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  "body": "\n## Definition\n\nLet $j \u003e -1$ be a real number and let $\\Gamma$ be the [[gamma function|0H5SYBSED9]].\nThe *complete Fermi–Dirac integral* of index $j$ is the function $F_j \\colon \\R \\to \\R$ given by\n$$F_j(x) = \\frac{1}{\\Gamma(j+1)} \\int_0^{\\infty} \\frac{t^j}{e^{t - x} + 1} \\, dt.$$\n\n## Remarks\n\nThe integral converges for $j \u003e -1$: near $t = 0$ the integrand is bounded by $t^j$, and for large $t$ it decays like $t^j e^{x - t}$.\n\nFor $x \u003c 0$, expanding $\\frac{1}{e^{t-x} + 1} = \\sum_{k \\ge 1} (-1)^{k-1} e^{-k(t - x)}$ and integrating term by term gives\n$$F_j(x) = \\sum_{k=1}^{\\infty} (-1)^{k-1} \\frac{e^{kx}}{k^{j+1}} = -\\operatorname{Li}_{j+1}(-e^x),$$\nwhere $\\operatorname{Li}_{j+1}$ is the [[polylogarithm|96CHDZKGQ0]], and the identity holds for all real $x$ by analytic continuation.\nIt also gives $F_j'(x) = F_{j-1}(x)$ for $j \u003e 0$.\n\nExample: for $j = 0$, $\\int_0^\\infty \\frac{dt}{e^{t-x} + 1} = \\log(1 + e^x)$, which matches $-\\operatorname{Li}_1(-e^x) = \\log(1 + e^x)$.\n\nNon-example: for $j = -1$ the integrand behaves like $\\frac{1}{t (e^{-x} + 1)}$ near $t = 0$, so the integral diverges; this is why $j \u003e -1$ is required, and $F_{-1}$ is defined instead through the polylogarithm form.\n\nSome authors omit the factor $1/\\Gamma(j+1)$; the normalized version above is the one with $F_j(x) \\sim e^x$ as $x \\to -\\infty$.\n",
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