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Re: numerical inversion of laplace transform

Your convolution expression
   G[t] = Integrate[F[t-x], {x,0,a}]
is a smoothing function. Inverting a smoothing operation is always
tricky, except if you have an analytic expression of what you're
trying to un-smooth. Since you do have such an expression, here's a
possible procedure. I'm glossing over any pathological subtleties,
assuming everything is reasonably behaved.

1) Fourier transform F(t) and G(t) formally:
   F[w] = Integrate[F[t]*Exp[i*w*t], {t,-Infinity,Infinity}]
   G[w] = Integrate[G[t]*Exp[i*w*t], {t,-Infinity,Infinity}]

2) formally express the convolution in terms of the Fourier transform:
   G[w] = F[w] * i * (1-Exp[i*w*a])/w

3) formally solve for F(w):
   F[w] = G[w] * (-i*w)/(1-Exp[i*w*a])

4) find the Fourier transform G(w) from your expression of G(t)

5) find F(w) from your expression of G(w) and the formula at 3)

6a) invert the Fourier transform:
   F[t] = Integrate[F[w]*Exp[-i*w*t], {t,-Infinity,Infinity}] / (2*Pi)

- or -

6b) since you say that your F(t) must have periodicity T=150,
inverting the Fourier transform should involve only those frequencies
that are multiples of 2*Pi/T:
   F[t] = Sum[F[w]*Exp[-i*w*t] /. w->n*2*Pi/T, {n,-Infinity,Infinity}]
(maybe missing some normalization factors here)

If everything goes as planned, then 6a) and 6b) should give the same
answer. This is clear: if F(t) has a certain periodicity T, then the
smoothed function G(t) must have this same periodicity, and its
Fourier transform G(w) will only have nonzero values for w any integer
multiple of 2*Pi/T.


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