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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?

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$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

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  • $\begingroup$ Thanks @marty cohen! This is excellent. I have to study a few steps more carefully to fully understand. I searched for your work on Q-functions but couldn't find it: can you offer a bit more info? I'd love to have a look at it $\endgroup$ – kevinkayaks Jul 25 at 23:41
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    $\begingroup$ It was done when I worked for a private company, and I don't have my work any more. Sorry. $\endgroup$ – marty cohen Jul 26 at 2:00
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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.

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