How do we find the inverse Laplace transform of $\frac{1}{(s^2+a^2)^2}$?

Question

How do we find the inverse Laplace transform of $\frac{1}{(s^2+a^2)^2}$? Do I need to use the convolution theory? It doesn't match any of the known laplace inverse transforms. It matches with the Laplace transform of $\sin(at)$ but I don't know if that helps or not. Also there seems to be a formula with limits and imaginary numbers. How do I just know to apply that here?

Answer 1

Well, we can use the 'time-domain integration' property of Laplace transform:

$$\mathscr{L}_t\left[\int_0^t\text{f}\left(\tau\right)\space\text{d}\tau\right]_{\left(\text{s}\right)}:=\int_0^\infty\left\{\int_0^t\text{f}\left(\tau\right)\space\text{d}\tau\right\}\cdot e^{-\text{s}t}\space\text{d}t=\frac{\text{F}\left(\text{s}\right)}{\text{s}}\tag1$$

Now, we know that (when $\Re\left(\text{s}\right)>0$):

$$\mathscr{L}_t\left[\cos\left(\omega t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\cos\left(\omega t\right)\cdot e^{-\text{s}t}\space\text{d}t=\frac{\text{s}}{\omega^2+\text{s}^2}\tag2$$

So, when we have:

$$\frac{1}{\omega^2+\text{s}^2}=\frac{1}{\text{s}}\cdot\frac{\text{s}}{\omega^2+\text{s}^2}\space\space\space\to\space\space\space\int_0^t\cos\left(\omega\tau\right)\space\text{d}\tau=\frac{\sin\left(\omega\tau\right)}{\omega}\tag3$$

Using the derivative we get:

$$\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\omega^2+\text{s}^2}\right\}=-\frac{2\text{s}}{\left(\omega^2+\text{s}^2\right)^2}\space\Longleftrightarrow\space-\frac{1}{2\text{s}}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\omega^2+\text{s}^2}\right\}=\frac{1}{\left(\omega^2+\text{s}^2\right)^2}\tag4$$

So, we get:

$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\left(\omega^2+\text{s}^2\right)^2}\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[-\frac{1}{2\text{s}}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\omega^2+\text{s}^2}\right\}\right]_{\left(t\right)}=$$ $$-\frac{1}{2}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\omega^2+\text{s}^2}\right\}\right]_{\left(t\right)}=-\frac{1}{2}\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\omega^2+\text{s}^2}\right\}\right]_{\left(\tau\right)}\space\text{d}\tau\tag5$$

Answer 2

Hint: Use $$\frac{d}{ds}\left(\frac{s}{s^2+a^2}\right)=2a^2\frac{1}{(s^2+a^2)^2}-\frac{1}{s^2+a^2}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{st} \over \pars{s^{2} + a^{2}}^{2}}\,{\dd s \over 2\pi\ic} = \lim_{s \to -a\ic}\partiald{}{s}\bracks{\expo{st} \over \pars{s - a\ic}^{2}} + \lim_{s \to a\ic}\partiald{}{s}\bracks{\expo{st} \over \pars{s + a\ic}^{2}} \\[5mm] = &\ 2\,\Re\bracks{-\,{-\ic\expo{-\ic at} + at\expo{-\ic at} \over 4a^{3}}} = \bbx{\ds{{\sin\pars{at} - at\cos\pars{at} \over 2a^{3}}}} \end{align}