I looked into this article http://en.wikipedia.org/wiki/Brownian_noise and it says that:
If we have a brownian motion $W(t) = \int _{0}^{t} dW(s)$, then given that the spectral density of white noise is constant
$S_0 = \left|\mathcal{F}\left[\frac{dW(t)}{dt}\right](\omega)\right|^2 = \text{const}$
Note that here $\mathcal{F}$ denotes the Fourier transform and $S_0$ is a constant. An important property of this transform is that the derivative of any distribution transforms as
$\mathcal{F}\left[\frac{dW(t)}{dt}\right](\omega) = i \omega \mathcal{F}[W(t)](\omega) $
from which we can conclude that the power spectrum of Brownian noise is
$S(\omega)= \left|\mathcal{F}[W(t)](\omega)\right|^2= \frac{S_0}{\omega^2}$
I don't understand this demonstration. Do you have a more detailed explanation or a link to a more detailed proof?
Thanks a lot for your help.
Best Answer
As mentioned above, the first equation about which you were confused is a property of the Fourier transform. Here is a very explicit derivation. First define the Fourier transform over a finite interval $(a,b)$ as $$ \mathcal{F}\left\{f(t)\right\} = \int_{(a,b)} f(t) e^{-i \omega t}\ dt. $$ With suitable technical considerations (if you care: that $f(t)$ is in the Sobolev space $W^{1,1}(a,b)$, which means that both $f$ and its derivative $f'$ are absolutely integrable over $(a,b)$) we can use our usual integration by parts formula: $\int u\ dv = uv|_{a}^b - \int v\ du$, where we will set $u = e^{-i \omega t}$ and $dv = f'(t) dt$. Then we have $$ \begin{aligned} \int e^{i \omega t}\frac{d}{dt} f(t)\ dt &= -i \omega e^{-i\omega t}f(t)\Big|_a^b - \int -i\omega e^{-i \omega t}f(t)\ dt \\ &= -i \omega \left(e^{-i \omega b}f(b) - e^{-i\omega a} f(a) \right) + i\omega \int f(t) e^{-i\omega t}\ dt\\ &= -i \omega \left(e^{-i \omega b}f(b) - e^{-i\omega a} f(a) \right) + i \omega \mathcal{F}\left\{ f(t) \right\}. \end{aligned} $$
If your function $f$ is well-behaved enough (if you somehow define a sequence of functions $f_n$ that converge to a limiting function and agree with $f$ on $(a,b)$, and if you can find a function $g$ so that, for any sequence of intervals $I_n$ converging to $\mathbb{R}$ you have $|f_n| \leq g$ for all $n$) then the constant term above cancels and you have the desired result.
All the formality seems a little contrived, and indeed from the point of view of a physicist it is a little bit---we just do this and don't worry about the formality. But if you want to get into the details, they are important. Suppose you were trying to do this with a random function: the Wiener process $\mathcal{W}(t)$, for example. All right, since $\mathcal{W}(t)$ is a.s. continuous everywhere then we can a.s. Riemann-Stieltjes integrate it as above. But its "derivative" is not well-defined in the traditional sense since $\mathcal{W}(t)$ is a.s. differentiable nowhere! Oops.
Everything does end up working out, and so even though the derivation you gave above is, technically-speaking, wrong, since $\frac{d}{dt}\mathcal{W}(t)$ is not defined, it is morally correct and, in physics, we use it all the time.