= Solution
For a piecewise continuous function of exponential order on $t\geq0$, its unilateral <Laplace transform> is
$$
\boxed{F(p)=\int_0^\infty e^{-pt}f(t)\,dt.}
$$
The <Bromwich inversion formula>, at a point of continuity, is
$$
\boxed{f(t)=\frac1{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{pt}F(p)\,dp.}
$$
The <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 $d\lambda=-e^{-t}dt$ and
$$
H(s)=\int_0^\infty e^{-st}h(e^{-t})\,dt.
$$
Thus 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 obtain
$$
\boxed{h(\lambda)=\frac1{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}\lambda^{-s}H(s)\,ds.}
$$
Here $\gamma$ 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.
Back to article page
