the Bromwich inversion integral
/ BROM-witch /
Once you have solved a problem with the Laplace transform, your answer F(s) lives in the s-world, and you must travel back to the time-world to read it. The Bromwich integral is the return ticket — the formula that recovers f(t) from F(s). Unlike the Fourier inverse, which integrates along the real axis, this one runs up a vertical line in the complex plane, which is what lets it handle the growth and decay that Laplace transforms encode.
The formula is f(t) = (1/(2*pi*i)) integral along the Bromwich contour of F(s) e^(s t) ds, where the contour is the vertical line Re(s) = c traversed upward, and c is chosen to the right of every singularity of F(s). In practice you almost never compute it as written; instead you close the contour into a big loop to the left and use the residue theorem. The poles of F(s) become the building blocks of f(t): a simple pole at s = a contributes a term like (residue) times e^(a t), and a branch cut (as in diffusion problems, where sqrt(s) appears) wraps the contour around the cut and produces an integral that gives error functions or algebraically decaying tails.
This is where the physics of a Laplace-solved PDE finally reveals itself. The poles of F(s) tell you the natural frequencies and decay rates of the system — a pole at s = -gamma means a mode dying like e^(-gamma t). For the telegrapher's and diffusion equations, the inversion contour wrapping a branch cut is exactly what produces the spreading, smearing, error-function behaviour in time. The honest caveat: doing the contour integral correctly — locating every pole and branch cut, and choosing how to close the contour so the added arcs vanish (Jordan's lemma) — is genuine complex-analysis work, and is usually the hardest step in the whole Laplace method.
If F(s) = 1/(s+a), the only singularity is a simple pole at s = -a; closing the contour left and taking the residue gives f(t) = e^(-a t). For a diffusion problem with F(s) = (1/s) e^(-sqrt(s/k) x), there is no pole but a branch cut along the negative real axis, and wrapping the contour around it reproduces the error-function solution.
Close the vertical contour to the left; poles give exponentials, branch cuts give diffusion tails.
The contour must lie to the right of all singularities of F(s); place it too far left and you will miss poles and get the wrong answer. The same contour is sometimes called the Mellin-Bromwich line, reflecting its kinship with the Mellin transform.