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The Complex Inversion Formula. Bromwich contour.

The Complex Inversion Formula. If L[F(t)] = f(s), then


Evaluating this formula provides a direct means for obtaining the inverse Laplace transform of a given function f(s). The primary device used in evaluating it is the Method of Residues of Complex Variable theory.

The integration in 1) is performed along the vertical line s = α in the complex plane where s = x + iy. The real number α is chosen so that the line s = α lies to the right of all singularities (branch points, poles, essential singularities, etc.). Otherwise the line is arbitrary.


The Method of Residues provides a means of computing the values of complex line integrals taken around simple closed curves in the complex plane. The technique is based on the Residue theorem:

Residue theorem. Let f(z) be analytic inside and on a simple closed curve C except at the isolated singularities a, b, c, ... inside C which have residues given by ar, br, cr ... . Then


See Fig. 1.

For information on methods and procedures for calculating the residues at the various singular points see Method of Residues.

The procedure followed in evaluating formula 1) above depends on whether the function f(s) does or does not contain branch points. We first consider the case in which f(s) contains no branch points.

Case 1. No branch points.


The Bromwich contour. The simple closed curve about which the integration is performed in evaluating formula 1) above is shown in Fig. 2 and is called the Bromwich contour. The curve C consists of two parts, C1 and C2, as shown in the figure. C1 is the portion of a circle of radius R, centered at the origin, shown in the figure. C2 is the vertical line AB located at a distance α to the right of the origin. Integration takes place in the counterclockwise direction on a limiting case of the curve shown in which the radius R is allowed to approach infinity. The integration corresponding to formula 1) takes place along the C2 portion of the curve. According to the residue theorem


            = sum of residues of all isolated singular points inside C

As R → ∞, the curve C will encompass all isolated singular points. In Fig. 2, points A and B have the complex coordinates α - iT and α + iT, respectively, where


Thus 2) becomes


            = sum of residues of all isolated singular points inside C

Now in certain circumstances the integral along curve C1 approaches zero as R → ∞. In such cases the integral along curve C2 ( i.e. the complex inversion integral) is equal to the sum of the residues of all isolated singular points inside C. The circumstances under which the integral along curve C1 approaches zero as R → ∞ is given by the following theorem:

Theorem 1. If we can find constants M > 0, k > 0 such that on C1 (where s = Re),


then the integral along C1 of est f(s) approaches zero as R → ∞, i.e.



It can be shown that the condition 4) always holds if f(s) = P(s)/Q(s) where P(s) and Q(s) are polynomials and the degree of P(s) is less than the degree of Q(s).

Example 1. Evaluate



by the method of residues.



The integrand has a simple pole at s = -1 and a double pole at s = 2. In addition, condition 4) of Theorem 1 is satisfied since the function is a quotient of two polynomials in which the degree of the numerator is less than that of the denominator. Consequently the integral along curve C1 of Fig. 2 is zero and the integral along C2 (the complex inversion integral) is equal to the sum of the residues of the two poles.

The residue at the simple pole s = -1 is


The residue at the double pole s = 2 is






Case 2. Branch points.


Modification of the Bromwich contour in case of branch points. The above procedure was based on the assumption that the function f(s) contained no branch points. If the function f(s) does contain branch points, a modification of the Bromwich contour must be made. In essence, a path of integration is then chosen that excludes the branch points. If, for example, f(s) has only one branch point located at s = 0, then we can use the contour shown in Fig. 3. In this case integrals along paths BDE and LNA will be zero if Theorem 1 is satisfied. However, integrals along paths EH, HJK, and KL will, in general, be non-zero and the integral about the entire contour will, by the residue theorem, be equal to the sum of the residues of all isolated singular points (poles, etc.) enclosed by the contour. By computing the residues and the integrals along paths EH, HJK, and KL we can obtain the value of the integral along path AB (i.e. the value of the complex inversion integral) by



Example 2. Find


by the complex inversion formula.

Solution. By the complex inversion formula, the required inverse Laplace transform is given by


Here the integrand contains a branch point at s = 0. It has no isolated singular points. We will use the Bromwich contour shown in Fig. 3. The integrand satisfies the conditions of Theorem 1 so that on taking the limit as R ole19.gif the integrals along BDE and LNA approach zero. Thus




We now evaluate the integrals along the paths EH, KL, and HJK.


Along EH,


and as s goes from -R to -ε, x goes from R to ε. Thus


Along KL,


and as s goes from -ε to -R, x goes from ε to R. Thus


Along HJK,

            s = εe



We thus obtain





Since the limit can be taken underneath the integral sign, we have


and so we obtain


This can be written as




  Murray R. Spiegel. Laplace Transforms. (Schaum)

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