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  "body": "\n## Definition\n\nLet $k \\in L^1(\\R)$, let $f \\colon \\R \\to \\C$ be a given function, and let $\\lambda \\in \\C$.\nA *convolution integral equation of the second kind* on $\\R$ is the equation\n$$\\varphi(x) - \\lambda \\int_{-\\infty}^{\\infty} k(x - t) \\, \\varphi(t) \\, dt = f(x), \\quad x \\in \\R,$$\nfor an unknown function $\\varphi$.\nA *convolution integral equation of the first kind* is the equation\n$$\\int_{-\\infty}^{\\infty} k(x - t) \\, \\varphi(t) \\, dt = f(x), \\quad x \\in \\R.$$\nIn both cases the kernel $(x, t) \\mapsto k(x - t)$ is a [[difference kernel|B5AB451TB7]], and the integral is the [[convolution|2Z29JN3VES]] $(k * \\varphi)(x)$.\nThe same names are used on $\\R^n$, and on an interval such as $[0, \\infty)$ when the integral runs only over that interval.\n\n## Remarks\n\nExample: take $k(s) = e^{-\\abs{s}}$, $\\lambda = \\frac{1}{4}$ and $f(x) = \\frac{1}{2} e^{-\\abs{x}}$.\nThe Fourier transform of $e^{-a\\abs{s}}$ is $\\frac{2a}{a^2 + \\xi^2}$ for $a \u003e 0$, so the transformed equation reads $\\hat\\varphi(\\xi) \\left(1 - \\frac{1}{2(1 + \\xi^2)}\\right) = \\frac{1}{1 + \\xi^2}$.\nHence $\\hat\\varphi(\\xi) = \\frac{2}{1 + 2\\xi^2} = \\frac{1}{1/2 + \\xi^2}$, and $\\varphi(x) = \\frac{1}{\\sqrt{2}} e^{-\\abs{x}/\\sqrt{2}}$.\n\nNon-example: the equation $\\varphi(x) - \\int_{-\\infty}^{\\infty} e^{-x^2 - t^2} \\varphi(t) \\, dt = f(x)$ is a linear integral equation, but its kernel $e^{-x^2 - t^2}$ is not a function of $x - t$ alone, so it is not of convolution type.\n\nThe equation on the half-line $\\varphi(x) - \\int_0^\\infty k(x - t) \\varphi(t) \\, dt = f(x)$, $x \u003e 0$, is the [[Wiener–Hopf equation|QYY4NMDY5H]].\nThe Volterra form $\\varphi(x) - \\int_0^x k(x - t) \\varphi(t) \\, dt = f(x)$ is the [[Volterra equation of convolution type|SNFYTC1TTA]].\n",
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