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
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