I remember there was a tongue-in-cheek rule in mathematical analysis saying that to obtain the Fourier transform of a function $f(t)$, it is enough to get its Laplace transform $F(s)$, and replace $s$ by $j\omega$. Because their formula is pretty much the same except for the variable of integration.
And I know that this is not necessarily true (that's why I used the term tongue-in-cheek) e.g. take $f(t)=1$ to see the obvious difference.
I read somewhere that although their formula is somehow similar, their result is not necessarily similar because Laplace transform is a function and Fourier transform is a distribution. For example, the Dirac delta, $\delta(\omega)$ is a distribution and not a function.
So this led me to wonder:
- What is the difference between a function and a distribution? (preferably in layman's terms)
- Why the Laplace transform is a function and Fourier transform is a distribution? I mean, they are both infinite integrals. So what am I missing?
Best Answer
A distribution is also a function (mapping), but its "input" are also functions and not "numbers". This is what a distribution is in layman's terms. A precise definition would be rather complicated (basically the topology on the test functions is rather difficult to define).
To clarify: when people say that "$\delta$ is not a function", then they mean there is no function $\delta:\mathbf{R}\rightarrow\mathbf{R}$ such that $$\int_{-\infty}^\infty \delta(x)f(x)dx=f(0)$$ for all $f\in\mathcal{F}$ where $\mathcal{F}$ is a certain vector space of functions. By definition $\delta$ is the function $\delta:\mathcal{F}\rightarrow\mathbf{R}$ by $\delta(f)=f(0)$. So this "not is a function" relies on a more narrow interpretation of "function", i.e. that functions are those whose domain is (say) a set of real numbers.
I think you mean that the Laplace transform of $1$ and the Fourier transform of $1$, right? Well the integral defining the Fourier transform of $1$ does not converge!