Mittag-Leffler function

The Mittag-Leffler function can be used to interpolate continuously between a Gaussian and a Lorentzian function.

In mathematics, the Mittag-Leffler function Eα,β is a special function, a complex function which depends on two complex parameters α and β. It may be defined by the following series when the real part of α is strictly positive:[1][2]

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

where ${\displaystyle \Gamma (x)}$ is the Gamma function. When ${\displaystyle \beta =1}$, it is abbreviated as ${\displaystyle E_{\alpha }(z)=E_{\alpha ,1}(z)}$. For ${\displaystyle \alpha =0}$, the series above equals the Taylor expansion of the geometric series and consequently ${\displaystyle E_{0,\beta }(z)={\frac {1}{\Gamma (\beta )}}{\frac {1}{1-z}}}$.

In the case α and β are real and positive, the series converges for all values of the argument z, so the Mittag-Leffler function is an entire function. This function is named after Gösta Mittag-Leffler. This class of functions are important in the theory of the fractional calculus.

For α > 0, the Mittag-Leffler function ${\displaystyle E_{\alpha ,1}(z)}$ is an entire function of order 1/α, and is in some sense the simplest entire function of its order.

The Mittag-Leffler function satisfies the recurrence property (Theorem 5.1 of [1])

${\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{z}}E_{\alpha ,\beta -\alpha }(z)-{\frac {1}{z\Gamma (\beta -\alpha ),}}}$

from which the Poincaré asymptotic expansion

${\displaystyle E_{\alpha ,\beta }(z)\sim -\sum _{k=1}{\frac {1}{z^{k}\Gamma (\beta -k\alpha )}}}$

follows, which is true for ${\displaystyle z\to -\infty }$.

Special cases

For ${\displaystyle \alpha =0,1/2,1,2}$ we find: (Section 2 of [1])

${\displaystyle E_{\frac {1}{2}}(z)=\exp(z^{2})\operatorname {erfc} (-z).}$

The sum of a geometric progression:

${\displaystyle E_{0}(z)=\sum _{k=0}^{\infty }z^{k}={\frac {1}{1-z}},\,|z|<1.}$
${\displaystyle E_{1}(z)=\sum _{k=0}^{\infty }{\frac {z^{k}}{\Gamma (k+1)}}=\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=\exp(z).}$
${\displaystyle E_{2}(z)=\cosh({\sqrt {z}}),{\text{ and }}E_{2}(-z^{2})=\cos(z).}$

For ${\displaystyle \beta =2}$, we have

${\displaystyle E_{1,2}(z)={\frac {e^{z}-1}{z}},}$
${\displaystyle E_{2,2}(z)={\frac {\sinh({\sqrt {z}})}{\sqrt {z}}}.}$

For ${\displaystyle \alpha =0,1,2}$, the integral

${\displaystyle \int _{0}^{z}E_{\alpha }(-s^{2})\,{\mathrm {d} }s}$

gives, respectively: ${\displaystyle \arctan(z)}$, ${\displaystyle {\tfrac {\sqrt {\pi }}{2}}\operatorname {erf} (z)}$, ${\displaystyle \sin(z)}$.

Mittag-Leffler's integral representation

The integral representation of the Mittag-Leffler function is (Section 6 of [1])

${\displaystyle E_{\alpha ,\beta }(z)={\frac {1}{2\pi i}}\int _{C}{\frac {t^{\alpha -\beta }e^{t}}{t^{\alpha }-z}}\,dt,\Re (\alpha )>0,\Re (\beta )>0,}$

where the contour C starts and ends at −∞ and circles around the singularities and branch points of the integrand.

Related to the Laplace transform and Mittag-Leffler summation is the expression (Eq (7.5) of [1], with m=0)

${\displaystyle \int _{0}^{\infty }e^{-tz}t^{\beta -1}E_{\alpha ,\beta }(\pm r\,t^{\alpha })\,dt={\frac {z^{\alpha -\beta }}{z^{\alpha }\mp r}},\Re (z)>0,\Re (\alpha )>0,\Re (\beta )>0.}$

1. Saxena, R. K.; Mathai, A. M.; Haubold, H. J. (2009-09-01). "Mittag-Leffler Functions and Their Applications". Cite journal requires |journal= (help)