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I'm wondering if the following series has a closed form:

$$S = e^{-x} \sum_{k=1}^{\infty} \left[\frac{x^k}{k!} \cdot \left(e^{-y} \sum_{l=0}^{k-1} \frac{y^l}{l!} \right) \right]$$

I occasionally stumble across it when I play around with integrals involving modified Bessel functions of the first kind, and it's becoming a bit of a nuisance.

I've checked Prudinkov's Integrals and Series, Vol. 2 and the NIST handbook, but I've had no luck.. If it helps, the inner series is a regularized incomplete gamma function, i.e. $\frac{\Gamma(k, y)}{\Gamma(k)} = e^{-y} \sum_{l=0}^{k-1} \frac{y^l}{l!}$.

All advice much appreciated!

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1 Answer 1

In case you meant something like:

$$\sum^{\infty}_{k=1}\left[\frac{x^{k}}{k!}\cdot\left(\sum_{j=0}^{k-1}\frac{y^{j}}{j!}\right)\right]$$

then that's obviously a different case. Sorry for bugging you about the notations, but the way you wrote it above simply does not imply any dependency. Anyway, this is equal to:

$$\sum_{k=1}^{\infty}\frac{x^{k}\cdot e^{y}\cdot\Gamma(k,y)}{k!\cdot\Gamma(k)}$$

and thus your sum simplifies to

$$S=e^{-x}\sum_{k=1}^{\infty}\frac{x^{k}\cdot Q(k,y)}{k!}.$$

which leaves us the last series... good luck in finding its closed form :-). To be honest, I haven't encountered it before.

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Thanks, but I think you missed the dependency of the second series on $k$, i.e. $\sum_{l=0}^{k-1} \cdots$. –  Donagh Dec 20 '12 at 14:18
    
Hm... you have written in the subscript $k=1$, which implies that there is no dependency. Do yo mean that the summation of the first series starts where the second one ends? If that's the case, I'll try to update the answer. –  Johnny Westerling Dec 20 '12 at 15:52
    
Yes, the outer series is indexed by $k$, which ranges from $1$ to $\infty$, but the inner series is indexed by $l$, which ranges from $0$ to $k - 1$. –  Donagh Dec 20 '12 at 15:58
    
I added an edit; please check whether this is what you meant. –  Johnny Westerling Dec 20 '12 at 16:08
    
Thanks again, but I think you might misunderstand what I wrote. The expression I want to evaluate is exactly as I wrote earlier. You can't simply replace the outer series with the lower regularized Gamma function, and the inner one with the upper regularized Gamma function, because the upper limit of the inner series has a dependency on the index of the outer one. It's not simply the product of two separable functions. –  Donagh Dec 20 '12 at 17:02

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