Applying the Bromwich integral to a cumulant generating function

In this note I want to quickly go through the motions of applying the Bromwich integral (i.e., inverse Laplace transform) to the cumulant generating function $\lambda(k)$ of an exponential variate $X$ with probability density function

$f(x) = \mu e^{-\mu x} \qquad \qquad \qquad (1)$

for $x \geq 0$ . The cumulant generating function is the log-Laplace transform of $f(x)$ , obtained as

$\lambda(k) = \ln \langle e^{kx} \rangle$

$= \ln\bigg(\int_0^{\infty} e^{kx} \mu e^{-\mu x} dx\bigg) = \ln\bigg(\frac{\mu}{\mu - k}\bigg) \qquad \qquad \qquad (2)$

Note that we need to assume $k < \mu$ for the integral to converge in this case.

We need to apply the Bromwich integral to the exponential of $\lambda(k)$ , i.e.,

$e^{\lambda(k)} = \frac{\mu}{\mu - k} \qquad \qquad \qquad (3)$

So, we have that taking the Laplace transform of the density function gives

$\mathcal{L}(\mu e^{-\mu x}) = \frac{\mu}{\mu - k} \qquad \qquad \qquad (4)$

How do we now apply the Bromwich integral to the right-hand side of (4) to deduce $\mu e^{-\mu x}$ ? One thing to notice is that this result is immediately obtainable from standard tables of Laplace transforms. For example, in most such tables one will find something like

$\mathcal{L}(e^{-at}) = \frac{1}{p+a} \qquad \qquad \qquad (5)$

where $p$ is the Laplace transform parameter such that

$F(p) = \mathcal{L}(f(t)) = \int_0^{\infty} f(t) e^{-pt} dt \qquad \qquad \qquad (6)$

Putting $a = \mu$ , $p = -k$ and $t = x$ in (5), we can then immediately deduce from (5) that

$\mathcal{L}^{-1} \bigg(\frac{\mu}{\mu-x}\bigg) = \mu \mathcal{L}^{-1} \bigg(\frac{1}{\mu-x}\bigg) = \mu e^{-\mu x} \qquad \qquad \qquad (7)$

Therefore, we will try to deduce our result by applying the Bromwich integral to the right-hand side of (5), where the Bromwich integral is given by

$f(t) = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} F(z) e^{zt} dz \qquad \qquad \qquad (8)$

for $t > 0$ . The basic procedure is simple. First, we convert $p$ on the right-hand side of (5) into a complex variable, say $p = x + iy \equiv z$ , so the function $F(z)$ in (8) is then

$F(z) = \frac{1}{z + a} \qquad \qquad \qquad (9)$

Then as long as $F(z)$ is of the form $F(z) = P(z)/Q(z)$ , where $P(z)$ and $Q(z)$ are polynomials such that the degree of $Q(z)$ is at least one greater than the degree of $P(z)$ , we can immediately conclude

$f(t) = \text{the sum of residues of } F(z) e^{zt} \text{ at all poles}$

This condition is certainly satisfied by (9), so all we need to do is work out the relevant residues. The function $\frac{e^{zt}}{z+a}$ has a pole of order 1 at $z = -a$ , so there is a single residue which can be obtained as

$\lim_{z \rightarrow -a} \bigg((z+a) \cdot \frac{e^{zt}}{(z+a)}\bigg) = e^{-at} \qquad \qquad \qquad (10)$

This then immediately confirms the result in (5), and hence our result as explained below (5). We could just finish here, but it is instructive to actually prove that this result is correct using contour integration.

We first construct a contour bearing in mind that we need the restriction $\lambda > k$ as per (1) above, which in the notation of (9) translates into the condition $Re z > -a$ . So, picking an arbitrary $c > -a$ , we have the following:

The contour $\Gamma$ is designed to enclose the singularity of the integrand in the contour integral

$\frac{1}{2 \pi i} \oint_{\Gamma} \frac{e^{zt}}{z+a} dz \qquad \qquad \qquad (11)$

We now split this contour integral into a straight part and an arc:

$\frac{1}{2 \pi i} \int_{c - iy}^{c + iy} \frac{e^{zt}}{z+a} dz + \frac{1}{2 \pi i} \int_{arc} \frac{e^{zt}}{z+a} dz = e^{-at} \qquad \qquad \qquad (12)$

We want to show that as $y \rightarrow \infty$ , the integral along the arc vanishes, and so

$\frac{1}{2 \pi i} \int_{c - i\infty}^{c + i\infty} \frac{e^{zt}}{z+a} dz = e^{-at} \qquad \qquad \qquad (13)$

as per the Bromwich integral. We use estimations, beginning by writing

$\int_{arc} \frac{e^{zt}}{z+a} dz \leq \pi y \sup_{arc} \bigg| \frac{e^{zt}}{z+a} \bigg| \qquad \qquad \qquad (14)$

We have along the arc that $|z| \geq y$ and also

$|z+a| = \sqrt{(z+a)(\bar{z}+a)}$

$= \sqrt{((x+a)+iy)((x+a)-iy)}$

$= \sqrt{(x+a)^2 + y^2}$

$\geq y$

In addition, for any $z$ on the arc, with $arg(z) = \phi$ , we must have $\cos(\phi) < 0$ since we are operating in the second and third quadrants, so