# Inverse Laplace transform of $\frac{1}{s^n+a}, n>2$

I was wondering if there are some formulas for such functions? I can't find it in any standard table. And if there aren't can someone explain to me why?

## 2 Answers

Write $a=|a|e^{i\phi}$ with $0\le \phi <2\pi$.

Then, note that $s^n+a=0$ for $s=s_k$ where

$$s_k=|a|^{1/n}e^{i\phi/n}e^{i(2k-1)\pi/n}$$

$k=1,\dots,n$.

We can then use partial fraction expansion to write

$$\frac{1}{s^n+a}=\sum_{k=1}^n \frac{a_k}{s-s_k} \tag 1$$

We can find the coefficients $a_k$ by multiplying both sides of $(1)$ by $s-s_k$ and letting $s\to s_k$. Proceeding we have

$$a_k=\lim_{s\to s_k}\frac{s-s_k}{s^n+a}=\frac{1}{ns_k^{n-1}}$$

The inverse Laplace Transform of $\frac{a_k}{s-s_k}$ is given by $a_ke^{s_k t}u(t)$ and hence we have

$$\mathscr{L}^{-1}\left(\frac{1}{s^n+a}\right)=\sum_{k=1}^n \frac{1}{ns_k^{n-1}}e^{s_k t}u(t)$$

As an example, suppose $n=3$ and $a=1$. Then, the terms $s_k$ for $k=1,2,3$ are

\begin{align} s_1&=e^{i\pi/3}=\frac12(1+i\sqrt 3)\\\\ s_2&=e^{i3\pi/3}=-1\\\\ s_3&=e^{i5\pi/3}=\frac12(1-\sqrt 3) \end{align}

The inverse Laplace Transform is

\begin{align} \mathscr{L}^{-1}\left(\frac{1}{s^3+1}\right)&=\sum_{k=1}^3 \frac{1}{3s_k^2}e^{s_k t}u(t)\\\\ &=\frac1{3 e^{i2\pi/3}}e^{t/2}(\cos(\sqrt 3 t/2)+i\sin(\sqrt 3 t/2) t)u(t)\\\\ &+\frac1{3 }e^{t/2}u(t)\\\\ &+\frac1{3 e^{i10\pi/3}}e^{t/2}(\cos(\sqrt 3 t/2)-i\sin(\sqrt 3t/2) t)u(t)\\\\ &=\frac13\left(e^{-t}-e^{t/2}\left(\sqrt{3}\sin(\sqrt{3}t/2)-\cos(\sqrt{3}t/2)\right) \right) \end{align}

• I don't quite understand $s_k$, How do i calculate it? Could you explain to me for example how would we use your example to find the inverse of $\frac{1}{s^3+1}$? – Scavenger23 May 21 '17 at 13:11
• I've edited to add the requested example. – Mark Viola May 21 '17 at 16:23

$\mathcal{L}^{-1}\left\{\dfrac{1}{s^n+a}\right\}$

$=\mathcal{L}^{-1}\left\{\dfrac{s^{-n}}{1+as^{-n}}\right\}$

$=\mathcal{L}^{-1}\left\{\sum\limits_{k=0}^{\infty}(-1)^ka^ks^{-(k+1)n}\right\}$

$=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^ka^kt^{(k+1)n-1}}{\Gamma((k+1)n)}$