Solved – the Fourier Transform of a brownian motion

I looked into this article 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.

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 $$ begin{aligned} int e^{i omega t}frac{d}{dt} f(t) dt &= -i omega e^{-iomega t}f(t)Big|_a^b – int -iomega e^{-i omega t}f(t) dt \ &= -i omega left(e^{-i omega b}f(b) – e^{-iomega a} f(a) right) + iomega int f(t) e^{-iomega t} dt\ &= -i omega left(e^{-i omega b}f(b) – e^{-iomega a} f(a) right) + i omega mathcal{F}left{ f(t) right}. end{aligned} $$

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.

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