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  "body": "\n## Definition\n\nFor $y \\in \\R$ and $x \u003e 0$ let $K_{iy}(x) = \\int_0^\\infty e^{-x\\cosh t} \\cos(yt)\\, dt$ be the [[modified Bessel function of the second kind|2ET068BZZH]] of order $iy$.\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable with $\\int_0^\\infty \\abs{f(x)}\\, K_0(x)\\, dx \u003c \\infty$.\nThe *Kontorovich–Lebedev transform* of $f$ is the function $g \\colon \\R \\to \\C$ given by the [[Lebesgue integral|DWJF7ABPR1]]\n$$g(y) = \\int_0^\\infty f(x)\\, K_{iy}(x)\\, dx.$$\n\n## Remarks\n\nThe integral converges absolutely because $\\abs{K_{iy}(x)} \\le \\int_0^\\infty e^{-x\\cosh t}\\, dt = K_0(x)$.\nThe function $K_{iy}(x)$ is real and even in $y$, so $g$ is even and is usually considered on $(0, \\infty)$.\nUnder suitable conditions on $f$, the inversion formula is\n$$f(x) = \\frac{2}{\\pi^2 x} \\int_0^\\infty g(y)\\, K_{iy}(x)\\, \\sinh(\\pi y)\\, y\\, dy.$$\n\nExample: $f(x) = 1$.\nSince $K_0(x)$ behaves like $-\\log x$ near $0$ and like $\\sqrt{\\pi/(2x)}\\, e^{-x}$ at infinity, $K_0$ is integrable and the condition holds.\nThe formula $\\int_0^\\infty K_\\mu(x)\\, dx = \\frac{\\pi}{2\\cos(\\pi\\mu/2)}$ for $\\abs{\\operatorname{Re} \\mu} \u003c 1$, with $\\mu = iy$, gives $g(y) = \\frac{\\pi}{2\\cosh(\\pi y / 2)}$.\n\nNon-example: $f(x) = e^x$.\nFor fixed $y$, $K_{iy}(x) \\sim \\sqrt{\\pi/(2x)}\\, e^{-x}$ as $x \\to \\infty$, so $f(x) K_{iy}(x) \\sim \\sqrt{\\pi/(2x)}$ and the integral diverges at infinity.\n",
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