Past exam of the short options course of the University of Oxford / 2023 / 5 / b

For real $p>0$, substitute $t=px$ in the Laplace transform:The Gamma function integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\mathrm{Re}p>0$: the modulus of the integrand is $x^{a-1}{e}^{-(\mathrm{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp (-a\mathrm{Log}p)$ whose logarithm is real on the positive axis, by holomorphic continuation of the two sides.

If “exists” permits conditionally convergent ordinary improper integrals, there are additional boundary points when $0<a<1$: for $p=i\omega \ne 0$, the Dirichlet test gives convergence, with value $\Gamma (a)|\omega {|}^{-a}{e}^{-i\mathrm{sgn}(\omega )\pi a/2}$. For $a\ge 1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\mathrm{Re}p<0$, the integral diverges. Thus $\mathrm{Re}p>0$ is the region of absolute convergence of a Laplace transform, with the stated conditional boundary qualification.

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 $p>0$,On the other hand, the substitution $t=xu$ gives $I(x)=x^{a+b-1}B(a,b)$, where $B(a,b)={\int }_0^1{u}^{a-1}(1-u{)}^{b-1}du$ 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 provesAll integrands used in the real-$p$ convolution argument are nonnegative, so Tonelli's theorem justifies the interchange; the endpoint singularities are integrable because $a,b>0$.