= Solution
{c}
For $p$ in a half-plane of absolute convergence, interchange the integrals using <Fubini's theorem>:
$$
\overline g(p)=\int_0^\infty f(s)\int_s^\infty e^{-pt}dt\,ds
=\boxed{\overline f(p)/p.}
$$
For positive real $p$, integration by parts gives $I_n=\int_0^\infty e^{-pt}t^n dt=nI_{n-1}/p$, starting from $I_0=1/p$. Hence $I_n=n!/p^{n+1}$, or
$$
\boxed{\mathcal L(t^n/n!)(p)=p^{-n-1}.}
$$
The formula extends throughout $\operatorname{Re}p>0$ by the same integration argument. Write $X=\mathcal Lx$. The integral term is the <convolution> of $t$ with $x(t)$, whose <Laplace transform> is $X/p^2$. The derivative transforms to $pX-x(0)=pX-1$. Therefore
$$
(p+p^{-2})X-1=p^{-2}+p^{-3}+p^{-5},
$$
and elementary factorization gives
$$
X=\frac{p^5+p^3+p^2+1}{p^3(p^3+1)}
=\frac{(p^2+1)(p^3+1)}{p^3(p^3+1)}
=\frac1p+\frac1{p^3}.
$$
Inverting the two polynomial transforms yields
$$
\boxed{x(t)=1+\tfrac12t^2.}
$$
Direct substitution verifies $x'=t$ and $\int_0^t(t-s)(1+s^2/2)ds=t^2/2+t^4/24$, as required. Uniqueness also follows without transform assumptions: differentiating the equation twice gives $x^{(3)}+x=1+t^2/2$ with initial data $x(0)=1$, $x'(0)=0$, $x''(0)=1$, a linear initial-value problem with a unique solution.
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