## Heaviside, Laplace and the Inversion Integral

The main purpose is to explain the complex inversion integral for Laplace transforms

### Heaviside

Oliver Heaviside (1850-1925) was a self-taught genius in electrical engineering who made many important contributions in the field. However, he was best known to engineers for his operational calculus, a tool for solving linear differential equations with constant coefficients, which he discovered around the turn of the century and which was popularized by Steinmetz in the United States. These equations appear in connection with transient phenomena, which Heaviside studied extensively. It happens that the operational calculus is very similar to Laplace transform procedures, which have now replaced it in electrical engineering. Heaviside knew nothing of Laplace transforms, which were a mathematical obscurity in his time. The operational calculus was severely limited by its lack of mathematical theory, which not only limited its field of application, but also created many uncertainties and ambiguities. I'll begin here by trying to give the flavor of Heaviside's methods, which is worth tasting.

In transient phenomena, a voltage E is suddenly applied to a circuit at time t = 0, and we wish to find the current thereafter as a function of time. Heaviside introduced the unit step function, U(t), which is zero for t < 0 and 1 for t > 0. A step at some other time t' is represented by U(t - t'). The derivative of U(t) is the Dirac delta function, δ(t), which is zero everywhere except at t = 0, where it is infinite in such a way that its integral is unity. These strange functions are very widely used in physics and engineering today, and avoid much cumbersome notation.

Heaviside used the operator p to represent differentiation: p = d/dt. Then, 1/p must represent integration from 0 to t, since (1/p)p f(t) = f(t). Operators are treated like algebraic symbols, and manipulated by the usual rules--we have just had an example. Then, pU(t) = δ(t), and (1/p)U(t) = ∫(0,t)U(t)dt = t. To make it easy to write in HTML, the limits of the integral in parentheses follow the integral sign. By repeated integration, we see that (1/pn)U(t) = tn/n!. The operators change one time function into another. Let's now solve a problem. A resistance R and inductance L are connected in series, and a voltage EU(t) applied. The differential equation for the current is L(di/dt) + Ri = EU(t), with the initial condition i(0) = 0. This is just Kirchhoff's Voltage Law, equating the drops to the rise. Heaviside wrote pLi + Ri = EU(t), or i = E/(pL + R) U(t). Put R/L = k; then i = (E/L)(p + k)-1 U(t). Expansion by the binomial theorem gives (p + k)-1 = 1/p - k/p2 + k2/p3 - .... If we let this operate on U(t), we get (t - kt2/2! + k2t3/3! - ...). Then, i = (E/R)[kt - (kt)2/2! + (kt)3/3! - ...] = (E/R)[1 - e-kt], which is the correct answer, obtained purely by algebra. The time constant τ = 1/k = L/R.

It is possible to build up a table of the result of operating on U(t) by various functions of p, so that it is not necessary to go through the series expansions for the usual problems. For example, [p/(p - a)]U(t) = eat. This table is equivalent to the table of Laplace transforms that is now used for the same purpose. In Heaviside's calculus, there is no concept of transform pairs, however. The powerful tool of convolution is also absent.

### Laplace

Pierre-Simon, marquis de Laplace (1749-1827), lived almost an exact century before Heaviside. He was a brilliant mathematical theoretician, making fundamental contributions to celestial mechanics, probability theory, field theory and other areas. The famous integral transform carries his name, but seems to have been used first by Denis Poisson (1781-1840) in 1815.

The Laplace transform replaces one function F(t) of t by another f(s) of the new variable s by the rule: f(s) = ∫(0,∞) e-stF(t)dt. f(s) is called the transform of F(t): f(s) = L[F(t)]. An easy integration by parts gives the transform of the derivative of F(t): L[F'(t)] = sf(s) - F(0). If we replace s by p, and F(0) = 0, we recover Heaviside's expression. The transform of U(t) is 1/s, as can be found by doing the easy integral. As far as F(t) is concerned, the transform uses only the values for t > 0, just as in the operational calculus.

Returning to our transient problem, we transform the equation L(dI/dt) + RI = EU(t) to find Lsi(s) + Ri(s) = E/s. This gives i(s) = E/[s(Ls + R)]. To find I(t), we must find the inverse transform of the function of s on the right-hand side. In the elementary use of the Laplace transform, this is done as follows. First, we note that 1/[s(s + k)] = (1/k)[1/s - 1/(s + k)]. This is a partial fraction expansion (used also by Heaviside). We already recognize one term: 1/s = L[U(t)]. As for the other term, it is the transform of e-ktU(t), which can be found by direct integration. If L-1[1/s] = 1 and L-1[1/(s + k)] = e-kt, then L-1[1/s(s+k)] = (1 - ekt)/k, and we find that I(t) = (E/R)(1 - ekt). It is usual to drop the U(t) when using the Laplace transform. Comparing the Heaviside and Laplace methods shows how very similar they are in practice, if not in theory.

The Laplace method, however, has much more powerful mathematics behind it. This includes the convolution theorem, that states that L-1 [f(s)g(s)] = F(t)◊G(t) = ∫(0,t) F(t - τ)G(τ)dτ, the convolution of the two functions F(t) and G(t). The symbol ◊ is used here because the browser does not support the usual symbol, a cross in a circle. It also shows that Laplace transforms are closely related to Fourier transforms, so that similar properties exist (such as convolution). The Fourier transform has an inverse transformation, and the transformation between time and frequency domains is equally convenient both ways, involving a simple integral along the real axes. The Laplace transformation is quite troublesome to invert. There is a formula involving real integrals, and another with repeated differentiation (see Churchill and Widder), but these are of little use because the integrals are too difficult to perform in practical cases. The only practical inversion integral uses analytic functions of a complex variable, and will be discussed below.

Most problems in elementary applications of the Laplace transform can be solved by reference to tables of transform pairs. This was also the usual method for the Heaviside calculus as well. Churchill gives a table with 122 transforms that is extensive enough for most purposes. It is curious that the Laguerre and Hermite polynomials, as well as Bessel functions, have relatively simple transforms. Students like Laplace transform because it seems to be a mechanical method that can be used to find answers without thinking. This is an erroneous impression.

### The Inversion Integral

If you are not familiar with analytic functions, poles, branch cuts and the residue theorem, please refer to Complex Variables, where these concepts are explained. Since undergraduates (and many instructors) are not usually well-prepared along these lines, the inversion integral is normally omitted in electrical engineering courses that introduce the Laplace transform and rely on tables of transforms. The simplified account presented here may make inversion more comprehensible. Most proofs and much algebra have been omitted; the omissions can be made good with the references.

The Laplace transform f(s) of a time function F(t) is an analytic function of the complex variable s, except at its singularities. The Laplace integral converges on the real axis only for s > c, where c is some constant. It may converge everywhere, c = -∞, or not at all, c = ∞. If it converges over some of the real axis, then f(s) can be extended by analytic continuation over the whole complex plane. Depending on the properties of F(t), f(s) behaves in certain ways as |s| → ∞, and its singularities are in certain parts of the complex plane.

Since f(s) is analytic, it can be expressed by Cauchy's Integral Formula as f(s) = (1/2πi)∫(C) f(z)dz/(z-s), where C is the boundary of a region where f(s) is analytic that contains the point s. The inverse transform can now be taken, so that F(t) = (1/2πi)∫(C)etzf(z)dz. This formula is valid if part of C is a line parallel to the imaginary axis from λ - i∞ to λ + i∞, where the infinity arises from a limiting process. This line must be to the right of all the singularities of f(s). For the functions F(t) of interest, this is always possible when t > 0. This line can be moved and deformed at will, so long as it does not leave the region in which f(s) is analytic, because of the Cauchy-Goursat theorem. Think of the path of integration as a stretchable cord, and the singularities as rigid posts, the branch cuts as high walls. You can go anywhere with the path of integration that the cord could go. Note that no transform ever goes to infinity as a positive power of s, or else it would have a singularity at infinity, which it must not. F(t) is not restricted in this way. Even eat is allowed, in which case the Laplace integral converges for s > a. For t > 0, we can consider the path of integration as part of a closed curve C that is closed to the left, perhaps by a large circle of radius R that approaches infinity, as shown in the diagram. For t < 0, C must be closed to the right, and because f(s) is analytic in this region, F(t) = 0. If f(s) is meromorphic--that is, if its only singularities are poles, which will lie to the left of some Re(s) = λ, it can be inverted easily by the residue theorem. In fact, F(t) = Σ ρn, where the ρn are the residues of f(s) at its poles. In the diagram, small circles around these poles are connected with the large circle to form one continuous contour within which F(s) is analytic. In the limit, the small circles shrink to an infinitesimal radius. It is clear that f(s) must have singularities somewhere in the finite plane, or F(t) will be identically zero. The line s = λ lies to the right of all the singularities of f(s); it can be proved that this is always possible for well-behaved F(t)'s.

In our transient problem, f(s) has simple poles at s = 0 and s = -k, with residues 1/k and -e-kt/k, respectively, as we see from the partial fractions expansion. This gives F(t) = (E/R)(1 - e-kt), as before. A pole in the left half-plane is seen to correspond to an F(t) that decreases exponentially with t. We find nothing that we did not know before from more elementary means, but gain confidence in the correctness of the inversion integral, and a generally more lofty appreciation. The inversion integral is simply your friend, a complex contour integral, and nothing to be afraid of. For a more interesting problem, we present the example given by Churchill in his Sec. 61. Here, the transform to be inverted is f(s) = e-√s/s. This function has a pole at s = 0, as well as a branch point, because of the √s. To keep f(s) single-valued and analytic, a branch cut must be made from the origin to -∞ along the negative real axis. In choosing the curve C, we must avoid the origin and must not cross the branch cut. A suitable C is shown in the diagram. Everywhere within C, f(s) is single-valued and analytic. We take the limit as R → ∞, r' → 0 and the small angle ε → 0.

Since the net value taken around the closed contour is zero by Cauchy's Theorem, F(t) = (1/2πi)[∫(AB) + ∫(BC) + ∫(CD) + ∫(DD') + ∫(D'C') + ∫(C'B') + ∫(B'A')], in the limit. The integral in each case is ∫ets - √sds/s. The first two and last two contributions give zero in the limit, since they turn out to be inversely proportional to R. On DD', If s = re, then ds/s = iθ and ∫(DD') = -2πi, as r → 0. The limits are π to -π. On CD, √s = -i√r and on D'C', √s = +i√r, so the integrals are together equal to 2i∫(0,R)(e-trsin √r/r)dr. Phases differ by -1 on the upper and lower edges of the branch cut. With a new variable u = √r, and going to the limit R → ∞, the integral is 4i∫(0,∞) exp(tu2)(sin u/u)du. This integral is given in Dwight, 861.22, and its value is 4i(π/2)erf(1/2√t). Therefore, F(t) = 1 - erf(1/2√t) is the desired inverse transform. The function erf(x) = (2/√π)∫(0,x)exp(-u2)du is the error function. This transform is useful in certain heat conduction problems. Note how the contribution to the inverse transform came from the neighbourhoods of the pole and the branch cut, which is typical. Once you discover the proper path of integration, the problem is all but solved for you.

The complex inversion integral allows the attack of much more difficult problems with the Laplace transform than the usual ones where we really already know the answers, and are merely looking for a shortcut. We are cut loose from the table of transforms, and can discover new things. Every singularity of f(s) will contribute something to F(t), so we have a means of finding out interesting things about the answer, even if a complete solution is not available.

### References

E. J. Berg, Heaviside's Operational Calculus, 2nd ed. (New York: McGraw-Hill, 1936).

R. V. Churchill, Modern Operational Mathematics in Engineering, (New York: McGraw-Hill, 1944).

D. V. Widder, Advanced Calculus, 2nd ed. (New York: Dover, 1989). Chapters 13 and 14.

H. B. Dwight, Tables of Integrals and Other Mathematical Data, 4th ed. (New York: Macmillan, 1961).