# PlanetPhysics/Heaviside Formula

Let $P(s)$ and $Q(s)$ be polynomials with the degree of the former less than the degree of the latter.

• If all complex zeroes $a_{1},\,a_{2},\,\ldots ,\,a_{n}$ of $Q(s)$ are simple, then

${\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}{\frac {P(a_{j})}{Q'(a_{j})}}e^{a_{j}t}.\end{matrix}}$ • If the different zeroes $a_{1},\,a_{2},\,\ldots ,\,a_{n}$ of $Q(s)$ have the multiplicities $m_{1},\,m_{2},\,\ldots ,\,m_{n}$ , respectively, we denote\, $F_{j}(s):=(s\!-\!a_{j})^{m_{j}}P(s)/Q(s)$ ;\, then

${\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}e^{a_{j}t}\sum _{k=0}^{m_{j}-1}{\frac {F_{j}^{(k)}(a_{j})t^{m_{j}\!-\!1\!-\!k}}{k!(m_{j}\!-\!1\!-\!k)!}}.\end{matrix}}$ A special case of the Heaviside formula (1) is $\displaystyle \mathcal{L}^{-1}\left\{\frac{Q'(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^ne^{a_jt}.\\$

Example. \, Since the zeros of the binomial $s^{4}\!+\!4a^{4}$ are\, $s=(\pm 1\!\pm \!i)a$ ,\, we obtain ${\mathcal {L}}^{-1}\left\{{\frac {s^{3}}{s^{4}\!+\!4a^{4}}}\right\}\;=\;{\frac {1}{4}}{\mathcal {L}}^{-1}\left\{{\frac {4s^{3}}{s^{4}\!+\!4a^{4}}}\right\}\;=\;{\frac {1}{4}}\sum _{\pm }e^{(\pm 1\pm i)at}\;=\;{\frac {e^{at}+e^{-at}}{2}}\cdot {\frac {e^{iat}+e^{-iat}}{2}}\;=\;\cosh {at}\,\cos {at}.$ \\

Proof of (1). \, Without hurting the generality, we can suppose that $Q(s)$ is monic.\, Therefore $Q(s)\;=\;(s\!-\!a_{1})(s\!-\!a_{2})\cdots (s\!-\!s_{n}).$ For\, $j=1,\,2,\;\ldots ,\,n$ ,\, denoting $Q(s)\;:=\;(s\!-\!a_{j})Q_{j}(s),$ one has\, $Q_{j}(a_{j})\neq 0$ .\, We have a partial fraction expansion of the form

${\begin{matrix}{\frac {P(s)}{Q(s)}}\;=\;{\frac {C_{1}}{s\!-\!a_{1}}}+{\frac {C_{2}}{s\!-\!a_{2}}}+\ldots +{\frac {C_{n}}{s\!-\!a_{n}}}\end{matrix}}$ with constants $C_{j}$ .\, According to the linearity and the formula 1 of the parent entry, one gets

${\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}C_{j}e^{a_{j}t}.\end{matrix}}$ For determining the constants $C_{j}$ , multiply (3) by $s\!-\!a_{j}$ .\, It yields ${\frac {P(s)}{Q_{j}(s)}}=C_{j}+(s\!-\!a_{j})\sum _{\nu \neq j}{\frac {C_{\nu }}{s\!-\!a_{\nu }}}.$ Setting to this identity \,$s:=a_{j}$ \, gives the value

${\begin{matrix}C_{j}\;=;{\frac {P(a_{j})}{Q_{j}(a_{j})}}.\end{matrix}}$ But since\, $Q'(s)={\frac {d}{ds}}((s\!-\!a_{j})Q_{j}(s))=Q_{j}(s)\!+\!(s\!-\!a_{j})Q_{j}'(s)$ ,\, we see that\, $Q'(a_{j})=Q_{j}(a_{j})$ ;\, thus the equation (5) may be written

${\begin{matrix}C_{j};=\;{\frac {P(a_{j})}{Q'(a_{j})}}.\end{matrix}}$ The values (6) in (4) produce the formula (1).