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) $a_1,a_2,\mathrm{\dots },a_n$ of $Q(s)$ are simple, then

  ${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{\displaystyle \frac{P(a_j)}{Q^{\prime }(a_j)}}{e}^{a_{j}t}.$(1)

• If the different zeroes $a_1,a_2,\mathrm{\dots },a_n$ of $Q(s)$have the multiplicities $m_1,m_2,\mathrm{\dots },m_n$, respectively,we denote  $F_j(s):={(s-a_j)}^{m_j}P(s)/Q(s)$;  then

  ${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{e}^{a_{j}t}{\displaystyle \sum _{k=0}^{m_j-1}}{\displaystyle \frac{F_j^{(k)}(a_j)t^{m_j-1-k}}{k!(m_j-1-k)!}}.$(2)

A special case of the Heaviside formula (1) is

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

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

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

Proof of (1).  Without hurting the generality, we cansuppose that $Q(s)$ is monic.  Therefore

$$Q(s)=(s-a_1)(s-a_2)\mathrm{\cdots }(s-s_n).$$

For  $j=1,2,\mathrm{\dots },n$,  denoting

$$Q(s):=(s-a_j)Q_j(s),$$

one has  $Q_j(a_j)\ne 0$.  We have a partial fractionexpansion of the form

${\displaystyle \frac{P(s)}{Q(s)}}={\displaystyle \frac{C_1}{s-a_1}}+{\displaystyle \frac{C_2}{s-a_2}}+\mathrm{\dots }+{\displaystyle \frac{C_n}{s-a_n}}$(4)

with constants $C_j$.  According to the linearity and theformula 1 of the parent entry (http://planetmath.org/LaplaceTransform),one gets

${ℒ}^{-1}\left\{{\displaystyle \frac{P(s)}{Q(s)}}\right\}={\displaystyle \sum _{j=1}^n}{C}_j{e}^{a_{j}t}.$(5)

For determining the constants $C_j$, multiply (3) by$s-a_j$.  It yields

$$\frac{P(s)}{Q_j(s)}=C_j+(s-a_j)\sum _{\nu \ne j}\frac{C_{\nu }}{s-a_{\nu }}.$$

Setting to this identity  $s:=a_j$  gives the value

$C_j={\displaystyle \frac{P(a_j)}{Q_j(a_j)}}.$(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) maybe written

$C_j={\displaystyle \frac{P(a_j)}{Q^{\prime }(a_j)}}.$(7)

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