# Numerical computation of two-sided (bilateral) Laplace transform

I need to compute the two-sided (bilateral) Laplace transform of a numerically given function $F$, $$I(t) = \int_{-\infty}^{+\infty} {dx} \, e^{-x} \, F(x + t) ~,$$ where $F(x)$ has some sharp features, e.g., at $x = \{ 0, x_p, \cdots \}$, has finite support, and vanishes sufficiently fast for $x < 0$ to prevent a divergence of the integral.

Due to the sharp features, I have to divide the integration region to some subregions, for instance, \begin{align} I(t) &= \int_{-\infty}^{-\Delta} {dx} \, e^{-x} \, F(x + t) + \int_{-\Delta}^{+\Delta} {dx} \, e^{-x} \, F(x + t) \\ &\quad + \int_{+\Delta}^{x_p - \Delta} {dx} \, e^{-x} \, F(x + t) + \int_{x_p - \Delta}^{x_p + \Delta} {dx} \, e^{-x} \, F(x + t) \\ &\quad + \int_{x_p + \Delta}^{+\infty} {dx} \, e^{-x} \, F(x + t) ~, \end{align} where $\Delta > 0$.

I'd like to know the best and most precise way to perform such an integral numerically (quadrature methods). Is there something like Gauss-Laguerre quadrature for such an integration?

I have also considered a transformation of the integration variable, $$u = e^{-x} ~,$$ to absorb the exponential factor, $$I(t) = \int_{0^+ = \, u(x \rightarrow +\infty)}^{+\infty = \, u(x \rightarrow -\infty)} {du} \, F(-\ln(u) + t) ~,$$ and performing the transformed integration by a Tanh-Sinh quadrature; yet I am not sure if this is the best method.

As an example, one can take $$f(x,\omega) = e^{-\alpha (x - \omega)} \, \Theta(x - \omega)$$ where $\Theta$ denotes the Heaviside step function and $\alpha > 0$. Hence, the ‘width’ of the peak at $\omega$ can be varied by $\alpha$. The two-sided Laplace transform of $f$ will be $$\mathcal{L}[f](\omega) := \int_{-\infty}^{+\infty} dx \, e^{-x} \, f(x, \omega) = \frac{ e^{-\omega} }{1 + \alpha} ~.$$

• The need to integrate high oscillatory kernels arises frequently in the evaluation of Sommerfeld integrals for dyadic Green's functions. The usual practice is to do a (numerical) rational fit of $F(x)$ as a collection of poles $\sum_i 1/(x+c_i)$ and to use a closed-form expression to convert the poles into the inverse transform. – Richard Zhang Mar 26 '16 at 5:25
• @RichardZhang : The integrand is not oscillatory here, as you can see, for instance, in the example provided. – AlQuemist Mar 26 '16 at 17:07
• My apologies, the intention of my comment wasn't to suggest that your kernel is oscillatory, only that difficult intergands can often be treated using a semi-analytical approach of rational fitting + closed-form transformation formulas. – Richard Zhang Mar 27 '16 at 14:06
• @RichardZhang: The integrands are only available numerically (obtained from another numerical procedure). No analytical expression exists for them. – AlQuemist Mar 27 '16 at 15:19
• Do you agree that a numerically evaluated $F(x)$ can be approximated to arbitrary accuracy (under regularity assumptions) as a rational function, $\tilde{F}(x)=\sum_i 1/(x+c_i)$? And do you agree that the bilateral Laplace transform has a closed form expression for $\tilde{F}(x)$? – Richard Zhang Mar 27 '16 at 15:25

With the substitution $$y=x+t$$, the $$t$$-dependency can be factored out:
$$I(t)=\int_{-\infty}^\infty e^{-(y-t)}F(y)\;dy = e^t \int_{-\infty}^\infty e^{-y}F(y)\;dy = C\cdot e^t$$
This still leaves you with the problem of computing $$C$$, but it is merely a constant number and the $$t$$-dependency is entirely described by the factor $$e^t$$. The two-sided Laplace transfrom of this function and its region of convergence (if it is non-empty) should be found in a table of Laplace transforms.
Edit: On a second thought, I doubt that the two-sided Laplace transform of $$I(t)$$ exists even for purely imaginary $$s$$, because $$I(t)\sim e^t$$ is not a tempered distribution.