Bromwich integral
/ BROM-witch /
Reading a transform table backward is convenient, but it is not a definition — it is recognition. The Bromwich integral is the honest, constructive definition of the inverse Laplace transform: a single integral formula that recovers f(t) from F(s) for any transform, not just the ones that happen to appear in a table.
The formula is a contour integral in the complex plane: f(t) = (1/(2 pi i)) integral along the vertical line Re s = c of e^{s t} F(s) ds, where the line, the Bromwich contour, is a vertical line placed to the right of every singularity of F(s) — that is, inside the region of convergence. You integrate up that line from c - i infinity to c + i infinity. This is the exact inverse of the defining Laplace integral, and it is sometimes called the Fourier-Mellin or Mellin inversion formula.
In practice you rarely integrate along the line directly. Instead you close the contour with a large arc to the left and apply the residue theorem: the integral equals the sum of the residues of e^{s t} F(s) at all the poles of F(s) to the left of the line. Each simple pole at s = a contributes a term proportional to e^{a t}, which is exactly why the inverse of a rational F(s) is a sum of exponentials — partial fractions are just residues in disguise. The Bromwich integral is the bridge from the engineer's table to genuine complex analysis, and it is what makes the inverse transform well-defined even when no table entry fits.
For F(s) = 1/(s - a), the only pole is at s = a; the residue of e^{s t}/(s - a) there is e^{a t}, so the Bromwich integral returns f(t) = e^{a t} — matching the table.
A single residue reproduces the simplest transform pair, showing partial fractions and residues are the same computation.
Closing the contour to the left works only when F(s) decays on the large arc (Jordan's lemma) and only for t > 0; for t < 0 you must close to the right, where there are no poles, recovering the causal fact that f(t) = 0 before t = 0.