Bromwich inversion formula

Let $f$ be of exponential order $a$, piecewise continuously differentiable on finite intervals of nonnegative time, and let $F$ be its unilateral Laplace transform. For $\gamma >a$ and $t>0$ at a point of continuity,At a jump the symmetric integral recovers $[f(t-)+f(t+)]/2$. The vertical line must lie in the defining region of absolute convergence of a Laplace transform; saying only that it avoids singularities of an analytic continuation does not supply that condition.^[ref]

To see the inversion mechanism, extend $e^{-\gamma t}f(t)$ by zero to negative time. Its Fourier transform at angular frequency $\omega$ is $F(\gamma +i\omega )$. The Fourier inversion theorem, followed by multiplication by $e^{\gamma t}$ and the substitution $s=\gamma +i\omega$, gives the displayed formula. Contour deformation is a subsequent step that needs separate growth estimates.

A Laplace transform $G_T(p)={\int }_0^T{e}^{-ps}g(s)ds$ can grow exponentially as $\mathrm{Re}p\to -\infty$. At an evaluation time $t<T$, multiplying by $e^{pt}$ does not automatically permit closing the inverse contour to the left. Either retain the growth-bearing contour pieces, use a causal time convolution, or justify a deformation with the time-truncated transform. Resolvent-pole information alone is insufficient.