Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Inverse Laplace transform of F(s)=exp(-as) as delta(t-a)

  1. Feb 17, 2017 #1
    This is mostly a procedural question regarding how to evaluate a Bromwich integral in a case that should be simple.

    I'm looking at determining the inverse Laplace transform of a simple exponential F(s)=exp(-as), a>0. It is known that in this case f(t) = delta(t-a). Using the Bromwich formula, with s = x+iw:
    $$
    2\pi f(t) = \int_{-\infty} ^\infty F(x+iw)exp(xt)exp(iwt)dw
    $$
    For this specific F(s):
    $$
    2\pi f(t) = \int_{-\infty} ^\infty exp((t-a)x)exp(iw(t-a))dw
    $$
    $$
    2\pi f(t) = exp((t-a)x) \int_{-\infty} ^\infty exp(iw(t-a))dw
    $$
    At t = a, this becomes:
    $$
    2\pi f(a) = \int_{-\infty} ^\infty dw
    $$
    Thus $$ f(a) -> \infty $$
    This is so far consistent with a delta(t-a) function.
    For all t <> a:
    $$
    \pi f(t) = lim_{T->\infty} exp((t-a)x) sin(T(t-a))/(t-a)
    $$
    which clearly does not converge rather than being zero as expected for f(t) = delta(t-a).

    Can anyone explain what I am doing wrong?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted