= Solution
The integral $I$ is the <convolution> of the positive-half-line kernels $x^{a-1}$ and $x^{b-1}$. The <convolution theorem for Laplace transforms> therefore gives, for $p>0$,
$$
\mathcal L I(p)=\Gamma(a)\Gamma(b)p^{-(a+b)}.
$$
On the other hand, the substitution $t=xu$ gives $I(x)=x^{a+b-1}B(a,b)$, where $B(a,b)=\int_0^1u^{a-1}(1-u)^{b-1}\,du$ is the <beta function>. Applying part (i) again yields $\mathcal L I(p)=B(a,b)\Gamma(a+b)p^{-(a+b)}$. Comparing the two expressions proves
$$
\boxed{B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}}.
$$
All integrands used in the real-$p$ convolution argument are nonnegative, so <Tonelli's theorem> justifies the interchange; the endpoint singularities are integrable because $a,b>0$.
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