Laplace transform using partial fractions

Question

I want to know how does $\dfrac{1}{(s^2+1)^2}$ is equal to $\mathcal{L}\{\sin t - t\cos t\}$ because my professor told me to use the partial fractions method but I'm not getting the exact answer.

Answer 1

For any $s>0$ we have: $$ \frac{1}{s^2+1} = \frac{1}{2i}\left(\frac{1}{s-i}-\frac{1}{s+i}\right)=\frac{1}{2i}\int_{0}^{+\infty}e^{-t(s-i)}-e^{-t(s+i)}\,dt\tag{1}$$ or $$ \frac{1}{s^2+1} = \int_{0}^{+\infty}\sin(t)e^{-ts}\,dt =\mathcal{L}(\sin t)\tag{2}$$ but we also have, by squaring the LHS of $(1)$, $$\begin{eqnarray*}\color{red}{\frac{1}{(s^2+1)^2}}&=&-\frac{1}{4}\left(\frac{1}{(s-i)^2}+\frac{1}{(s+i)^2}-\frac{2}{s^2+1}\right)\\&=&-\frac{1}{4}\int_{0}^{+\infty}te^{-t(s-i)}+t e^{-t(s+i)}-i e^{-t(s-i)}+ie^{-t(s+i)}\,dt\\&=&\frac{1}{2}\int_{0}^{+\infty}(\sin t-t\cos t)e^{-ts}\,dt=\color{red}{\mathcal{L}\left(\frac{\sin t-t\cos t}{2}\right)}.\tag{3}\end{eqnarray*}$$ In the opposite direction, $$\forall s>0,\quad\int_{0}^{+\infty}\frac{\sin t-t\cos t}{2}e^{-ts}\,dt = \frac{1}{(s^2+1)^2}\tag{4} $$ is easy to prove by integration by parts.

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{c} > 0}$ and $\ds{t > 0}$:

\begin{align} \int_{c - \infty\ic}^{c + \infty\ic}{\expo{ts} \over \pars{s^{2} + 1}^{2}}\,{\dd s \over 2\pi\ic} & = \mrm{Res}\pars{\expo{ts} \over \bracks{s^{2} + 1}^{2}}_{\ s\ =\ -\ic} + \mrm{Res}\pars{\expo{ts} \over \bracks{s^{2} + 1}^{2}}_{\ s\ =\ \ic} \\[5mm] & = \partiald{}{s}\bracks{{\expo{ts} \over \pars{s - \ic}^{2}}}_{\ s\ =\ -\ic} + \partiald{}{s}\bracks{{\expo{ts} \over \pars{s + \ic}^{2}}}_{\ s\ =\ \ic} \\[5mm] & = 2\,\Re\pars{{1 \over 4}\,\expo{-\ic t}\bracks{-t + \ic}} = \bbox[8px,border:1px groove navy]{{1 \over 2}\bracks{-t\cos\pars{t} + \sin\pars{t}}} \end{align}