I have strong feeling that the function, $$ f_\alpha (x) = \sum_{n=0}^\infty \frac{x^n}{n!\Gamma(1+n\alpha)}, $$ is a known special function (here $\Gamma(x)$ is the usual extension of the factorial). Is this the case?

1$\begingroup$ Without the factorial, this would be the MittagLeffler function. Dividing on that factorial corresponds to a kind of Laplace transform. I don't think that Laplace transform of a MittagLeffler function is a "known function". $\endgroup$– Alexandre EremenkoApr 10, 2014 at 3:14

$\begingroup$ You are right in that sense. I am wondering if there are other ways to identify this. The problem shows up in number theory my initial inclination is Bessel Functions. $\endgroup$– Daniel ParryApr 10, 2014 at 3:25

$\begingroup$ When $\alpha>1$ is irrational, I can prove that it does not satisfy any linear ODE with polynomial coefficients. So it is unlikely to be related to Bessel or to any special function. $\endgroup$– Alexandre EremenkoApr 10, 2014 at 3:33

$\begingroup$ How about $\alpha \in (1,0)?$ if I don't mind asking. $\endgroup$– Daniel ParryApr 10, 2014 at 3:35

$\begingroup$ If $\alpha<0$ my arguments are not valid, but then all depends on the arithmetic nature of alpha, how close its multiples can approximate integers. Still does not look like anything familiar. $\endgroup$– Alexandre EremenkoApr 10, 2014 at 3:49
2 Answers
This is an entire function of order $1/\alpha$ when $\alpha>1$. So for irrational $\alpha$ it cannot satisfy any linear differential equation with polynomial coefficients. If $0<\alpha<1$, the order is $1$ but the type is minimal, so again it cannot satisfy any such equation. This excludes most special functions. (But does not exclude their compositions with some irrational power inside).
Entire solutions of linear differential equations with polynomial coefficients have rational order and normal type.
One can obtain an integral representation of this function by taking the integral representation of the MittagLeffler function and then a sort of Laplace transform of it.
Edit. If $\alpha=1$ it is expressed in terms of a Bessel function as the comment below shows.

$\begingroup$ It is related to special functions! dlmf.nist.gov/10.46. Regardless, you helped me get in the right direction so I shall accept the answer anyway. $\endgroup$ Apr 10, 2014 at 3:45

2$\begingroup$ The reference you gave shows relation to Bessel functions when $\alpha=1$ only. $\endgroup$ Apr 10, 2014 at 3:53

$\begingroup$ What do you mean by rational order and normal type? $\endgroup$– joaopaApr 10, 2014 at 8:53

$\begingroup$ For the definition of order and type, see any book which has "Entire functions" in the title, the best one is Levin, Lectures on entire functions, or another book of the same author. $\endgroup$ Apr 10, 2014 at 13:47
This is socalled generalized MittagLeffler function, more exactly the Wright function (as series ) or the Fox function (as inverse Mellin transform). A lot is known about them
Start with 1. http://en.wikipedia.org/wiki/Fox%E2%80%93Wright_function You may use inet search with these names.
Other useful references are:
A.Kilbas, M.Saigo. Htransforms: theory and applications. http://books.google.ru/books?id=SLHqdvUYzEC&pg=PA352&lpg=PA352&dq=Kilbas+saigo&source=bl&ots=xuQJ79z6c&sig=RrYNqEKUIYv64RuK6urTemqUC28&hl=ru&sa=X&ei=eG5LU9bKEoa7ygOt4oCACQ&ved=0CD8Q6AEwAg#v=onepage&q=Kilbas%20saigo&f=false
A.M. Mathai, Ram Kishore Saxena, Hans J. Haubold. The HFunction: Theory and Applications. http://www.springer.com/physics/theoretical,+mathematical+%26+computational+physics/book/9781441909152
Papers of V.Kiryakova, e.g. Multiple (multiindex) Mittag–Leffler functions and relations to generalized fractional calculus: http://www.sciencedirect.com/science/article/pii/S0377042700002922 (and much more her papers on the subject).
MULTIPARAMETRIC MITTAGLEFFLER FUNCTIONS AND THEIR EXTENSION. Anatoly A. Kilbas , Anna A. Koroleva, Sergei V. Rogosin: http://link.springer.com/article/10.2478/s1354001300249
and so on... For sure you will find enough in these references, hope it will be useful!

1$\begingroup$ Hey Sergi. Some work is coming out of this answer and I would like to try to give you an acknowledgement. Could you PM me your full name so you can get credit? $\endgroup$ Sep 26, 2014 at 23:15
