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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable, let $a \u003e 0$ and let $g(x) = f(ax)$ for $x \u003e 0$.\nLet $s \\in \\C$.\nThen $g$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$ if and only if $f$ is.\nIn that case\n$$\\mathcal{M}g(s) = a^{-s} \\, \\mathcal{M}f(s),$$\nwhere $\\mathcal{M}$ is the [[Mellin transform|N5PVGRJE1T]] and $a^{-s} = e^{-s \\log a}$ is a [[complex power|0RNGDMHT7M]].\n\n## Remarks\n\nThe proof is the substitution $u = ax$:\n$$\\int_0^\\infty x^{s-1} f(ax) \\, dx = \\int_0^\\infty \\left(\\frac{u}{a}\\right)^{s-1} f(u) \\, \\frac{du}{a} = a^{-s} \\int_0^\\infty u^{s-1} f(u) \\, du.$$\nThe same computation with $\\abs{f}$ and $\\operatorname{Re} s$ gives the equivalence of convergence.\nExample: $f(x) = e^{-x}$ gives $\\int_0^\\infty x^{s-1} e^{-ax} \\, dx = a^{-s} \\Gamma(s)$ for $\\operatorname{Re} s \u003e 0$.\nApplied to $a = n$ and summed over $n$, this is the step that links the gamma function with Dirichlet series.\nThe hypothesis $a \u003e 0$ is needed so that $x \\mapsto ax$ maps $(0, \\infty)$ onto itself.\n\n## Explanations\n\n### L1 human\n\nStretching or squeezing the input of $f$ by a factor $a$ multiplies its Mellin transform by $a^{-s}$.\nSo the Mellin transform turns the scaling of the variable into a simple factor, in the same way the Fourier transform turns a shift into a phase factor.\n\n### L2 human\n\nThis is the reason the Mellin transform suits problems with scale invariance, such as homogeneous kernels and harmonic sums $\\sum_k \\lambda_k f(\\mu_k x)$.\nFor a harmonic sum the rule gives $\\mathcal{M}\\left[\\sum_k \\lambda_k f(\\mu_k x)\\right](s) = \\left(\\sum_k \\lambda_k \\mu_k^{-s}\\right) \\mathcal{M}f(s)$, when the sum and integral can be exchanged.\nThe factor in brackets is a Dirichlet series, and its poles drive the asymptotic behaviour of the sum.\n",
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