# closed form solution to a sum of functions

I need to find a closed form expression for the following:

$$\sum_{n=0}^{\infty}\tfrac{a^n}{(n!)(c-bn)}e^{(c-bn)t} \text{ with } a,b,c<1$$

By closed form expression, I mean a formula that can be evaluated in a finite number of standard operations.

Thanks

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What count as "standard operations"? And why do you believe one exists? –  Alex Becker Mar 12 '12 at 22:35
I'm not sure how I'd define standard operations, but an infinite sum wouldn't help my case, so I guess I'm looking for a formula with a finite number of terms. As for why it exists, I believe that based on theorem 3.42 in little Rudin. –  johnny israeli Mar 12 '12 at 22:42
I don't have my little Rudin handy, but I was not aware that it said anything about expressing sums or integrals in finite terms. –  Robert Israel Mar 12 '12 at 23:04

Let $f(t)$ be the sum of your series.

Differentiating term by term yields: $$\begin{split} f^\prime (t) &= \sum_{n=0}^\infty \frac{a^n}{n!}\ e^{(c-bn)t} \\ &= e^{ct}\ \sum_{n=0}^\infty \frac{1}{n!}\ \left( \frac{a}{e^{bt}}\right)^n\\ &= e^{ct}\ \exp \left( ae^{-bt}\right)\\ &= \exp \left( ct+a\ e^{-bt}\right) \end{split}$$ hence $f^\prime$ has a nice elementary expression.

Neverthless $f$ do not possess an elementary expression: in fact, as Raymond shows in his answer, $f$ can be expressed in terms of incomplete gamma functions which aren't elementary.

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I fear that your $e^c$ term should be $e^{ct}$ (I made the same mistake see my correction). Sorry... –  Raymond Manzoni Mar 13 '12 at 21:46
@RaymondManzoni LOL, twin posts sharing the same error... Thanks a lot for making me notice it. –  Pacciu Mar 13 '12 at 22:44

Let's define : $$f(t)=\sum_{0}^{\infty}\tfrac{a^n}{(n!)(c-bn)}e^{(c-bn)t}$$

then $$f'(t)=\sum_{0}^{\infty}\tfrac{a^n}{n!}e^{(c-bn)t}=e^{ct}\sum_{0}^{\infty}\tfrac{(a e^{-bt})^n}{n!}=e^{ct+ae^{-bt}}$$

EDIT (the $t$ was missing in $ct$ !)

for $u=ae^{-bt}$ that is $t=-\frac{\log(\frac ua)}b$ we have : $$\int e^{ct+ae^{-bt}} dt= -\frac 1b \int \frac{e^{-\frac{c\log(\frac ua)}b}e^{u}}u du=-\frac 1b \int \left(\frac ua\right)^{-\frac cb}\frac{e^{u}}u du$$ $$=-\frac {a^{\frac cb}}b \int u^{-1-\frac cb}e^{u} du=-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-u\right)$$

with $\gamma$ the 'lower incomplete gamma function' getting : $$f(t)=C(a,b,c)-\frac {(-a)^{\frac cb}}b\gamma\left(-\frac cb,-ae^{-bt}\right)$$

where $C(a,b,c)=0$ I think.

This is a clearly non elementary result that we may rewrite as : $$f(t)=\frac {(-a)^{\frac cb}}b\left[\Gamma\left(-\frac cb,-ae^{-bt}\right)-\Gamma\left(-\frac cb\right)\right]$$

The same result was obtained by Alpha ('Alternate form' assuming $a,b,c,t$ positive) :

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