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Complex inversion formula

4. Complex inversion formula

The complex inversion formula is a very powerful technique for computing the
−1
inverse of a Laplace transform, f(t) = L F(s). The technique is based on the
methods of contour integration and requires that we consider our parameter s to
be a complex variable.

Fourier-Mellin formula

For a continuous function f possessing a Laplace transform, let us extend f to
(−∞, ∞) by taking f(t) = 0 for t < 0. Then for s = x + iy,

∫ ∫
∞ ∞
L(f(t)) = F(s) = e −st f(t)dt = e −iyt (e −xt f(t))dt = F(x, y).
0 −∞

In this form F(x, y) represents the Fourier transform of the function

g(t) = e −xt f(t). The Fourier transform is one of the most useful tools in
mathematical analysis; its principal virtue is that it is readily inverted.
Towards this end, we assume that f is continuous on [0, ∞), f(t) = 0 for t <
0, f has exponential order α, and f is piecewise continuous on [0, ∞). Then by
′
Theorem 1, L(f(t)) converges absolutely for Re(s) = x > α, that is,

∫ ∫
∞ ∞
|e −st f(t)|dt = e −xt |f(t)|dt < ∞, x > α. (4.1)
0 −∞

This condition means that g(t) = e −xt f(t) is absolutely integrable, and we may
thus invoke the Fourier inversion theorem, which asserts that g(t) is given by the
integral
∫
1 ∞
iyt
g(t) = e F(x, y)dy, t > 0.
2π
−∞
This leads to the representation for f,
∫ ∞
1
xt iyt
f(t) = e e F(x, y)dy, t > 0 (4.2)
2π
−∞
Transforming (4.2) back to the variable s = x + iy, since x > α is fixed, we have
dy = (1/i)ds and so f is given by

∫ x+i∞ ∫ x+iy
1
ts
ts
f(t) = e F(s)ds = lim e F(s)ds. (4.3)
2πi y→∞
x−i∞ x−iy
Here the integration is to be performed along a vertical line at x > α (Figure 4.1).
The expression (4.3) is known as the complex (or Fourier–Mellin)

inversion formula, and the vertical line at x is known as the Bromwich line.

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