Explicit solution for equation The claim is that this equation has an explicit solution.
$$\frac{\partial}{\partial t}c(x,t)=\frac{a}{\pi}\int_{\mathbb{R}}\frac{c(y,t)-c(x,t)}{(y-x)^2}dy.$$
What can one do to find this solution? Fourier-transformation or similar transformations? Find the operator semigroup?
 A: This isn't totally rigorous.  The Fourier transform of $x\mapsto \frac1{x^2}$ is $\xi\mapsto -\sqrt{\frac \pi2}|\xi|$.  (Here I am using $\hat f(w) = \frac1{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) e^{i w t}\,dt$.)  So realizing that in your formula you need to subtract the $c(x,t)$ inside the integral to avoid an otherwise unworkable singularity at the origin, you see that the Fourier transform of your equation is
$$ \frac{\partial}{\partial t} \hat c(\xi,t) = -a \frac1{\sqrt{2\pi}} |\xi| \hat c(\xi,t) .$$
The solution to this is
$$ \hat c(\xi,t) = \exp\left(-a t \frac1{\sqrt{2\pi}} |\xi|\right) \hat c(\xi,0) .$$
So taking the inverse Fourier transform, we get the convolution
$$ c(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{2at}{a^2t^2  + 2\pi (y-x)^2} c(y,t) \, dy .$$
So, to make it semi-rigorous, we need to show that the inverse Fourier transform of $\xi \mapsto -|\xi|$ is the distribution $\mu$ which acts on test functions $\phi$ by
$$ \langle \mu,\phi\rangle = \sqrt{\frac 2\pi} \int_{-\infty}^\infty \frac{\phi(x) - \phi(0)}{x^2} \, dx.$$
Now, $$ -|\xi| = \lim_{a\to 0^+} \frac{e^{-a|\xi|} -1}a.$$  And the inverse Fourier transforms of $\xi\mapsto e^{-a|\xi|}$ and $1$ are $x\mapsto \sqrt{\frac 2\pi} \frac a{a^2+x^2}$ and $\frac1{\sqrt{2\pi}} \delta$ respectively.  So the inverse Fourier transform of $\xi \mapsto -|\xi|$ acting on the test function $\phi$ is the limit as $a\to 0^+$ of
$$ \frac1a \left( \sqrt{\frac 2\pi} \int_{-\infty}^\infty \frac a{a^2+x^2} \phi(x) \, dx - \frac1{\sqrt{2\pi}} \phi(0) \right) =
\sqrt{\frac 2\pi} \int_{-\infty}^\infty \frac {\phi(x) - \phi(0)}{a^2+x^2}  \, dx ,$$
which converges to the desired quantity.
Finally, it is worth noting that if $a = \sqrt{2\pi}$, then the formula for $c(x,t)$ is convolution next to the Poisson kernel, and $c(x,t)$ is a harmonic function of $x$ and $t$.  So the distribution $\mu$ is actually the negative square root of $-\frac{\partial^2}{\partial x^2}$.
