2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 $\frac{1}{s^3+1}$? $\endgroup$ – Scavenger23 May 21 '17 at 13:11
  • $\begingroup$ I've edited to add the requested example. $\endgroup$ – Mark Viola May 21 '17 at 16:23
0
$\begingroup$

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.