use laplace transform to solve the given integral equation

Question

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

Answer 1

I have answered it in the paper , I think that it's correct

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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\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}