The unilateral Laplace transform is in its region of absolute convergence of a Laplace transform. Under the usual inversion hypotheses, the Bromwich inversion formula is
The vertical Bromwich contour is directed upward, with inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.
For a piecewise continuously differentiable with an integrable transformed derivative, integration by parts gives the Laplace transform of a derivative:
Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.
TODO: the original closed-contour drawing is absent from the source. The following independent Bromwich contour deformation establishes the required discontinuity relation without asserting the missing drawing's exact geometry. Orient the vertical line upward, as required by the positive imaginary answer in the next part, and use the principal square root, .
Close a truncated vertical line to the left with a rectangle, slit it along the negative real axis, and exclude a small circle about zero. The integrand is holomorphic inside that slit domain, so the boundary contour integral is zero. Horizontal boundary integrals are : integrating along gives a bounded factor while . The left boundary contributes , and the small circle contributes . All vanish in the indicated limits.
At with , the upper and lower bank values are respectively and . The boundary follows the upper bank toward zero and the lower bank away from zero. Their combined contribution is . Hence
The vertically improper integral converges conditionally; these estimates also justify the contour limit. Reversing the vertical orientation reverses its sign.
Figure 1.
An independently chosen slit-rectangle contour relates the upward Bromwich integral to the square-root branch jump
. Details.
Substitute into the cubic denominator. Its leading terms are
The nonzero large root obeys , so
The sum of the three roots is exactly . Combining this with the two slow-root expansions gives the refinement .
For physical radiative sign , the fast pole lies high in the upper half-plane. Its homogeneous mode is , a runaway. It prevents a bounded retarded response with the real-axis Fourier prescription even though the two slow poles are damped. If , the fast pole decays but the slow pair grow. No nonzero real sign puts every pole in the lower half-plane.
There are two distinct boundary choices. A strictly causal initial-value Green function exists and includes the growing mode for ; it must be represented by a Laplace/Bromwich contour above that pole rather than the stated ordinary Fourier integral. Alternatively the real-axis inverse gives a bounded stationary no-runaway response and preacceleration: for , upper closure encloses the fast pole and yields a nonzero response before the force. Its leading contribution is
This follows from the upper-pole residue and the factor from upper contour closure. Eliminating runaway growth by a future boundary condition therefore sacrifices strict retardation on the short scale .
The singular root also shows why a perturbative expansion around only misses the problem. For a low-frequency effective description, order reduction of a radiation-reaction oscillator uses in the small correction and gives . For positive this approximate second-order equation has a causal damped response and discards the fast mode outside its intended regime; it does not make the original third-order equation simultaneously bounded and retarded.