Usually, we can find the inverse of the Laplace transform by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.
Consider the piece-wise differentiable function that vanishes for . We can express the function by the complex Fourier representation of
for any value of the real constant , where the integral
exists. By multiplying both sides of first equation by and bringing it inside the first integral
With the substitution , where is a new, complex variable of integration,
The quantity inside the square brackets is the Laplace transform . Therefore, we can express in terms of its transform by the complex contour integral
Theorem (Bromwich’s integral). The following line integral runs along the line parallel to the imaginary axis and units to the right of it, the so-called Bromwich contour
Example. Let us invert
Solution. From Bromwich’s integral,
where is a semicircle of infinite radius in either the right or left half of the -plane and is the closed contour that includes and Bromwich’s contour.
Our first task is to choose an appropriate contour so that the integral along vanishes. By Jordan’s lemma, this requires a semicircle in the right half-plane if and a semicircle in the left half-plane if . Consequently, by considering these two separate cases, we force the second integral to zero and the inversion simply equals the closed contour.
Consider the case first. Because Bromwich’s contour lies to the right of any singularities, there are no singularities within the closed contour and .
Consider now the case . Within the closed contour in the left half-plane, there is a second-order pole at and a simple pole at . Therefore,
Taking our earlier results into account, the inverse equals
Source: Green’s Functions with Applications by Dean G. Duffy