Closed form for an infinite series involving lower incomplete gamma functions I need to evaluate the inverse Laplace transform
$$Q(t) = \mathcal{L}^{-1}\big\{\frac{e^{b/s}}{s(s-a)}\big\}(t).$$
Using the identity $\mathcal{L}^{-1}\{\frac{f(s)}{s-a}\}(t)= e^{at}\int_0^tdu e^{-au}\mathcal{L}^{-1}\{f(s)\}(u)$ with knowledge of the inverse transform $\mathcal{L}^{-1}\{\frac{e^{b/s}}{s}\}(u) = I_0(2\sqrt{bu})$, the series representation of the modified Bessel function $I_0(z) = \sum_{k=0}^\infty \frac{1}{k!k!}\big(\frac{z}{2}\big)^{2k}$, and the definition of the lower incomplete gamma function $ \gamma(k,x) = \int_0^x  t^{k-1}e^{-t}dt$ provides $Q(t)$ in the form
$$ Q(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!k!}\gamma(k+1,at).$$
Is this as good as it gets? Is there an approach I could use to evaluate this sum? So far I have tried expressing the incomplete gamma function in terms of hypergeometric functions, but this does not seem to provide any traction.
One option is to introduce the identity 
$$\gamma(k+1,at) = k!(1-e^{-at}
\sum_{l=0}^k \frac{(at)^k}{k!})$$
obtaining
$$ Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].$$
The second term of this resembles a Humbert series
$$ \Phi_3(\beta,\gamma,x,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(\beta)_m}{(\gamma)_{m+n}m!n!}x^my^n$$ with the wrong summation limits.
Does anyone see a path here? I suppose taking negative values in the Pockhammer symbols might produce a correspondence. 
In any case I expect some hypergeometric function representation of this sum. Can anyone offer guidance? 
I have found several related problems Closed-form Solution for series involving incomplete Gamma Function 
and 
Any way to simplify integral of Confluent Hypergeometric Function of the First Kind?
 A: $Q(t) = \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big].
$
I'll blindly try
to reverse the order of summation
and see what happens.
$\begin{array}\\
S(u, v)
&=\sum_{k=0}^\infty \sum_{l=0}^k \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\sum_{k=l}^\infty  \frac{u^lv^k}{k!l!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=l}^\infty  \frac{v^k}{k!}\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}(e^v-\sum_{k=0}^{l-1}  \frac{v^k}{k!})\\
&=\sum_{l=0}^\infty\frac{u^l}{l!}e^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1}  \frac{v^k}{k!}\\
&=e^ue^v-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l-1}  \frac{v^k}{k!}\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}(\sum_{k=0}^{l}  \frac{v^k}{k!}-\frac{v^l}{l!})\\
&=e^{u+v}-\sum_{l=0}^\infty\frac{u^l}{l!}\sum_{k=0}^{l}  \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{u^l}{l!}\frac{v^l}{l!}\\
&=e^{u+v}-\sum_{l=0}^\infty\sum_{k=0}^{l}\frac{u^l}{l!}  \frac{v^k}{k!}+\sum_{l=0}^\infty\frac{(uv)^l}{l!^2}\\
&=e^{u+v}-S(v, u)+I_0(2\sqrt{uv})
\\
\end{array}
$
where
$I_0$
is the modified Bessel function
of the first kind.
So this isn't a evaluation
but we get the relation
$S(u, v)+S(v, u)
=e^{u+v}+I_0(2\sqrt{uv})
$.
Then
$\begin{array}\\
Q(t) 
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}\sum_{k=0}^\infty \sum_{l=0}^k \frac{(at)^l(b/a)^k}{k!l!}\Big]\\
&= \frac{e^{at}}{a}\Big[e^{b/a}-e^{-at}S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-S(at, b/a)\Big]\\
&= \frac{1}{a}\Big[e^{at+b/a}-(e^{at+b/a}-S(b/a, at)+I_0(2\sqrt{(at)(b/a)}))\Big]\\
&= \frac{1}{a}\Big[S(b/a, at)-I_0(2\sqrt{tb})\Big]\\
\end{array}
$
Again,
not an evaluation,
but a possibly useful
alternative expression.
This reminds me
very much
of some work I did
over forty years ago
on the Marcum Q-function.
You might look that up
and follow the references.
You can start here:
https://en.wikipedia.org/wiki/Marcum_Q-function
A: To recap my findings from @martycohen's guidance, I got to this result for the inverse Laplace transform I need: 
$$ \mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{e^{at}}{a}\sum_{k=1}^\infty \frac{(b/a)^k}{k!}\frac{\gamma(k+1,at)}{\Gamma(k+1)}.$$
The book "An Introduction to the Classical Functions of Mathematical Physics" by Temme (1996) provides the definition 
$$Q_\mu(u,v) = 1- e^{-u}\sum_{k=0}^\infty\frac{u^k}{k!}\frac{\gamma(\mu+k,v)}{\Gamma(\mu+k)}$$
for the non-central $\chi^2$ distribution, also known as the "generalized Marcum $Q$-function", or the just the "Marcum $Q$-function" when $\mu=1$.
Marty's suggestion provides 
$$\mathcal{L}^{-1}\Big\{\frac{1}{s(s-a)}e^{b/s}\Big\}(t) = \frac{1}{a}e^{at+b/a}[1-Q_1(b/a,at)]. $$
There is a representation of this function as an infinite superposition of modified Bessel functions of the first kind, zeroth order:
$$ Q_\mu(u,v) = 1-\int_0^v \Big(\frac{z}{u}\Big)^{\frac{1}{2}(\mu-1)}e^{-z-x}I_{\mu-1}(2\sqrt{xz}).$$
This makes perfect sense in context of the problem which led to the need for this inverse Laplace transform.
Thanks Marty! This helps my research. 
