Laplace transform of a Cauchy's problem I have to determine the Laplace transform of $u$, where $u$ is the solution to this Cauchy's problem
$$ u'(t) + \int_0^t e^{t-s} \left( \int_0^s u(r) ~ dr \right) ~ ds = 0, t > 0, u(0) = 1$$
$L[u](z) = ~? $
I know the the Laplace transform is $ \int_0^{+\infty} u(t)e^{-zt} \ dt$ but I can't determine the $u$ function. Can you please help me? :(
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\mrm{u}'\pars{t} + \int_{0}^{t}\expo{t - s}
\bracks{\int_{0}^{s}\mrm{u}\pars{r}\,\dd r}\dd s = 0\,,\quad t > 0.
\qquad\mrm{u}\pars{0} = 1}$.


\begin{align}
0 & =
\mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\expo{-s}\bracks{s < t}
\int_{0}^{\infty}\mrm{u}\pars{r}\bracks{r < s}\dd r\,\dd s
\\[5mm] & =
\mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\mrm{u}\pars{r}
\int_{0}^{\infty}\expo{-s}\bracks{r < s < t}\dd s\,\dd r
\\[5mm] &=
\mrm{u}'\pars{t} + \expo{t}\int_{0}^{\infty}\mrm{u}\pars{r}\bracks{r < t}
\pars{\expo{-r} - \expo{-t}}\,\dd r
\\[5mm] & =
\mrm{u}'\pars{t} + \expo{t}\int_{0}^{t}\mrm{u}\pars{r}\expo{-r}\,\dd r -
\int_{0}^{t}\mrm{u}\pars{r}\,\dd r
\end{align}


Multiply both members by $\ds{\expo{-st}}$ and integrate over $\ds{t > 0}$
  $\ds{\pars{~\mbox{note that}\ \hat{\mrm{f}}\pars{s} \equiv \int_{0}^{\infty}\mrm{f}\pars{t}\expo{-st}\,\dd t ~}}$:

\begin{align}
0 & =
-\ \overbrace{\mrm{u}\pars{0}}^{\ds{=\ 1}}\ + s\,\hat{\mrm{u}}\pars{s} +
\int_{0}^{\infty}\expo{-st}\expo{t}\int_{0}^{t}\mrm{u}\pars{r}\expo{-r}
\,\dd r\,\dd t -
\int_{0}^{\infty}\expo{-st}\int_{0}^{t}\mrm{u}\pars{r}\,\dd r\,\dd t
\\[5mm] & =
-1 + s\,\hat{\mrm{u}}\pars{s} +
\int_{0}^{\infty}\mrm{u}\pars{r}\expo{-r}
\int_{r}^{\infty}\expo{-\pars{s - 1}t}\dd t\,\dd r - \int_{0}^{\infty}\mrm{u}\pars{r}\int_{r}^{\infty}\expo{-st}\dd t\,\dd r 
\\[5mm] & =
-1 + s\,\hat{\mrm{u}}\pars{s} + {1 \over s - 1}\int_{0}^{\infty}\mrm{u}\pars{r}\expo{-sr}\dd r -
{1 \over s}\int_{0}^{\infty}\mrm{u}\pars{r}\expo{-sr}\dd r
\\[5mm] & =
-1 + s\,\hat{\mrm{u}}\pars{s} + \pars{{1 \over s - 1} -
{1 \over s}}\hat{\mrm{u}}\pars{s} \implies
\bbx{\ds{\,\hat{\mrm{u}}\pars{s} = {s^{2} - s \over s^{3} - s^{2} + 1}}}
\end{align}

$\ds{s^{3} - s^{2} + 1 = 0}$ has one negative real root
  $\ds{s_{1} \approx -0.7549}$ and two complex roots
  $\ds{a \pm b\ic\approx 0.8774 \pm 0.7449\ic}$.


With $\ds{c > a}$, $\ds{\mrm{u}\pars{t}}$ is given by:
\begin{align}
\mrm{u}\pars{t} & =
\int_{c - \infty\ic}^{c + \infty\ic}{s^{2} - s \over s^{3} - s^{2} + 1}\,
\expo{st}\,{\dd s \over 2\pi\ic} =
{s_{1} - 1 \over 3s_{1} - 2}\,\expo{-\verts{s_{1}}t} +
2\,\Re\pars{{a + b\ic - 1 \over 3\bracks{a + b\ic} - 2}\,
\expo{\bracks{a + b\ic}t}}
\end{align}
