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  "body": "\n## Definition\n\nLet $P$ and $Q$ be complex polynomials with $Q \\ne 0$ and $\\deg P \u003c \\deg Q$, and let $F = P/Q$ be the [[rational function|0WCZH4AFMT]].\nLet $a_1, \\ldots, a_r$ be the distinct roots of $Q$, with multiplicities $m_1, \\ldots, m_r$.\nThe [[partial fraction decomposition|030VFEREKG]] writes $F$ uniquely as\n$$F(s) = \\sum_{j=1}^{r} \\sum_{k=1}^{m_j} \\frac{c_{j k}}{(s - a_j)^k}$$\nwith complex constants $c_{j k}$.\n*Inversion by partial fractions* is the computation of the [[inverse Laplace transform|0JRW4X1MDC]] of $F$ as\n$$f(t) = \\sum_{j=1}^{r} \\sum_{k=1}^{m_j} c_{j k} \\, \\frac{t^{k-1}}{(k-1)!} \\, e^{a_j t}, \\qquad t \\ge 0.$$\nIt rests on the identity $\\mathcal{L}\\{t^{k-1} e^{a t}/(k-1)!\\}(s) = 1/(s - a)^k$ for $\\operatorname{Re} s \u003e \\operatorname{Re} a$.\n\n## Remarks\n\nThe identity follows from $\\mathcal{L}\\{t^{k-1}\\}(s) = (k-1)!/s^k$ and the [[frequency shift theorem|CD35MSBNE7]].\nExample: $F(s) = \\frac{s + 3}{(s + 1)(s + 2)} = \\frac{2}{s + 1} - \\frac{1}{s + 2}$, so $f(t) = 2 e^{-t} - e^{-2t}$; check $f(0) = 1 = \\lim_{s \\to \\infty} s F(s)$.\nNon-example: $F(s) = s/(s + 1)$ has $\\deg P = \\deg Q$, so the method does not apply as stated.\nIndeed $F(s) \\to 1$ as $s \\to \\infty$, while the Laplace transform of a function [[tends to $0$ at infinity|N2JFN9VHD9]]; here $F = 1 - 1/(s+1)$ is the transform of the distribution $\\delta - e^{-t}$.\nFor real $P$ and $Q$, the complex roots come in conjugate pairs, and the two terms of a simple pair combine into $e^{\\alpha t}(A \\cos \\beta t + B \\sin \\beta t)$.\nThe coefficients of simple roots are quickly found with the [[Heaviside cover-up method|SRQ5P3077W]].\n",
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