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

  (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

  (2)

  
  

A special case of the Heaviside formula (1) is $$\mathcal{L}^{-1}\left\{\frac{Q'(s)}{Q(s)}\right\} = \sum_{j=1}^ne^{a_jt}.\\$$

Example. Since the zeros of the binomial $s^4\!+\!4a^4$ are $s = (\pm1\!\pm\!i)a$ , 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

(3)

(4)

(5)

(6)

Bibliography

1

K. V¨AISÄLÄ: Laplace-muunnos. Handout Nr. 163.    Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).