# Contour Deformation in the Laplace Inversion Formula

Following Szpankowski - Average Case Analysis of Algorithms on Sequences the exponential generating function of $$g(n)$$ which is thought to be analytic in $$n$$ is defined as $$G(z)=\sum_{n=0}^{\infty} g(n) \, \frac{z^n}{n!}$$ and can be written as a Laplace Inversion Integral $$G(z)=\frac{1}{2\pi i} \int_{\delta-i\infty}^{\delta+i\infty} g^{*}(s) \, \exp\left(ze^{s}\right) \, {\rm d}s$$ where $$\delta >0$$ and $$g^{*}(s)$$ is the Laplace transform of $$g(n)$$.

Szpankowski now claims that the line of integration can be replaced by some contour $${\cal L}_\epsilon$$ which is defined by $$y={\rm sgn}(y)\tan\left(\frac{\pi}{2}+\theta\right)(x-\epsilon) \tag{1}$$ where $$\theta,\epsilon$$ I think (he is using $$\theta$$ in the context of a linear cone opening around the x-axis with angle $$\theta<\pi/2$$) are just parameters here and $$z=x+iy$$.

He also supplies a plot of the image of $${\cal L}_{\epsilon}$$ for the specific case $$\epsilon=0.1$$ and says that its parametrization reads $$e^{-|t|/2+\epsilon} \begin{pmatrix} \cos(t) \\ \sin(t) \end{pmatrix} \qquad t\in (-\infty,\infty) \, .$$ Even though I can not relate this plot or parametrization to (1) in any way, I am also confused about how this contour arises from the original one.

Hoping for input.

• How is the complex integral you wrote an Inverse Laplace Transform? – Ron Gordon Oct 17 '18 at 14:31
• Are you talking about the fact the there is a factor $e^{ze^{s}}$ in the integrand and not $e^{zs}$ ? – Diger Oct 17 '18 at 16:55
• yes ${}{}{}{}{}{}$ – Ron Gordon Oct 17 '18 at 17:23
• Write $g(n)$ as an inverse Laplace transform, interchange integral and summation order and carry out the sum. – Diger Oct 17 '18 at 18:02