Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$

Question

How to compute the Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$? I know that to compute Inverse Laplace $\frac 1 {(s^2+a^2)^2}$, the convolution Theorem is useful. but is there an interesting idea for general case?

Answer 1

We could, also, use the definition \begin{align} \mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s^2 + a^2)^n}ds\\ &= \sum\text{Res} \end{align} We have $n$ poles at $s = \pm ia$ so $$ \mathcal{L}^{-1}\{F(s)\} = \lim_{s\to ia}\frac{1}{(n - 1)!}\frac{d^{n-1}}{ds^{n-1}}(s-ia)^n\frac{e^{st}}{(s^2+a^2)^n} + \lim_{s\to -ia}\frac{1}{(n - 1)!}\frac{d^{n-1}}{ds^{n-1}}(s+ia)^n\frac{e^{st}}{(s^2+a^2)^n} $$

Answer 2

You can use the convolution theorem inductively:

$$\mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^n} \right] = \mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^{n-1}} \right] * \mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)} \right]$$

and we have the base case.

Hence

$$\mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^n} \right] = \underbrace{\frac{\sin(ax)}{a} * \cdots * \frac{\sin(ax)}{a}}_{n \text{ times}}$$

Answer 3

Case $1$: $n>0$

According to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/LapInv4.pdf,

$\mathcal{L}_{s\to t}^{-1}\left\{\dfrac{1}{(s^2+a^2)^n}\right\}=\dfrac{\sqrt\pi t^{n-\frac{1}{2}}J_{n-\frac{1}{2}}(at)}{2^{n-\frac{1}{2}}a^{n-\frac{1}{2}}\Gamma(n)}$

Case $2$: $n\in\mathbb{Z}^-$

$\mathcal{L}_{s\to t}^{-1}\left\{\dfrac{1}{(s^2+a^2)^n}\right\}=\mathcal{L}_{s\to t}^{-1}\{(s^2+a^2)^{-n}\}=\mathcal{L}_{s\to t}^{-1}\left\{\sum\limits_{k=0}^{-n}C_k^{-n}a^{-2n-2k}s^{2k}\right\}=\sum\limits_{k=0}^{-n}C_k^{-n}a^{-2n-2k}\delta^{(2k)}(t)$