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 $$ \mathcal{L}[f](\omega) := \int_{-\infty}^{+\infty} dx \, e^{-x} \, f(x, \omega) = \frac{ e^{-\omega} }{1 + \alpha} ~. $$

  • 1
    $\begingroup$ 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. $\endgroup$ Commented Mar 26, 2016 at 5:25
  • $\begingroup$ @RichardZhang : The integrand is not oscillatory here, as you can see, for instance, in the example provided. $\endgroup$
    – AlQuemist
    Commented Mar 26, 2016 at 17:07
  • $\begingroup$ 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. $\endgroup$ Commented Mar 27, 2016 at 14:06
  • $\begingroup$ @RichardZhang: The integrands are only available numerically (obtained from another numerical procedure). No analytical expression exists for them. $\endgroup$
    – AlQuemist
    Commented Mar 27, 2016 at 15:19
  • $\begingroup$ 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 $\tilde{F}(x)$? $\endgroup$ Commented Mar 27, 2016 at 15:25

1 Answer 1


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.


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