Probablity of Cauchy jump between two position I have some doubts about how to calculate the probability $P(x,t)$ of finding a particle with a certain initial uniform distribution $ P(x,0)=\rho (x) $ and typical displacement $ x^*=Dt $.
My idea was compute this integral:
$$\int dx_0 \int dx \rho(x) P(x-x_0,t) x\; x_0 $$
In particular the probability distribution i take into account is the Cauchy one-dimensional distribution.
 A: If the interval in which your particle can move is unbounded it is not possible to assume an initial uniform distribution. It sounds like you mean an initial Cauchy distribution.
The Cauchy distribution has the PDF
$$
P(x;x_0,\gamma)=\frac{\gamma/\pi}{(x-x_0)^2+\gamma^2}\,.
$$
This function has a maximum at $x_0$ and $2\gamma$ is the full width at half maximum:
$$
P_{max}=P(x_0;x_0,\gamma)=\frac{1}{\pi\gamma}\,,~~P(x_0\pm\gamma;x_0,\gamma)=\frac{1}{2}P_{max}\,.
$$
The Gaussian heat kernel is a one-parameter semigroup. The Cauchy distribution is a two-parameter convolution semigroup:
$$
\int_{-\infty}^{+\infty}P(x;x_0,\gamma)\,P(y-x;y_0,\delta)\,dx=P(y,x_0+y_0,\gamma+\delta)\,.
$$
This allows to think of the parameter $\gamma$ as being time $t\,.$ Let's assume the mean of your particle is at zero for all times: $x_0=y_0=0$ and drop those paremeters from the notation.
If the initial distribution was $\rho(x)=P(x;0)$ then, by the semigroup property, the distribution at time $t$ is simply
$$
\int_{-\infty}^{+\infty}P(x;0)\,P(y-x;t)\,dx=P(y,t)=\frac{t/\pi}{y^2+t^2}\,.
$$
The probability to find the particle at time $t$ in the interval $[a,b]$ is calculated with the known Cauchy CDF as
$$
\frac{\arctan(b/t)-\arctan(a/t)}{\pi}\,.
$$
