The formula for the Inverse Laplace Transform is (Bromwich Intergal):

$$f_{(t)}=\frac{1}{2\pi i}\lim_{x\to\infty}\int_{\alpha-x i}^{\alpha+x i} \left(e^{st}\cdot F_{(s)}\right) \text{d}s$$

My questions are:

1) What is $\alpha$? Is $\alpha$ a real value? $\left(\alpha \in \mathbb{R}\right)$;

2) If $\alpha$ is real, than $\lim_{x\to\infty}\left(\alpha+x i\right)=\infty i$?


Now I know that $\alpha$ is a real number, now I would like to calculate:

$$\frac{1}{2\pi i}\lim_{x\to\infty}\int_{\alpha-x i}^{\alpha+x i} \left(e^{st}\cdot \frac{c}{s}\right) \text{d}s$$

And I am realy interested in the conditions, when the integral converges (if there are any)?!

Bromwich Integral: http://mathworld.wolfram.com/BromwichIntegral.html


To answer your second question, which largely seems to be about how to evaluate an integral of this type.

By the integral's domain, $s \in \{ \alpha + x'i : x' \in [-x, x]\}$

Set $s' = (s-\alpha)i = -x', \textrm{d}s' = i\textrm{d}s = -\textrm{d}x'$

Transforming the integral, you end up with \begin{equation} \underset{{x\rightarrow \infty}}{\lim} \textrm{$$} \frac{c}{2\pi i}\int_{-x}^x e^{-\alpha t}\cdot \frac{e^{-ix't}}{ix' + \alpha} i\textrm{d}x' \end{equation}

Run your good old Riemann integral treating $x$ as a constant, and then try to take the limit on the result. The values of $\alpha$ where this may or may not exist will pop out of that.


Yes, $\alpha$ is a real number, such that $\alpha + it$ is in the half-plane of convergence (strip if you are working with the two-sided Laplace transform).

The second question seems meaningless. What do you mean by $\infty i$?

  • $\begingroup$ If $\lim_{x\to\infty}\alpha+xi=\infty i$ than we can replace that in the intergral?! $\endgroup$ Sep 11 '15 at 19:57
  • 1
    $\begingroup$ No, you should integrate along a vertical line in the complex plane. $\endgroup$
    – mrf
    Sep 11 '15 at 19:59

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