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  "body": "\n## Statement\n\nLet $c \u003e 0$ and let $y \\in \\C$ with $\\operatorname{Re} y \u003e 0$.\nLet $y^{-s} = e^{-s \\log y}$, with the principal branch of $\\log y$.\nThen\n$$e^{-y} = \\frac{1}{2\\pi i} \\int_{c - i\\infty}^{c + i\\infty} \\Gamma(s) \\, y^{-s} \\, ds,$$\nwhere $\\Gamma$ is the [[gamma function|0H5SYBSED9]] and the integral converges absolutely.\n\n## Remarks\n\nFor real $y \u003e 0$ this is the [[inverse Mellin transform|CHESR1W1CY]] of $\\Gamma = \\mathcal{M}[e^{-x}]$, so it follows from the Mellin inversion theorem.\nFor complex $y$, put $\\abs{\\arg y} = \\theta \u003c \\pi/2$; then $\\abs{y^{-c - it}} = \\abs{y}^{-c} e^{t \\arg y} \\le \\abs{y}^{-c} e^{\\theta \\abs{t}}$, while $\\abs{\\Gamma(c + it)}$ decays like $\\abs{t}^{c - 1/2} e^{-\\pi \\abs{t}/2}$ by Stirling's formula.\nSo the integral converges absolutely and locally uniformly, it is holomorphic in $y$, and the identity extends from $y \u003e 0$ by the identity theorem.\nA direct proof moves the line of integration to the left: by the [[residue theorem|032HKYS329]], the poles of $\\Gamma$ at $s = -k$ with residues $(-1)^k / k!$ give $\\sum_{k \\ge 0} (-1)^k y^k / k! = e^{-y}$.\nThe hypothesis $c \u003e 0$ is needed: for $-1 \u003c c \u003c 0$ the line passes to the left of the pole at $0$, and the integral equals $e^{-y} - 1$.\n\n## Explanations\n\n### L1 human\n\nThe exponential function $e^{-y}$ can be written as an integral of powers $y^{-s}$ against the gamma function along a vertical line.\nIt is the inverse of the fact that the gamma function is the Mellin transform of $e^{-x}$.\n\n### L2 human\n\nThis integral is the prototype of [[Mellin–Barnes integrals|1PDGBGEHM2]] and of the method of shifting contours.\nIn analytic number theory it is applied to $y = n x$ and summed over $n$: the sum $\\sum_n a(n) e^{-n x}$ then becomes $\\frac{1}{2\\pi i} \\int \\Gamma(s) F(s) x^{-s} \\, ds$ with $F$ the Dirichlet series of the $a(n)$.\nShifting the line past the poles of $\\Gamma(s) F(s)$ gives asymptotic expansions of the exponential sum as $x \\to 0^+$.\nIt is the smooth counterpart of Perron's formula, which uses the kernel $x^{s}/s$ instead of $\\Gamma(s) x^{-s}$ and has worse convergence.\n",
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  "name": "Cahen–Mellin integral",
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