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  "body": "\n## Statement\n\nLet $\\sigma \\in \\R$ and let $f, g \\colon (0, \\infty) \\to \\C$ be measurable.\nWrite $\\mathcal{M}$ for the [[Mellin transform|N5PVGRJE1T]].\n\n1. Assume that $x^{\\sigma - 1} f(x)$ is [[integrable|X7GTF3D98F]] on $(0, \\infty)$ and that $\\int_0^\\infty x^{2\\sigma - 1} \\abs{f(x)}^2 \\, dx \u003c \\infty$. Then\n$$\\int_0^\\infty x^{2\\sigma - 1} \\abs{f(x)}^2 \\, dx = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} \\abs{\\mathcal{M}f(\\sigma + i\\tau)}^2 \\, d\\tau.$$\n2. Assume moreover that $x^{-\\sigma} g(x)$ is integrable on $(0, \\infty)$ and that $\\int_0^\\infty x^{1 - 2\\sigma} \\abs{g(x)}^2 \\, dx \u003c \\infty$. Then\n$$\\int_0^\\infty f(x) g(x) \\, dx = \\frac{1}{2\\pi} \\int_{-\\infty}^{\\infty} \\mathcal{M}f(\\sigma + i\\tau) \\, \\mathcal{M}g(1 - \\sigma - i\\tau) \\, d\\tau.$$\n\n## Remarks\n\nThe integrability hypotheses say that $f$ is [[Mellin convergent|SGZZ4Y0D36]] at $\\sigma$ and $g$ at $1 - \\sigma$, so the transforms on the right are defined pointwise.\nPut $h_f(u) = e^{\\sigma u} f(e^{u})$ and $h_g(u) = e^{(1 - \\sigma) u} g(e^{u})$.\nThe substitution $x = e^{u}$ turns the hypotheses into $h_f, h_g \\in L^1(\\R) \\cap L^2(\\R)$ and the left sides into $\\int \\abs{h_f}^2 \\, du$ and $\\int h_f h_g \\, du$.\nIt also turns $\\mathcal{M}f(\\sigma + i\\tau)$ and $\\mathcal{M}g(1 - \\sigma - i\\tau)$ into Fourier transforms of $h_f$ and $h_g$ at $\\mp \\tau / 2\\pi$.\nThe [[Plancherel theorem|HPZHRZ0502]] for the Fourier transform on $\\R$ then gives both formulas.\nExample: $f(x) = e^{-x}$ and $\\sigma = 1/2$ give $\\int_0^\\infty e^{-2x} \\, dx = \\frac{1}{2} = \\frac{1}{2\\pi} \\int_{\\R} \\abs{\\Gamma(\\tfrac12 + i\\tau)}^2 \\, d\\tau$; since $\\abs{\\Gamma(\\tfrac12 + i\\tau)}^2 = \\pi / \\cosh(\\pi \\tau)$, the right side is $\\frac12 \\int_{\\R} \\frac{d\\tau}{\\cosh \\pi \\tau} = \\frac12$.\nBy density, part 1 extends $\\mathcal{M}$ to a unitary map, up to the factor $2\\pi$, from $L^2((0, \\infty), x^{2\\sigma - 1} dx)$ onto $L^2(\\R)$.\n\n## Explanations\n\n### L1 human\n\nThe total energy of a function, measured with the weight $x^{2\\sigma - 1}$, equals the total energy of its Mellin transform along the line $\\operatorname{Re} s = \\sigma$.\nThe second formula is the same fact for products of two functions.\nIt is the Mellin version of the Plancherel and Parseval identities for Fourier series and integrals.\n\n### L2 human\n\nThe formula is used to evaluate integrals of products: one knows $\\mathcal{M}f$ and $\\mathcal{M}g$, and computes $\\int_0^\\infty f g \\, dx$ by a single integral on a vertical line, often by residues.\nPart 2 can also be read as the convolution rule for the Mellin convolution of $f$ and $x \\mapsto g(1/x)/x$, combined with Mellin inversion at the point $x = 1$.\nOn the critical line $\\sigma = 1/2$ the weight disappears, and part 1 says that $f \\mapsto \\mathcal{M}f(\\tfrac12 + i\\tau) / \\sqrt{2\\pi}$ preserves the norm of $L^2(0, \\infty)$.\n",
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