Post's inversion formula

Remarks

By differentiation in the transform variable, $\frac{(-1)^k}{k!} \left(\frac{k}{t}\right)^{k+1} F^{(k)}\!\left(\frac{k}{t}\right) = \int_0^\infty f(u) \, \frac{1}{k!} \left(\frac{k}{t}\right)^{k+1} u^k e^{-k u / t} \, du$. The kernel in this integral is a probability density in $u$ (a gamma density) with mean $t (k+1)/k$ and variance $t^2 (k+1)/k^2$. As $k \to \infty$ it concentrates at $u = t$, so the integral tends to $f(t)$ at points of continuity. Example: for $f(t) = e^{-t}$, $F(s) = 1/(s + 1)$ and $\frac{(-1)^k}{k!} F^{(k)}(s) = (s+1)^{-k-1}$, so the expression is $\left(\frac{k/t}{k/t + 1}\right)^{k+1} = \left(1 + \frac{t}{k}\right)^{-k-1} \to e^{-t}$.

Going deeper

For a student

The formula matters in theory, for example in the theory of semigroups, where it gives the exponential formula for $T(t)$. Numerically it is poor: high derivatives amplify small errors, which reflects the ill-posedness of real Laplace inversion. The method of proof, a kernel that concentrates at one point, is the same as in many approximation theorems.