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  "body": "\n## Definition\n\nLet $\\alpha, \\beta \\in \\C$ with $\\operatorname{Re} \\alpha \u003e 0$, and let $\\Gamma$ be the [[gamma function|0H5SYBSED9]], with the convention $1/\\Gamma(-m) = 0$ for $m = 0, 1, 2, \\ldots$.\nThe *two-parameter Mittag-Leffler function* is the [[power series|0MZB06ST2B]]\n$$E_{\\alpha, \\beta}(z) = \\sum_{k=0}^{\\infty} \\frac{z^k}{\\Gamma(\\alpha k + \\beta)}, \\qquad z \\in \\C.$$\nThe *Mittag-Leffler function* is $E_\\alpha = E_{\\alpha, 1}$.\n\n## Remarks\n\nThe series has infinite radius of convergence, because $\\Gamma(\\alpha k + \\beta)$ grows faster than any power of $k$ when $\\operatorname{Re} \\alpha \u003e 0$, so $E_{\\alpha, \\beta}$ is an [[entire function|0G4F0XB9X2]].\n\nExample: for $\\alpha = \\beta = 1$, $\\Gamma(k + 1) = k!$, so $E_{1,1}(z) = e^z$.\nAlso $E_2(z) = \\sum_k \\frac{z^k}{(2k)!} = \\cosh \\sqrt{z}$, and $E_{1/2}(z) = e^{z^2} \\operatorname{erfc}(-z)$.\n\nNon-example: for $\\alpha = 0$ and $\\beta = 1$ the series is $\\sum_k z^k$, which is not entire; it converges only for $\\abs{z} \u003c 1$, so the condition $\\operatorname{Re} \\alpha \u003e 0$ cannot be dropped.\n\nLaplace transform: for $\\operatorname{Re} s$ large enough,\n$$\\int_0^\\infty e^{-st} \\, t^{\\beta - 1} E_{\\alpha, \\beta}(\\lambda t^\\alpha) \\, dt = \\frac{s^{\\alpha - \\beta}}{s^\\alpha - \\lambda},$$\nwhich follows by integrating the series term by term.\n",
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