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I want to know under what conditions does the Mittag-Leffler function ${E_{\alpha ,1}}(z),(0 < \alpha < 1)$ has no real zero, where

${E_{\alpha ,1}}(z) = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{\Gamma (\alpha k + 1)}}}$.

${E_{1,1}}(z) = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{\Gamma (k + 1)}}} {\text{ = }}{{\text{e}}^z}$ or ${E_{0,1}}(z) = \frac{1}{{1 - z}}$ has no zero, how about $\alpha \in (0,1)$?

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None of these functions have real zeroes, because they can be written as moment generating functions of certain random variables $Y_\alpha$. More precisely, for $0<\alpha <1$ $$E_{\alpha,1}(z)=\mathbb{E} e^{zY_\alpha}$$ where $$X_\alpha:=(1/Y_{\alpha})^{1/\alpha}$$ has the extreme positive stable distribution with index $\alpha$ (Levy-distribution), i.e- Laplace transform $$Ee^{-pX_\alpha}=e^{-p^\alpha}$$ for $p\geq 0$. See W. Feller, Intro. Prob. Theo. Appl. II, (1971), p.453.

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EDIT: at the time of this answer the OP did not specify that the zero should be real.

The order (as an entire function) of $E_{\alpha,1}$ is $\frac{1}{\alpha}$. It so happens that entire functions with non-integer order take all complex values infinitely often. So in general $E_{\alpha,1}$ will have infinitely many zeroes. I don't know offhand about the case $\frac{1}{\alpha}\in \mathbb{N}$ though. In that case you can use the functional equation of $\Gamma$.

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  • $\begingroup$ For $1/\alpha=2$, I think $E_{1/2,1}(z)=e^{z^2}(1+\operatorname{erf}(z))$, so the question is about solutions of $\operatorname{erf}(z)=-1$ (certainly not real). $\endgroup$
    – Pietro Majer
    Commented Mar 31, 2014 at 17:26
  • $\begingroup$ Indeed when I mentionned the functional equation of $\Gamma$ I intented it in order to factor out an exponential term $\exp(z^{1/\alpha})$. As to whether the factor vanishes, I'd bet it does (when $\alpha\neq1$), but don't have a proof. Yet surely the right people must know that. $\endgroup$
    – Loïc Teyssier
    Commented Mar 31, 2014 at 18:48
  • $\begingroup$ The Maple code $$ restart; RootFinding:-Analytic(erf(z)+1, z = -3-10*I .. 3+10*I) $$ outputs $$1.354810128-1.991466843i, 2.177044906-2.691149024i, 2.784387613-3.235330868i, 1.354810128+1.991466843i, 2.177044906+2.691149024i, 2.784387613+3.235330868i. $$ $\endgroup$
    – user64494
    Commented Mar 31, 2014 at 18:58

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