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