- #1

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## Main Question or Discussion Point

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?

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?