Are Laplace Transforms a Special Case of Fourier Transforms?

Question

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."

Answer 1

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 $(-\infty, 0)$).

Answer 2

According to this Youtube video. that is indeed the case that a Laplace transform is a generalized version of the Fourier transform.

Fourier series work well for functions that approach zero as t goes to $(-\infty, and, \infty)$. It does not work for functions that go to a large number as t goes to either $(-\infty, or, \infty)$. La place's solution for managing the second type of function is to multiply it by a negative exponential as a dampener, to transform the function to F(t). This transformation creates an F(t) that is amenable to Fourier methods.

Thus, a Laplace transform of f(t) is equivalent to the Fourier transform of F(t).