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A Laplace Transform is based on the integral:

$F(\xi) = \int_0^{\infty} f(x) e^ {-\xi x}\,dx.$

In a roundabout way, a Fourier transform can get to $\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx,$
Also, they both seem to use convolutions and transposes in "indirect" forms of "multiplication."

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Your formula for the Laplace transform is wrong. It should be $F(\xi) = \int_0^\infty f(x) e^{-\xi x}\, dx$. But yes, when $\xi$ is imaginary you have (up to normalization) the Fourier transform of $f$ (considered as a function on $(-\infty, \infty)$ which is 0 on $(-\infty, 0)$).

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Fixed the Laplace transform per your text. –  Tom Au Aug 8 '11 at 16:53
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@Robert: Could you elaborate? I've often wondered why these two transforms seem to be treated as quite separate things when there's an obvious connection between them. –  joriki Aug 8 '11 at 17:29
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As Robert said. But the properties of the two transforms are so different that (to me) the answer to the question in your title is a resounding: No. –  Did Aug 8 '11 at 17:31
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For example, the Bromwich integral for inverting the Laplace transform is really an inverse Fourier transform. See chapter 9 of Dettman, "Applied Complex Variables" books.google.ca/books?id=vmZ6PVtaexwC –  Robert Israel Aug 8 '11 at 18:27
    
See also the Paley-Wiener theorems (e.g. Rudin, "Real and Complex Analysis", sec. 19.1 and 19.2). Remarkably, Rudin does not mention the name Laplace. –  Robert Israel Aug 10 '11 at 7:52
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