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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 $t \neq a$:

$$ \pi f(t) = lim_{T->\infty} \exp((t-a)x) \sin(T(t-a))/(t-a) $$

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?

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    I doubt anybody can read this, please take the time to improve the readability otherwise your question will get no attention if not closed soon. – 2017-02-16
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    Zaid, edited accordingly. – 2017-02-16
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    You can also use the slash '\' to create better looking syntax. For example $sin(x)$ becomes $\sin(x)$ and $lim$ becomes $\lim$ . – 2017-02-16
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    Another user made some very useful comments on this at https://www.physicsforums.com/threads/inverse-laplace-transform-of-f-s-exp-as-as-delta-t-a.904385/ – 2017-02-19

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