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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\C$ be a Lebesgue [[integrable function|X7GTF3D98F]].\nThe *cosine transform* of $f$ is the function on $[0, \\infty)$ given by\n$$\\mathcal{F}_c f(\\xi) = \\int_0^\\infty f(x) \\cos(2\\pi x \\xi)\\, dx.$$\nThe *sine transform* of $f$ is the function on $[0, \\infty)$ given by\n$$\\mathcal{F}_s f(\\xi) = \\int_0^\\infty f(x) \\sin(2\\pi x \\xi)\\, dx.$$\n\n## Remarks\n\nBoth integrals converge absolutely because $\\abs{\\cos}$ and $\\abs{\\sin}$ are at most $1$.\nExample: for $f(x) = e^{-x}$, two integrations by parts give $\\mathcal{F}_c f(\\xi) = \\frac{1}{1 + a^2}$ and $\\mathcal{F}_s f(\\xi) = \\frac{a}{1 + a^2}$ with $a = 2\\pi \\xi$.\nNon-example: the constant function $f = 1$ is not integrable on $[0, \\infty)$, and $\\int_0^\\infty \\cos(2\\pi x \\xi)\\, dx$ diverges for every $\\xi$.\n\nLet $f_e$ and $f_o$ be the even and the odd extensions of $f$ to $\\R$, and let $\\hat g$ denote the [[Fourier transform|DPVM34ADZV]] $\\int_\\R g(x) e^{-2\\pi i x \\xi}\\, dx$.\nThen $\\hat f_e(\\xi) = 2 \\mathcal{F}_c f(\\xi)$ and $\\hat f_o(\\xi) = -2i\\, \\mathcal{F}_s f(\\xi)$.\nConsequently, if $f$ is continuous and $\\mathcal{F}_c f$ is integrable, then $f(x) = 4 \\int_0^\\infty \\mathcal{F}_c f(\\xi) \\cos(2\\pi x \\xi)\\, d\\xi$ for $x \\ge 0$.\n\nOther conventions use $\\cos(x \\xi)$ and $\\sin(x \\xi)$, often with a factor $\\sqrt{2/\\pi}$ that makes the transforms their own inverses on $L^2(0, \\infty)$.\n",
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