Laplace transform inversion by contour integration
/ lah-PLAHSS /
The Laplace transform is the standard tool for solving differential equations and analyzing circuits and control systems: it turns the hard operation of differentiating in time into the easy operation of multiplying by s, so an ordinary differential equation becomes an algebra problem in F(s). But after you have solved that algebra problem you hold F(s), and you need the actual time-domain answer f(t) back. Inverting the transform is that last step, and complex analysis gives a clean, almost automatic recipe for it.
The inversion formula is f(t) = (1 / (2 pi i)) times (integral along the Bromwich line of e^(s t) F(s) ds), the line being a vertical line Re(s) = c lying to the right of all singularities of F. The practical method is not to evaluate this line integral head-on but to close the contour with a large left-hand semicircle and apply the residue theorem. Each pole s_k of F(s) of order m contributes a residue of e^(s t) F(s); for a simple pole that residue is just e^(s_k t) times the limit as s -> s_k of (s - s_k) F(s). So the inverse transform is literally the sum of residues, and the answer comes out as a sum of exponentials e^(s_k t) (times polynomials in t when poles are repeated). Jordan's lemma certifies that the large arc vanishes for t > 0, which is what makes the closing step legal.
This is why rational transforms invert so painlessly: factor F(s), do partial fractions or residues, and read off the exponentials. The method also handles branch points — say F(s) contains the square root of s — by deforming the Bromwich line into a keyhole that wraps a branch cut along the negative real axis, turning that part of the inversion into a real integral that represents a continuous spectrum (typical of diffusion and heat problems). The candid limits: the formula assumes f is of exponential order so the transform converges, and choosing c too far left (left of some pole) silently gives the wrong, non-causal answer.
Solve y'' + y = 0 with y(0) = 0, y'(0) = 1. The transform gives (s^2 + 1) Y(s) = 1, so Y(s) = 1 / (s^2 + 1). Its poles are at s = plus or minus i; summing the residues of e^(s t) / (s^2 + 1) there gives y(t) = sin t. The differential equation never had to be solved directly — algebra plus two residues did it.
Poles of Y(s) become the exponential (here sinusoidal) modes of y(t).
The residue recipe assumes you can close the contour and that the arc vanishes; if F(s) does not decay enough on the big arc (for instance F(s) = e^(a s) for a > 0, which corresponds to a time shift), closing fails and you must invert by other means.