) >0 , \,  b >0 ),$$ where $J_{n}(x)$ is the Bessel function of the first kind of order $n$.
But the result I get using an integral representation of $J_{n}(bx)$ is off by a factor of $ \displaystyle \frac{1}{b}$, and I don't understand why.
$$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx &=  \frac{1}{2 \pi} \int_{0}^{\infty} \int_{-\pi}^{\pi}  e^{i(n \theta -bx \sin \theta)} e^{-ax} \, d \theta \, dx \\ &=  \frac{1}{2 \pi} \int_{-\pi}^{\pi} \int_{0}^{\infty} e^{i n \theta} e^{-(a+ib \sin \theta)x} \, dx \, d \theta \\ &=  \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta}}{a + ib \sin \theta} \, d \theta  \\  &= \frac{1}{2 \pi} \int_{|z|=1} \frac{z^{n}}{a+\frac{b}{2} \left(z-\frac{1}{z} \right)} \frac{dz} {iz} \\ &= \frac{1}{i\pi} \int_{|z|=1} \frac{z^{n}}{bz^{2}+2az-b} \, dz  \end{align}$$
The integrand has simple poles at $\displaystyle z= -\frac{a}{b} \pm \frac{\sqrt{a^{2}+b^{2}}}{b}$. 
But only the pole at $\displaystyle z= -\frac{a}{b} + \frac{\sqrt{a^{2}+b^{2}}}{b}$ is inside the unit circle.
Therefore,
$$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-ax} \, dx &= \frac{1}{i \pi} \, 2 \pi i \ \text{Res} \left[ \frac{z^{n}}{bz^{2}+2az-b}, -\frac{a}{b} + \frac{\sqrt{a^{2}+b^{2}}}{b} \right] \\ &= {\color{red}{b}} \  \frac{(\sqrt{a^{2}+b^{2}}-a)^{n}}{b^{n}\sqrt{a^{2}+b^{2}}} . \end{align}$$

Everything is correct up until your computation of the residue. Write
$$bz^2+2az-b=b(z-z_+)(z-z_-)$$
where
$$z_\pm=-\frac{a}{b}\pm\frac{\sqrt{a^2+b^2}}{b}$$
as you have determined. Now,
$${\rm Res}\Bigg(\frac{z^n}{b(z-z_+)(z-z_-)};\quad z=z_+\Bigg)=\lim_{z\to z_+} (z-z_+)\frac{z^n}{b(z-z_+)(z-z_-)}$$
and here you get the desired factor of $1/b$.
