Different methods to find Inverse Laplace Transform Find Laplace Transform of:$$F(s)=\frac{1}{s^4(s^2+1)}$$
It was the Bonus point question in my exam. I solved it with this Lemma :
Let $F(s)=\mathcal{L}\{f(t)\}$, we have $\frac{F(s)}{s}=\mathcal{L}\{\int_o^tf(x)dx\}$
 and I evaluate 4 integrals and I get $\frac{t^3}{6}+\sin(t)-t$,correctly.
Here is my Question:
1.Is There any other method to evaluate it?(By the way I could solve it by using partial fractions. and convolution.)
2.In such questions that We can find Inverse Laplace Transform directly wtih respect to $t$ parameter or Solve it with Convolution. which method is better and worth more mathematically ? I mean when we use Convolution We have integral in our final answer.(I think writing the answer directly It has only variable $t$ is more propriety and better in math.)
Thanks in advance!
 A: If you want to find the inverse Laplace transform of $\frac{1}{s^4(s^2+1)}$ you just have to notice that there is a pole of order $4$ at the origin and simple poles at $\pm i$, so for some constants
$$ \frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{B}{s^3}+\frac{C}{s^2}+\frac{D}{s}+\frac{E}{s-i}+\frac{F}{s+i} $$
holds and it is very simple to compute $\mathcal{L}^{-1}$ of any term in the RHS. In order to find such constants you may notice that
$$ \frac{1}{s^4(s^2+1)}=\frac{1}{s^2}\cdot\left(\frac{1}{s^2}-\frac{1}{s^2+1}\right)=\frac{1}{s^4}-\frac{1}{s^2}+\frac{1}{s^2+1} $$
so $(\mathcal{L}^{-1} F)(t)$ is given by $\frac{t^3}{6}-t+\sin(t)$.
A: By the theorem of convolution you have :
$$f(t)=I=\dfrac {1}{3!}\int_0^t\sin(t-\tau) \tau ^3 d\tau$$
You can evaluate this integral of course. Integrate by part.You will get the same answer as yours. 
A first integration 
$$-6I=\int_0^t\sin(\tau-t) \tau ^3 d\tau$$
$$-6I=-\cos(\tau-t)\tau^3\bigg |_0^t+ \int_0^t-\cos(\tau-t)3\tau ^2 d\tau$$
$$-6I=-t^3+ 3\int_0^t\cos(\tau-t)\tau ^2 d\tau$$
$$-6I=-t^3- 6\int_0^t\sin(\tau-t)\tau  d\tau$$
$$-6I=-t^3+ 6t+6\int_0^t\cos(\tau-t)  d\tau$$
$$-6I=-t^3+ 6t-6\sin(t)$$
Finally :
$$ \boxed {f(t)=\dfrac {t^3}6-t+\sin(t)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\mrm{f}\pars{t}\right\vert_{\, t\ >\ 0} & =
\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{1 \over s^{4}\pars{s^{2} +1}}\,\expo{ts}\,{\dd s \over 2\pi\ic}
\\[3mm]\ & =
2\pi\ic\bracks{{\expo{t\ic} \over 2\ic} + {\expo{t\pars{-\ic}} \over -2\ic}}\,{1 \over 2\pi\ic} +
2\pi\ic\,{1 \over 3!}\lim_{s \to 0}
\partiald[3]{}{s}\pars{{\expo{ts} \over s^{2} +1}\,{1 \over 2\pi\ic}}
\\[3mm]\ & =
\sin\pars{t} +
{1 \over 6}\lim_{s \to 0}
\partiald[3]{}{s}\braces{\bracks{1 + ts +
{1 \over 2}\pars{ts}^{2} + {1 \over 6}\pars{ts}^{3}}\pars{1 - s^{2}}}
\\[3mm]\ & =
\sin\pars{t} +
{1 \over 6}\lim_{s \to 0}
\partiald[3]{}{s}\pars{-ts^{3} + {1 \over 6}\,t^{3}s^{3}}
\\[3mm] & =
\bbx{\sin\pars{t} + {1 \over 6}\,t^{3} - t}
\end{align}
A: An easy approach to split the fraction $F(s)=\frac{1}{s^4(s^2+1)}$ by partial fraction:
$$\frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{B}{s^3}+\frac{C}{s^2}+\frac{D}{s}+\frac{Es+F}{s^2+1}$$
Because $F(s)$ is even function and $F(s)=F(-s)$ so $B=D=E=0$:
$$\frac{1}{s^4(s^2+1)} = \frac{A}{s^4}+\frac{C}{s^2}+\frac{F}{s^2+1}$$
$A=1,C=-1,F=1$
