\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{description}
\item[(i)]
Find the Laplace transform of the convolution integral

$$F(t)=\ds\int_0^1 x^2 \exp(-x)\cos(t-x)dx.$$

\item[(ii)]
Find the inverse Laplace transform of

$$\exp(-3p)\ds\frac{\overline{f}(p)}{p^3}$$

\end{description}

\medskip

{\bf Answer}

\begin{description}
\item[(i)]
Use convolution result: If $f(t)=\ds\int_0^t d\xi g(\xi)h(t-\xi)$

Laplace: $\bar{f}(p)=\bar{g}(p)\bar{h}(p)$

Thus if $F(t)=dx\ \ds\int_0^t x^2e^{-x}\cos(t-x)$

$L[F(t)]=L[x^2e^{-x}]\times L[\cos x]$

\begin{eqnarray*} L[x^2e^{-x}] & = & \ds\int_0^{\infty}dx\ x^2
e^{-x}e^{-px}\\ & = & \ds\int_0^{\infty} dx\ x^2 e^{-x(p+1)}\\ & =
& \ds\frac{2!}{(p+1)^3} \end{eqnarray*}

from integral definition of factorial function

\begin{eqnarray*} L[\cos t] & = & \ds\int_0^{\infty}dt\ \cos
te^{-pt}\\ & = &
\ds\frac{1}{2}\ds\int_0^{\infty}e^{-(p+i)t}+\ds\frac{1}{2}\ds\int_0{\infty}e{-(p+i)t}\\
& = & \ds\frac{1}{2(p-i)}+\ds\frac{1}{2(p-i)}\\ & = &
\ds\frac{1}{2}\ds\frac{(p+i+p-i)}{p^2+1}\\ & = &
\ds\frac{p}{(p^2+1)}\end{eqnarray*}

so $L[F(t)]=\ds\frac{2}{(p+1)^3} \times
\ds\frac{p}{(p^2+1)}=\ds\frac{2p}{(p+1)^3(p^2+1)}$

\item[(ii)]
\begin{eqnarray*}
L^{-1}\left[\ds\frac{e^{-3p}\bar{f}(p)}{p^3}\right] & = &
L^{-1}\left[e^{-3p}\bar{f}(p)\times \ds\frac{1}{p^3}\right]\\ & =
& L^{-1} \left[L[\undb{u(t-3)f(t-3)}]\times
L\left[\undb{\ds\frac{t^2}{2}}\right]\right] \end{eqnarray*}

from $L[u(t-a)f(t-a)]=\bar{f}(p)e^{-ap}$ and
$L[t^n]=\ds\frac{n!}{p^{n+1}}$

So we have a $L^{-1}$[of product of $L$] $\Rightarrow$ inverse
transform is a convolution:

\begin{eqnarray*}
L^{-1}\left[\ds\frac{e^{-3p}\bar{f}(p)}{p^3}\right] & = &
\ds\int_0^t dx\ u(x-3)f(x-3)\ds\frac{(x-t)^2}{2}\\ & = &
\ds\frac{3}{t}dx\ f(x-3)\ds\frac{(x-t)^2}{2}\\ \rm{or} & = &
\ds\frac{1}{2}\ds\int_3^t dx\ f(t-x)(x-3)^2 \end{eqnarray*}

\end{description}

\end{document}