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  "body": "\n## Statement\n\nLet $\\mathcal{M}$ be the [[Mikusiński field of operators|82YAXZMMQ1]], the field of fractions of the convolution ring of locally integrable functions on $[0, \\infty)$.\nLet $h$ be the constant function $1$ on $[0, \\infty)$, and let $s = 1/h$ be its inverse in $\\mathcal{M}$.\nFor a complex number $c$, write $c$ also for the numerical operator $c \\cdot 1$ of $\\mathcal{M}$, which corresponds to the function $c\\,h$ divided by $h$.\nLet $f$ be continuously differentiable on $[0, \\infty)$.\nThen\n$$s f = f' + f(0)$$\nin $\\mathcal{M}$.\nMore generally, if $f$ is $n$ times continuously differentiable on $[0, \\infty)$, then\n$$s^n f = f^{(n)} + \\sum_{k=0}^{n-1} f^{(k)}(0) \\, s^{n-1-k}.$$\n\n## Remarks\n\nBy the [[fundamental theorem of calculus|0NHGY0N0JB]], $\\int_0^t f'(\\tau) \\, d\\tau = f(t) - f(0)$.\nBy the [[integration operator result|BDBDJSP6DF]] this reads $h f' = f - f(0) h$ in the convolution ring.\nMultiplying both sides by $s$ gives $f' = s f - f(0)$, which is the first formula.\nThe general formula follows by induction on $n$.\nFor example, with $f(t) = e^{a t}$, one gets $s f = a f + 1$, so $f = 1/(s - a)$ in $\\mathcal{M}$.\nThe operator $s$ is not a function: it is a quotient of functions, and $s f$ equals the function $f'$ only when $f(0) = 0$.\n\n## Explanations\n\n### L1 human\n\nIntegration from $0$ is multiplication by the constant function $1$.\nIts inverse in the field of operators is an operator $s$ that acts like differentiation.\nIt differs from plain differentiation by a term that remembers the starting value $f(0)$.\n\n### L2 human\n\nThis formula has the same shape as the rule $\\mathcal{L}\\{f'\\}(s) = s F(s) - f(0)$ for the Laplace transform.\nMikusiński's calculus gets it with no condition on the growth of $f$ and no convergence of integrals.\nA linear differential equation with constant coefficients and initial values becomes an algebraic equation in $s$ in the field $\\mathcal{M}$.\nOne then solves for the unknown and reads off a function, often after a partial fraction expansion in $s$.\n",
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