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  "body": "\n## Definition\n\nLet $a \u003c b$ and let $K \\colon [a, b] \\times [a, b] \\to \\C$ be a [[continuous kernel|ZM5G144PZG]].\nFor $n \\ge 1$ and points $x_1, \\dots, x_n \\in [a, b]$, let $\\det[K(x_i, x_j)]_{i,j=1}^{n}$ be the [[determinant|00KDZRW4C0]] of the $n \\times n$ matrix with entries $K(x_i, x_j)$.\nFor $n \\ge 1$ let\n$$c_n = \\frac{1}{n!} \\int_a^b \\cdots \\int_a^b \\det[K(x_i, x_j)]_{i,j=1}^{n} \\, dx_1 \\cdots dx_n,$$\nand let $c_0 = 1$.\nThe *Fredholm determinant* of $K$ is the function $D \\colon \\C \\to \\C$ given by\n$$D(\\lambda) = \\sum_{n=0}^{\\infty} (-1)^n c_n \\lambda^n.$$\nThe series converges for every $\\lambda \\in \\C$, so $D$ is an [[entire function|0G4F0XB9X2]].\n\n## Remarks\n\nExample: for $K(x, y) = xy$ on $[0, 1]$, $c_1 = \\int_0^1 x^2 \\, dx = 1/3$.\nFor $n \\ge 2$ the matrix $[x_i x_j]$ has rank at most $1$, so its determinant is $0$ and $c_n = 0$.\nHence $D(\\lambda) = 1 - \\lambda/3$.\n\nNon-example: the definition does not apply when $K$ is only square-integrable, because the diagonal values $K(x, x)$ enter $c_1 = \\int_a^b K(x, x) \\, dx$ and are not defined for an element of $L^2$.\nFor such kernels one uses a modified determinant in which the diagonal terms are removed.\n\nThe convergence for all $\\lambda$ follows from Hadamard's inequality: if $\\abs{K} \\le M$, then $\\abs{c_n} \\le n^{n/2} M^n (b - a)^n / n!$.\nThe coefficient $c_1$ is the integral $\\int_a^b K(x, x) \\, dx$, the trace of the kernel, and $D(\\lambda) = 1 - c_1 \\lambda + \\cdots$.\nFor a degenerate kernel $K(x, y) = \\sum_{i=1}^{m} a_i(x) b_i(y)$, one has $D(\\lambda) = \\det(I_m - \\lambda A)$, where $A_{ij} = \\int_a^b b_i(y) a_j(y) \\, dy$.\n",
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