PlanetPhysics/Heaviside Formula

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

$$\begin{matrix} \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^n\frac{P(a_j)}{Q'(a_j)}e^{a_jt}. \end{matrix}$$ "$\begin{matrix} \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^ne^{a_jt}\sum_{k=0}^{m_j-1}\frac{F_j^{(k)}(a_j)t^{m_j\!-\!1\!-\!k}}{k!(m_j\!-\!1\!-\!k)!}. \end{matrix}$|undefined"
 * If all complex zeroes $$a_1,\,a_2,\,\ldots,\,a_n$$ of $$Q(s)$$ are simple, then
 * 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

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 $$\begin{matrix} \frac{P(s)}{Q(s)} \;=\; \frac{C_1}{s\!-\!a_1}+\frac{C_2}{s\!-\!a_2}+\ldots+\frac{C_n}{s\!-\!a_n} \end{matrix}$$ with constants $$C_j$$.\, According to the linearity and the formula 1 of the parent entry, one gets $$\begin{matrix} \mathcal{L}^{-1}\left\{\frac{P(s)}{Q(s)}\right\} \;=\; \sum_{j=1}^nC_je^{a_jt}. \end{matrix}$$ 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).