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  "body": "\n## Statement\n\nLet $a \\colon \\set{1, 2, 3, \\dots} \\to \\C$ and let $F(s) = \\sum_{n \\ge 1} a(n) n^{-s}$ be its [[Dirichlet series|8MWTEF6462]].\nLet $\\sigma_a$ be the abscissa of [[absolute convergence|0YBDSJARWN]] of $F$, assume $\\sigma_a \u003c \\infty$, and let $c \u003e \\max(0, \\sigma_a)$.\nFor $x \u003e 0$, let $A^*(x) = \\sum_{n \u003c x} a(n) + \\frac{1}{2} a(x)$, where the last term is present only when $x$ is an integer.\nThen for every $x \u003e 0$,\n$$A^*(x) = \\lim_{T \\to \\infty} \\frac{1}{2\\pi i} \\int_{c - iT}^{c + iT} F(s) \\, \\frac{x^{s}}{s} \\, ds.$$\n\n## Remarks\n\nThe proof rests on the discontinuous integral.\nFix $c \u003e 0$, and for $y \u003e 0$ let $I(y) = \\lim_{T \\to \\infty} \\frac{1}{2\\pi i} \\int_{c - iT}^{c + iT} \\frac{y^{s}}{s} \\, ds$.\nThen $I(y) = 1$ for $y \u003e 1$, $I(1) = 1/2$, and $I(y) = 0$ for $0 \u003c y \u003c 1$.\nFor $y \u003e 1$ one closes the path to the left and picks up the residue $1$ at $s = 0$; for $y \u003c 1$ one closes it to the right, where there is no pole.\nOne applies this with $y = x/n$ to each term $a(n) n^{-s}$ and exchanges sum and limit.\nThe exchange uses the explicit error bound of the discontinuous integral and the absolute convergence of $F$ on $\\operatorname{Re} s = c$.\nThe limit cannot be replaced by an absolutely convergent integral in general.\nFor $F = 1$, that is $a(1) = 1$ and $a(n) = 0$ otherwise, the integrand $x^{s}/s$ has modulus of order $1/\\abs{s}$ on the line.\nExample: take $a = \\Lambda$, the [[von Mangoldt function|0NK8JTE1NN]], so that $F(s) = -\\zeta'(s)/\\zeta(s)$, and take $c \u003e 1$.\nThen the formula expresses the Chebyshev function $\\psi(x)$ as a contour integral.\nShifting the contour to the left and collecting the residue $x$ at $s = 1$ is the route to the [[prime number theorem|0EHBY5Y84N]].\n\n## Explanations\n\n### L1 human\n\nPerron's formula recovers the partial sums of the coefficients of a Dirichlet series from the values of the series on a vertical line.\nIt works because the integral of $x^{s}/s$ along a vertical line acts as a switch: it is $1$ when $x \u003e 1$ and $0$ when $x \u003c 1$.\nSo each term $a(n)$ is counted exactly when $n \u003c x$.\n\n### L2 human\n\nThe formula is an inverse Mellin transform: by the Mellin representation of Dirichlet series, $F(s)/s$ is the Mellin transform at $-s$ of the summatory function.\nIn practice one uses a truncated version with an explicit error term in $T$ and $x$, then moves the line of integration to the left of $\\operatorname{Re} s = 1$.\nThe poles of $F(s) x^{s}/s$ then give the main terms, and bounds for $F$ in the critical strip control the error.\nThis is how the prime number theorem and its explicit formulas are derived from properties of the zeta function.\n",
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