Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
1  
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

2 Answers 2

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.

share|improve this answer
    
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) :

pict

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.