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  "body": "\n## Definition\n\nLet $f, g \\colon (0, \\infty) \\to \\C$ be measurable.\nLet $x \u003e 0$ be such that $y \\mapsto f(x/y) g(y) / y$ is [[integrable|X7GTF3D98F]] on $(0, \\infty)$.\nThe *Mellin convolution* of $f$ and $g$ at $x$ is the [[Lebesgue integral|DWJF7ABPR1]]\n$$(f \\star g)(x) = \\int_0^\\infty f\\left(\\frac{x}{y}\\right) g(y) \\, \\frac{dy}{y}.$$\n\n## Remarks\n\nThe substitution $y \\mapsto x/y$ shows that $f \\star g = g \\star f$.\nAssume that $f$ and $g$ are both Mellin convergent at $s$.\nThen $f \\star g$ is defined at almost every $x \u003e 0$, it is Mellin convergent at $s$, and $\\mathcal{M}(f \\star g)(s) = \\mathcal{M}f(s) \\, \\mathcal{M}g(s)$.\nThe proof is Fubini's theorem and the substitution $x = yu$, which splits the double integral.\nExample: $f = g = \\mathbf{1}_{(0, 1]}$.\nThe integrand $\\mathbf{1}_{(0,1]}(x/y) \\mathbf{1}_{(0,1]}(y) / y$ is $1/y$ for $x \\le y \\le 1$ and $0$ otherwise, which is integrable for each $x \u003e 0$.\nSo $(f \\star g)(x) = \\int_x^1 dy/y = -\\log x$ for $0 \u003c x \\le 1$ and $0$ for $x \u003e 1$.\nAs a check, $\\int_0^1 x^{s-1} (-\\log x) \\, dx = 1/s^2 = (1/s)^2$ for $\\operatorname{Re} s \u003e 0$.\nNon-example: $f = g = 1$ on $(0, \\infty)$; the integrand $1/y$ is not integrable on $(0, \\infty)$, so $(f \\star g)(x)$ is defined at no $x$.\nSome authors use the other normalization $\\int_0^\\infty f(xy) g(y) \\, dy$, whose Mellin transform is $\\mathcal{M}f(s) \\mathcal{M}g(1 - s)$.\n",
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