 Solution

For a piecewise continuous function of exponential order on $t\ge 0$, its unilateral Laplace transform isThe Bromwich inversion formula, at a point of continuity, isThe Bromwich contour is a straight vertical line, traversed upwards, with $\gamma$ to the right of the abscissa of convergence and hence of all singularities of the transform. Under the usual piecewise smoothness assumptions, at a jump the symmetric inversion limit gives the mean of the two one-sided values.

In the transform in the question, put $\lambda =e^{-t}$. Then $d\lambda =-e^{-t}dt$ andThus the restricted Mellin transform on the unit interval is the Laplace transform of $f(t)=h(e^{-t})$. Apply the Bromwich inversion formula with $t=-\log \lambda$, where $0<\lambda <1$, to obtainHere $\gamma$ lies in the convergence half-plane of $H$. At a discontinuity this formula has the same mean-value interpretation. The real natural logarithm defines ${\lambda }^{-s}$ unambiguously.