# An incomplete double series: Is there a special functions representation?

Consider the sum $$Q(v,u) = \sum_{l=0}^\infty\sum_{k=0}^{l}\frac{u^l}{l!} \frac{v^k}{k!}$$ which arises from the inverse Laplace transform of $$f(s) = \frac{1}{s(s-a)}e^{b/s}.$$

Is there a means to express $$Q(v,u)$$ in terms of some special functions? It seems to be some sort of incomplete Humbert series like $$\Phi_3(\beta,\gamma,x,t) = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(\beta)_m}{(\gamma)_{m+n}m!n!}x^my^n$$ in that it has the wrong summation limits. Any thoughts are greatly appreciated!

As I showed in my answer to your other question Closed form for an infinite series involving lower incomplete gamma functions,

$$Q(u, v)+Q(v, u) =e^{u+v}+I_0(2\sqrt{uv})$$.

I also suggested that you research the Marcum Q-function.

• Thanks @marty. This is an alternate representation which helps me a lot. I did not find much of an analogue with the Marcum Q yet, but will research more today. Jul 26, 2019 at 17:52
• Update: Temme 1996 special functions book carefully outlines this function and highlights its relation to the $\chi^2$ distribution. This is definitely going to be the way forward. Thanks Marty. I will update with a solution shortly in case anyone else comes across such a problem. Jul 26, 2019 at 19:04

$$\sum\limits_{l=0}^\infty\sum\limits_{k=0}^l\dfrac{u^lv^k}{l!k!}$$

$$=\sum\limits_{k=0}^\infty\sum\limits_{l=k}^\infty\dfrac{u^lv^k}{l!k!}$$

$$=\sum\limits_{k=0}^\infty\sum\limits_{l=0}^\infty\dfrac{u^{l+k}v^k}{(l+k)!k!}$$

$$=\Phi_3(1,1;u,uv)$$ (according to https://en.wikipedia.org/wiki/Humbert_series)