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?
Best Answer
$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