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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable and let $c \\in \\R$.\nAssume that $f$ is [[Mellin convergent|SGZZ4Y0D36]] at $c$, so that the [[Mellin transform|N5PVGRJE1T]] $\\mathcal{M}f(s) = \\int_0^\\infty x^{s-1} f(x) \\, dx$ is defined on the line $\\operatorname{Re} s = c$.\nAssume also that $y \\mapsto \\mathcal{M}f(c + iy)$ is [[integrable|X7GTF3D98F]] on $\\R$.\nThen, for every $x \u003e 0$ at which $f$ is continuous,\n$$f(x) = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} x^{-s} \\, \\mathcal{M}f(s) \\, ds.$$\nThe right side is the [[inverse Mellin transform|CHESR1W1CY]] of $\\mathcal{M}f$ along $\\operatorname{Re} s = c$.\n\n## Remarks\n\nPut $x = e^{u}$ and $g(u) = e^{cu} f(e^u)$.\nMellin convergence at $c$ says that $g$ is integrable on $\\R$, and $\\mathcal{M}f(c + iy) = \\int_{\\R} g(u) e^{iyu} \\, du$ is a Fourier transform of $g$.\nThe [[Fourier inversion theorem|T4Z8YQZKZV]] then recovers $g$ almost everywhere, hence at every point where $g$ is continuous, and this is the formula above.\nExample: $f(x) = e^{-x}$ with $c \u003e 0$ gives $e^{-x} = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} \\Gamma(s) x^{-s} \\, ds$, the Cahen–Mellin integral.\nContinuity at $x$ is needed: if $f(x) = e^{-x}$ is changed at the single point $x = 1/2$, its Mellin transform is still $\\Gamma$.\nSo the formula gives $e^{-1/2}$ there and not the changed value.\nFor $f = \\mathbf{1}_{(0, 1]}$, the transform $\\mathcal{M}f(s) = 1/s$ is not integrable on vertical lines, so the integrability hypothesis fails.\nThe formula still holds at $x \\ne 1$ as a limit of integrals over $[c - iT, c + iT]$, but that needs a separate argument.\n\n## Explanations\n\n### L1 human\n\nThe Mellin transform loses no information: a reasonable function can be rebuilt from its transform on any single vertical line where the transform is defined.\nThe rebuilding is an integral of the powers $x^{-s}$ against the transform.\n\n### L2 human\n\nThe theorem is the Fourier inversion theorem written on the multiplicative group $(0, \\infty)$.\nIt is how one passes from a Mellin transform back to a function, as in Perron's formula and in [[Barnes integrals|1PDGBGEHM2]].\nThere the transform is often a product of gamma functions or a Dirichlet series.\nIn practice one then shifts the line of integration and collects residues, which turns the inversion integral into series or asymptotic expansions.\nWithout integrability of $\\mathcal{M}f$ on the line, as for jump functions, the integral must be taken as a symmetric limit.\nFor functions of bounded variation near a jump, this limit is the average of the one-sided limits.\n",
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