Inverse Laplace transform of $F(s) = \frac{s}{(b+s)(a+b+s)^c}$? After much laboring, sparked by the incredibly helpful comment on my previous question:
Showing that the pmf of a complicated expression sums to 1 (i.e. it converges)
I obtained a closed form for the summation of the pmf (i.e. it was not 1)...
So, based on the context of my problem, I started looking over Integral equations of the first kind, and found that I can use Inverse Laplace transforms.
My question now is something I could not get in Mathematica (again).
How to compute the inverse Laplace transform of $$F(s) = \dfrac{s}{(b+s)(a+b+s)^c}$$
where $a,b > 0$ and $c$ is a whole number?
If I can get this, then I am definitely done with my problem!! Because the pmf will sum to 1!
 A: Using the convolution theorem:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{b}+\text{s}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}\tag1$$
Now, use:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{b}+\text{s}}\right]_{\left(t\right)}=\delta\left(t\right)-\frac{\text{b}}{e^{\text{b}t}}\tag2$$
And:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\frac{t^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot t}\cdot\Gamma\left(\text{c}\right)}\tag3$$
So:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\left(\delta\left(t\right)-\frac{\text{b}}{e^{\text{b}t}}\right)*\frac{t^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot t}\cdot\Gamma\left(\text{c}\right)}\tag4$$
In order to undo the convolution, solve:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\int_0^t\left(\delta\left(\tau\right)-\frac{\text{b}}{e^{\text{b}\tau}}\right)\cdot\frac{\left(t-\tau\right)^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot\left(t-\tau\right)}\cdot\Gamma\left(\text{c}\right)}\space\text{d}\tau\tag5$$
