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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be integrable on every compact subinterval of $(0, \\infty)$.\nLet $a, b \\in \\R$ with $b \u003c a$, and assume that $f(x) = O(x^{-a})$ as $x \\to \\infty$ and $f(x) = O(x^{-b})$ as $x \\to 0^+$, in [[big O notation|0TGV80E6A4]].\nLet $S = \\set{s \\in \\C : b \u003c \\operatorname{Re} s \u003c a}$.\nThen:\n\n1. For every $s \\in S$, the function $x \\mapsto (\\log x) f(x)$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$.\n2. The [[Mellin transform|N5PVGRJE1T]] $\\mathcal{M}f$ is complex differentiable at every $s \\in S$, with\n$$(\\mathcal{M}f)'(s) = \\int_0^\\infty x^{s-1} (\\log x) f(x) \\, dx = \\mathcal{M}[(\\log x) f(x)](s).$$\n\nIn particular $\\mathcal{M}f$ is [[holomorphic|0NAHXR1FRH]] on the open strip $S$.\n\n## Remarks\n\nThe derivative of $x^{s-1}$ in $s$ is $x^{s-1} \\log x$.\nFix $s \\in S$ and $\\varepsilon \u003e 0$ with $b \u003c \\operatorname{Re} s - 2\\varepsilon$ and $\\operatorname{Re} s + 2\\varepsilon \u003c a$.\nFor $\\abs{z - s} \\le \\varepsilon$, the bound $\\abs{\\log x} \\le \\varepsilon^{-1} (x^{\\varepsilon} + x^{-\\varepsilon})$ dominates $\\abs{x^{z-1} (\\log x) f(x)}$ by a function that is integrable by the estimates of the fundamental strip.\nSo [[differentiation under the integral sign|0AR06W1E1B]] applies.\nIterating gives $(\\mathcal{M}f)^{(n)}(s) = \\mathcal{M}[(\\log x)^n f(x)](s)$ for all $n \\ge 0$.\nExample: for $f(x) = e^{-x}$ the statement gives $\\Gamma'(s) = \\int_0^\\infty x^{s-1} (\\log x) e^{-x} \\, dx$ for $\\operatorname{Re} s \u003e 0$, and at $s = 1$ this is $\\Gamma'(1) = -\\gamma$, with $\\gamma$ the Euler–Mascheroni constant.\nThe open strip is needed: for $f(x) = 1/(1 + x)$, $\\mathcal{M}f(s) = \\pi / \\sin(\\pi s)$ on $0 \u003c \\operatorname{Re} s \u003c 1$, and this function has poles at $s = 0$ and $s = 1$, the edges of the strip.\n\n## Explanations\n\n### L1 human\n\nUnder the usual bounds on $f$ at $0$ and at $\\infty$, the Mellin transform is a smooth function of $s$ in the complex sense.\nIts derivative is the Mellin transform of $f$ multiplied by $\\log x$.\n\n### L2 human\n\nHolomorphy on the strip is what makes Mellin methods work: one can continue $\\mathcal{M}f$ analytically beyond the strip, shift lines of integration, and pick up residues.\nFor instance the gamma function is first defined on $\\operatorname{Re} s \u003e 0$ as a Mellin transform, holomorphic there by this result, and then continued meromorphically to $\\C$.\nThe same holds for the completed zeta function $\\pi^{-s/2} \\Gamma(s/2) \\zeta(s)$, which is a Mellin transform of a theta series.\n",
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