The Bromwich inversion integral is used to find inverse Laplace Transforms of functions. It is a complicated piece of mathematics, and requires a good knowledge of contour integration to be applied. It is included here for completeness: in most practical cases inverse Laplace Transforms are looked up in tables, and are seldom worked out by hand.

#### Statement of result

Let f(t) be a function with Laplace Transform F(p). Then we have the relation

f(t) = 1/(2πi) integral(F(p)exp(pt)dp, p=γ-i∞...γ+i∞)

where γ is a large real number chosen so that the path of integration lies to the right of all the singularities of F(p).

#### Proof

Recall that if F(p) is the Laplace Transform of a function f(t), then

F(p) = integral(f(t)exp(-pt)dt, t=0...inf).

We assume f(t)=0 for t

integral(F(p)exp(pt)dp, p=γ-i∞...γ+i∞)
= integral(exp(pt)integral(f(τ)exp(-pτ)dτ, τ=-∞...∞)dp, p=γ-i∞...γ+i∞)
= i integral(exp(γt)exp(iy(t-τ))f(τ)exp(-γτ)dydτ, τ,y=-∞...∞)
= i integral(exp(iy(t-τ))f(τ)exp(γ(t-τ))dydτ, τ,y=-∞...∞).

We now make the substitution u=t-τ and define g(u)=exp(γu)f(t-u).

= i integral(exp(iyu)g(u)dudy, u,y=-∞...∞)
= i limit(integral(integral(exp(iyu)g(u)du, u=-∞...∞)dy, y=-R...R), R→∞)
= i limit(integral(integral(exp(iyu)dy, y=-R...R)g(u)du, u=-∞...∞), R→∞)
= i limit(integral(2sin(Ru)/u g(u)du, u=-∞...∞), R→∞).
= i limit(integral(2sin(Ru)/u g(u)du, u=-ε...ε) + O(1/R), R→∞).

We now make use of continuity of g(u).

= i limit(g(0)integral(2sin(Ru)/u du, u=-ε...ε) + O(1/R), R→∞)
= i limit(g(0)integral(2sin(s)/s ds, s=-Rε...Rε) + O(1/R), R→∞)
= 2i g(0) integral(sin(s)/s ds, s=-∞...∞).

The integral of sin(s)/s can be proved using contour integration to be π, which gives us the required result.

= 2πi g(0).

Hence

1/(2πi) integral(F(p)exp(pt)dp, p=γ-i∞...γ+i∞) = 1/(2πi) 2πi g(0) = f(t).

#### Example

We consider the simple example of f(t) = 1 for t>0. The Laplace Transform is given by

F(p) = integral(f(t)exp(-pt), p=0...∞)
= integral(exp(-pt), p=0...∞)
= 1/p.

We wish to verify the inversion integral. The only singularity of F(p) is when p=0, so set γ=1. We are therefore interested in the integral

1/(2πi) integral(exp(pt)/p dp, p=1-i∞...1+i∞).

Make the substitution p=1+iq to obtain

1/(2πi) integral(exp(t(1+iq))/(1+iq) idq, q=-∞...∞)
= 1/(2πi) integral(exp(t(1+iq))/(q-i) dq, q=-∞...∞).

We evaluate this using the techniques of contour integration. For t>0, we close the contour in the upper half plane with a large semicircle: by Jordan's lemma the contribution of this semicircle to the integral vanishes. Hence the integral is equal to the sum of the enclosed residues.

= 1/(2πi) 2πi residue(exp((1+iq)t)/(q-i), q=i)
= exp(0t)
= 1.

which is the expected result. For negative t we close the contour in the lower half plane. Since there are no enclosed singularities the integral reduces to zero. Hence we have recovered our original f(t).