White noise and Fourier transform I try to solve a Langevin equation in the Fourier space. My understanding of the white noise in the Fourier space seems to be wrong.
Suppose I have a particle with its time evolution of the position given by the stochastic differential equation.
$$
\frac{d f(t)}{dt}=g(f)+\xi(t)
$$
$f$ is the position, $\xi$ is the white noise, $g$ some function of the position, and $t$ the time.
The particle will move according to the differential equation but because of the noise, every time I redo the experiment I will observe a different trajectory. What is of interest is not the position of the particle at time $t_0$ but the average over a large number of realizations, all evaluated at time $t_0$.
I write this "average over all realization" with $<>$.
The noise, at any time $t$, is zero in average: $<\xi(t)>=0 \quad \forall ~t$.
This means: $$\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N}\xi_j(t)=0 \quad \forall ~t $$
with "$j$" the different realizations.
I want to Fourier transform the noise. Given one particular realization "$j$" of the noise $\xi_j(t)$ I can compute its Fourier transform:
$
\hat\xi_j(\omega)=\int \xi_j(t) e^{i\omega x}dt
$
Now, I want to compute the average over all realization:
$$
<\hat\xi(\omega)>=\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \hat\xi_j(\omega) \\
\quad \quad \quad \quad ~= \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \int \xi_j(t) e^{i\omega t}dt \\
\quad \quad \quad \quad ~= \int \left( \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum\limits_{j=1}^{N} \xi_j(t) \right) e^{i\omega t}dt \\
\quad \quad \quad \quad ~= \int <\xi(t)> e^{i\omega x}dt \\
\quad \quad \quad \quad ~= 0
$$
Which is wrong since the Fourier spectrum of a white noise is a constant function. What am I missing?
 A: The OP is correct in stating that the Fourier transform $$\xi(\omega) = \int\mathrm{d}t\, \mathrm{e}^{\mathrm{i}\omega t} \xi(t), $$
vanishes upon averaging over realisations, $\langle \xi(\omega)\rangle  = 0$, so long as we assume that the noise is also zero on average in the time domain, $\langle \xi(t)\rangle = 0 $. 
However, the noise is not only characterised by its first moment, but also by its auto-correlation function:
$$ \langle \xi(t) \xi(t^\prime)\rangle = \eta\delta(t - t^\prime).$$
This last equation characterises the fluctuations of $\xi(t)$ in time; the presence of the delta function on the right-hand side is what actually defines white noise. The Fourier transform of the auto-correlation function gives the power spectrum: how noise power is distributed over different frequencies. For white noise this clearly takes the constant value $\eta$ in frequency space (up to a choice of normalisation for the Fourier transform). This means that the fluctuations contain equal contributions from all frequencies, i.e. fast and slow fluctuations contribute equally. 

As a sidenote, it is worth mentioning that we could easily consider white noise with non-zero average $\langle \xi(t) \rangle = \xi_0.$ This simply means that the noise has a constant (i.e. non-random) component. In this case we have that $\langle\xi(\omega)\rangle = 2\pi\xi_0 \delta(\omega)$, and the white noise condition is
$$\langle \xi(t) \xi(t^\prime)\rangle  =\xi_0^2+ \eta\delta(t - t^\prime).$$
The choice $\xi_0 = 0$ is merely a convention that simplifies these expressions. We can always get zero-mean white noise by the shift $\xi(t) \to \xi(t) -\xi_0$.
