The problem occurs when the first element of the summation includes $\Gamma(0)$ in the denominator which is not defined?
By the recurrence property of the Mittag-Leffler extended function, for a>0, z \in R,
E_{1,a}(z) =z E_{1,1+a}(z) + {\Gamma(a)}^{-1}
Thus, for all real z, E_{1,a}(z) tends to z \exp(z) as a>0 tends to 0
Hence the function can be defined for \alpha = 1, \beta = a = 0 by continuous extension as beta goes to 0^+. And this holds for all positive \alpha.
Thus it sufficies simply to replace {\Gamma(0)}^{-1} by 0, if the continuity with respect to \beta is assumed.
Regards
NOTE. The series when started from k=1 is convergent to z \exp(z)
Dear Walid,
I would like to make a short remark and continue the fine answer given by Professor Joachim Domsta given above. The series representation defining the two-parameter Mittag-Leffler function is given e.g. in Wikipedia
https://en.wikipedia.org/wiki/Mittag-Leffler_function
Specializing now \alpha=1 and separating the constant term 1/\Gamma(\beta) in the resulting series you will get a series representation
(1) E_{1,\beta}(z)=1/\Gamma(\beta)+\sum_{k=1}^\infty 1/\Gamma(k+\beta)z^k.
This actually proves the recurrence property given in the first answer. Now, using the property
(2) \Gamma(z+1)=z\Gamma(z)
of the Gamma function allows us to write the constant term on the right hand side of (1) in a form
\beta/\Gamma(\beta+1).
Now the both terms on the right hand side of (1) depend analytically on \beta, so you can extend the definition of the function E_{1,\beta} to cover also the case \beta=0 by defining
E_{1,0}(z)=0+\sum_{k=1}^\infty 1/\Gamma(k) z^k.
But the series is equal to z\exp(z) (since \Gamma(k)=(k-1)!) as it was emphasized in the answer of Professor Domsta. Therefore you will end up to an elementary function by using the analytic continuation of a known formula to cover a wider range of parameters.
Note that the continuation works not only for \beta=0 but also for negative \beta by using the property (2) of the Gamma function. For example for \beta=-1 separate the first two terms of the series expansion. Since the Gamma function has poles at non-positive integers, two first two terms will disappear in the limit and you will end up to an elementary function z^2\exp(z).
Hopefully you will find this procedure interesting.
Best regards, Jukka
Thanks to all participants of this interesting discussion. However there are still some natural questions.The basic one is the following:
Is the continuous extension over the complex plane (by the limit procedure) ianalytic; if yes, then with respect to the system of the independent variable and the parameters or just with respect to each single variable (i.e. with the other kept constant)? I guess that the stronger analyticity should hold due to the relation with hypergeometric function(s). Possibly, someone of RGaters has found a detailed elaborated work on this problem?!
Regards, Joachim
Dear Joachim,
you posed an interesting question. I have to say that I am not familiar with the theory of functions of several complex variables (here also the parameter \beta is allowed to be complex). A quick look on the web concerning multidimensional analytic functions reveals that the concept of analyticity in the multidimensional case means just being analytic with respect to each variables separately.
Therefore, since the reciprocal Gamma function and the series appearing in the representation are entire functions (with the other variable being fixed), I would say that the extension is analytic even regarded as a two-dimensional complex function. But this is just heuristic reasoning, which should be checked if one would like to be sure on the statement. Perhaps someone else can give an answer to this question immediately.
Best regards, Jukka
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