# 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:

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

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 $E_{\alpha ,1}(z)$ 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 )

$E_{\alpha ,\beta }(z)={\frac {1}{z}}E_{\alpha ,\beta -\alpha }(z)-{\frac {1}{z\Gamma (\beta -\alpha ),}}$ from which the Poincaré asymptotic expansion

$E_{\alpha ,\beta }(z)\sim -\sum _{k=1}{\frac {1}{z^{k}\Gamma (\beta -k\alpha )}}$ follows, which is true for $z\to -\infty$ .

## Special cases

For $\alpha =0,1/2,1,2$ we find: (Section 2 of )

$E_{\frac {1}{2}}(z)=\exp(z^{2})\operatorname {erfc} (-z).$ The sum of a geometric progression:

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

$E_{1,2}(z)={\frac {e^{z}-1}{z}},$ $E_{2,2}(z)={\frac {\sinh({\sqrt {z}})}{\sqrt {z}}}.$ For $\alpha =0,1,2$ , the integral

$\int _{0}^{z}E_{\alpha }(-s^{2})\,{\mathrm {d} }s$ gives, respectively: $\arctan(z)$ , ${\tfrac {\sqrt {\pi }}{2}}\operatorname {erf} (z)$ , $\sin(z)$ .

## Mittag-Leffler's integral representation

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

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

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