Bromwich justification of the Heaviside calculus

Remarks

Under suitable growth conditions on $\varphi$, the Laplace transform of $x$ is $\varphi(s)/s$, so this interpretation agrees with the Laplace transform reading of the Heaviside operational calculus. By Cauchy's theorem, the value does not depend on $\gamma > c$. When $\varphi$ is meromorphic and the line can be closed to the left, $x(t)$ is the sum of the residues of $\varphi(s) e^{st} / s$.

Example: $\varphi(p) = p^{-n}$ with $n \ge 0$. For $t > 0$, the line can be closed to the left, and the only pole of $e^{st}/s^{n+1}$ is at $0$, with residue $t^n / n!$. So $x(t) = t^n/n!$, which is Heaviside's rule $p^{-n} H = t^n H / n!$.

Non-example: $\varphi(s) = e^{-s^2}$ is entire, but on the line $s = \gamma + iy$ one has $\abs{\varphi(s)} = e^{y^2 - \gamma^2}$. So the integrand grows without bound, the integral diverges for every $t$, and $\varphi(p) H$ has no Bromwich interpretation.