Each of the Maxwell's equations has both an integral and differential form. Schroedinger's equation is a differential equation and, apparently (I haven't studied it yet), Feynman's path integrals are an equivalent formulation. Newton has got a differential equation $F = \dot p$. Is there an equivalent integral formulation of Newton's law? Would it even be useful if there were?
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$\begingroup$ Well Newton's formula for gravity is identical to Guass's Law for electrostatics: Poisson's Equation. I only comment because that's not exactly what you are asking about. $\endgroup$– JaredCommented Mar 23, 2015 at 6:20
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The impulse momentum theorem would be your best answer:
$p(t) = \int_0^tf(t')dt'$
answered Mar 23, 2015 at 4:34
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$\begingroup$ I would suggest that the it should be from $t_0$ to $t$ rather than $0$ to $t$--but I'm not sure it matters. $\endgroup$– JaredCommented Mar 23, 2015 at 6:17
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$\begingroup$ Sure, writing the lower limit as $t_0$ generalises the integral form as this can later be set to 0 or whatever, as necessary. $\endgroup$– DaiCommented Mar 23, 2015 at 6:52
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1$\begingroup$ Or &p(t) = \int_{t_0}^{t} m(t')a(t') dt$ to be even more general... $\endgroup$– DaiCommented Mar 23, 2015 at 6:54
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$\begingroup$ @CStarAlgebra can you define your variables? Is $t' = t$? Is $f$ the resultant force, that is $\sum \vec F= m \vec a$? $\endgroup$ Commented May 27, 2016 at 13:47