So, I am confused as to how I would evaluate the integral: $\int_s^\infty J_n(t)e^{-t}dt$
Of course, I would a step by step explanation that shows how I would compute this integral.
Yes, this integral was loosely inspired by the Laplace Transform, but the main difference is the fact that the results of this seed ($\mathcal V${$f(t)$}($s$)=$\int_s^\infty f(t)e^{-t}dt$) are different from the Laplace transform's seed.
For example: if I take the $\mathcal{V}${$1$}($s$), I get $e^{-s}$, as opposed to $\frac1s$ that the Laplace transform gives me.
Therefore, many of the results are different from the Laplace Transform.
I'll post it as an answer though it is incomplete, but too long for a comment: $$ \begin{align} \int_{s}^{\infty} J_{n}(bx) e^{-ax} \, dx &= \frac{1}{2 \pi} \int_{s}^{\infty} \int_{-\pi}^{\pi} e^{i(n \theta -bx \sin \theta)} e^{-ax} \, d \theta \, dx \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \int_{s}^{\infty} e^{i n \theta} e^{-(a+ib \sin \theta)x} \, dx \, d \theta \\ &= \frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta}}{a + ib \sin \theta}e^{-(sa+isb\sin \theta)} \, d \theta \\ &= \frac{e^{-sa}}{2 \pi} \int_{-\pi}^{\pi} \frac{e^{i n \theta} e^{-isb\sin \theta}}{a + ib \sin \theta} \, d \theta \\ &= \frac{e^{-sa}}{2 \pi} \int_{|z|=1} \frac{z^{n}e^{-{sb \over 2} \left(z-\frac{1}{z} \right) }}{a+\frac{b}{2} \left(z-\frac{1}{z} \right)} \frac{dz} {iz} \\ &= \frac{e^{-sa}}{i \pi} \int_{|z|=1} \frac{z^{n}e^{-{sb \over 2} \left(z-\frac{1}{z} \right) }}{bz^{2}+2az-b} dz \end{align}$$
Now you just have to evaluate the two residues and you're done.