# PlanetPhysics/Heaviside Formula

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

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

${\displaystyle {\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}{\frac {P(a_{j})}{Q'(a_{j})}}e^{a_{j}t}.\end{matrix}}}$

• If the different zeroes ${\displaystyle a_{1},\,a_{2},\,\ldots ,\,a_{n}}$ of ${\displaystyle Q(s)}$ have the multiplicities ${\displaystyle m_{1},\,m_{2},\,\ldots ,\,m_{n}}$, respectively, we denote\, ${\displaystyle F_{j}(s):=(s\!-\!a_{j})^{m_{j}}P(s)/Q(s)}$;\, then

${\displaystyle {\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}e^{a_{j}t}\sum _{k=0}^{m_{j}-1}{\frac {F_{j}^{(k)}(a_{j})t^{m_{j}\!-\!1\!-\!k}}{k!(m_{j}\!-\!1\!-\!k)!}}.\end{matrix}}}$

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

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

${\displaystyle {\begin{matrix}{\frac {P(s)}{Q(s)}}\;=\;{\frac {C_{1}}{s\!-\!a_{1}}}+{\frac {C_{2}}{s\!-\!a_{2}}}+\ldots +{\frac {C_{n}}{s\!-\!a_{n}}}\end{matrix}}}$

with constants ${\displaystyle C_{j}}$.\, According to the linearity and the formula 1 of the parent entry, one gets

${\displaystyle {\begin{matrix}{\mathcal {L}}^{-1}\left\{{\frac {P(s)}{Q(s)}}\right\}\;=\;\sum _{j=1}^{n}C_{j}e^{a_{j}t}.\end{matrix}}}$

For determining the constants ${\displaystyle C_{j}}$, multiply (3) by ${\displaystyle s\!-\!a_{j}}$.\, It yields ${\displaystyle {\frac {P(s)}{Q_{j}(s)}}=C_{j}+(s\!-\!a_{j})\sum _{\nu \neq j}{\frac {C_{\nu }}{s\!-\!a_{\nu }}}.}$ Setting to this identity \,${\displaystyle s:=a_{j}}$\, gives the value

${\displaystyle {\begin{matrix}C_{j}\;=;{\frac {P(a_{j})}{Q_{j}(a_{j})}}.\end{matrix}}}$

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

${\displaystyle {\begin{matrix}C_{j};=\;{\frac {P(a_{j})}{Q'(a_{j})}}.\end{matrix}}}$

The values (6) in (4) produce the formula (1).

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## References

1. {\sc K. V\"ais\"al\"a:} Laplace-muunnos .\, Handout Nr. 163.\quad Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).