# 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 (http://planetmath.org/Zero) $a_{1},\,a_{2},\,\ldots,\,a_{n}$ of $Q(s)$ are simple, then

 $\displaystyle\mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\}=\sum_{j=1}^{n}% \frac{P(a_{j})}{Q^{\prime}(a_{j})}e^{a_{j}t}.$ (1)
• 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

 $\displaystyle\mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\}=\sum_{j=1}^{n}e^% {a_{j}t}\sum_{k=0}^{m_{j}-1}\frac{F_{j}^{(k)}(a_{j})t^{m_{j}\!-\!1\!-\!k}}{k!(% m_{j}\!-\!1\!-\!k)!}.$ (2)

A special case of the Heaviside formula (1) is

 $\displaystyle\mathcal{L}^{-1}\left\{\frac{Q^{\prime}(s)}{Q(s)}\right\}\;=\;% \sum_{j=1}^{n}e^{a_{j}t}.$ (3)

Example.  Since the zeros of the binomial $s^{4}\!+\!4a^{4}$ are  $s=(\pm 1\!\pm\!i)a$,  we can calculate by (3) as follows:

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

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

 $\displaystyle\frac{P(s)}{Q(s)}=\frac{C_{1}}{s\!-\!a_{1}}+\frac{C_{2}}{s\!-\!a_% {2}}+\ldots+\frac{C_{n}}{s\!-\!a_{n}}$ (4)

with constants $C_{j}$.  According to the linearity and the formula 1 of the parent entry (http://planetmath.org/LaplaceTransform), one gets

 $\displaystyle\mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\}=\sum_{j=1}^{n}C_% {j}e^{a_{j}t}.$ (5)

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

 $\displaystyle C_{j}=\frac{P(a_{j})}{Q_{j}(a_{j})}.$ (6)

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

 $\displaystyle C_{j}=\frac{P(a_{j})}{Q^{\prime}(a_{j})}.$ (7)

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

## References

• 1 K. Väisälä: Laplace-muunnos.  Handout Nr. 163. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).
Title Heaviside formula HeavisideFormula 2014-03-19 9:14:46 2014-03-19 9:14:46 pahio (2872) pahio (2872) 11 pahio (2872) Topic msc 44A10 Heaviside expansion formula inverse Laplace transform of rational function HyperbolicFunctions ComplexSineAndCosine