For instance, by introducing a weighting function $a(t)$ such that $$\langle U_\alpha(\vec{x},t)\rangle=\int^\infty_{-\infty}U_\alpha (\vec{x},t+s)a(s)ds$$ where $\int^\infty_{-\infty}a(t)dt=1$. If we choose the weighting function to be: $$a(t)=\frac{1}{\pi t}\sin\left(\frac{\pi t}{\tau}\right)$$ Then the above equation will average out those frequencies which are greater than $\omega=\pi/\tau$.
I am trying to understand this intuitively or mathematically but I can't find anywhere to start from. How does the integral above filter out the high frequency components?
As an illustration of how the filter works we can consider the function $U(x)=\frac{1}{\pi}e^{-\alpha x^2}\cos ax$ and integral $$I(a,b)=\int_{-\infty}^\infty U(x)\frac{\sin bx}{x}dx=\frac{1}{\pi}\int_{-\infty}^\infty e^{-\alpha x^2}\cos ax\frac{\sin bx}{x}dx$$ We consider $\alpha$ as a small parameter ($\alpha\ll a,b$), and the function $e^{-\alpha x^2}$ is a smooth envelope function ensuring the convergence of the integral at infinity. At $\alpha\to0 \,\,e^{-\alpha x^2}\to 1$.
Using $\frac{\sin bx}{x}=\frac{1}{2}\int_{-b}^b e^{itx}dt\,\,$ we can write $$I(a,b)=\frac{1}{2\pi}\Re\int_{-\infty}^\infty e^{-\alpha x^2+iax}\int_{-b}^b e^{itx}dt\, dx$$ Changing the order of integration, making full square in the exponent power and integrating over $x$ $$I(a,b)=\frac{1}{2\pi}\Re\int_{-b}^b dt\int_{-\infty}^\infty e^{-\alpha x^2+iax+itx} dx=\frac{1}{2\pi}\Re\int_{-b}^b dt\int_{-\infty}^\infty e^{-\alpha\big( x^2-ix(a+t)/\alpha-(a+t)^2/4\alpha^2+(a+t)^2/4\alpha^2\big)} dx$$ $$=\frac{1}{2\pi}\Re\int_{-b}^b dt\int_{-\infty}^\infty e^{-\alpha\big( x-i(a+t)/2\alpha\big)^2}e^{-(a+t)^2/4\alpha} dx$$ $$=\sqrt{\frac{1}{4\pi\alpha}}\int_{-b}^be^{-(a+t)^2/4\alpha}dt=\sqrt{\frac{1}{4\pi\alpha}}\int_{a-b}^{a+b}e^{-x^2/4\alpha}dx$$ At $\alpha\to 0$ the function $f(x)=\sqrt{\frac{1}{4\pi\alpha}}e^{-x^2/4\alpha}\,$ is almost zero everywhere, except for a narrow interval near $x=0$, where it has a sharp peak. It is easy to check that $\int_{-\infty}^{\infty}f(x)dx=1$, so at $\alpha\to 0\,\,f(x)\to \delta(x)$ - delta-function.
It follows that, up to exponentially small correction terms, $I(a,b)=1$, as soon as $a\in[-b, b]$, and is equal to zero otherwise.
Therefore, we have a filter that cuts out all oscillations with the frequencies higher than $b$. At $\alpha\ll1$ (but $\neq0$) the filter is not ideal, so it not only cuts out the higher frequencies, but also changes the amplitude of the oscillations with the frequencies near the boundary $b$ (at $a\approx b$) - the perfect step-function becomes blurry at the edges.
At $\alpha=0$ the filter is ideal, and $$I(a,b)=\frac{1}{\pi}\int_{-\infty}^{\infty}\cos ax\frac{\sin bx}{x}dx=1 \,\text{at} \,a\in[-b;b] \,\text {, and zero otherwise}$$ The last result can be obtained using the straightforward integration.