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

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1. Feb 17, 2017

### cg78ithaca

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?