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?