use laplace transform to solve the given integral equation use Laplace transform to solve the given integral equation

I don't know how start because it differences on other Laplace question I see before  
 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}
&\overbrace{\int_{0}^{\infty}\mrm{f}\pars{t}\expo{-st}\dd t}
^{\ds{\equiv\ \hat{\mrm{f}}\pars{s}}}\ +\
\int_{0}^{\infty}
\bracks{\int_{0}^{t}\pars{t - \tau}\dd\tau}\expo{-st}\dd t =
\int_{0}^{\infty}t\expo{-st}\dd t
\\[5mm] &
\hat{\mrm{f}}\pars{s} + \int_{0}^{\infty}\mrm{f}\pars{\tau}\
\overbrace{\int_{\tau}^{\infty}
\pars{t - \tau}\expo{-st}\dd t}^{\ds{\expo{-s\tau} \over s^{2}}}\,\dd\tau =
{1 \over s^{2}}
\\[5mm] &
\implies
\hat{\mrm{f}}\pars{s} + {\hat{\mrm{f}}\pars{s} \over s^{2}} = {1 \over s^{2}}
\implies
\bbx{\hat{\mrm{f}}\pars{s} = {1 \over s^{2} + 1}}
\end{align}

\begin{align}
\mrm{f}\pars{t} & = \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}
{\expo{st} \over s^{2} + 1}\,{\dd s \over 2\pi\ic} =
\lim_{s \to -\ic}\bracks{\pars{s + \ic}{\expo{st} \over s^{2} + 1}} +
\lim_{s \to \ic}\bracks{\pars{s - \ic}{\expo{st} \over s^{2} + 1}}
\\[5mm] & =
{\expo{-\ic t} \over -2\ic} + {\expo{\ic t} \over 2\ic} = \bbx{\sin\pars{t}}
\end{align}
A: I have answered it in the paper , I think that it's correct  

