Let $\mathcal{T}$ be a (pseudo-)differential operator that admits the following kernel representation:
\begin{equation} \mathcal{T}f(x) = \int_{-\infty}^{\infty} K(x,t)f(t)dt. \end{equation}
What can be said about the kernel representation of $\mathcal{T}^{n} f(x)$ for $n \geq 2$? Is there a better way of computing the kernel for $\mathcal{T}^{n}$ than the following approach:
For instance the action of $\mathcal{T}^{2}$ can be computed as:
\begin{equation} \mathcal{T}^2f(x)=\mathcal{T}\mathcal{T}f(x)= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} K(x,t_{1})K(t_{2},t_{1})f(t_{2})dt_{1}dt_{2}. \end{equation}
This approach gets really complicated if $n$ is a large number.
Edit: It is suggested in the comment section to be more specific. I am concerned with the translation-variant kernels in general. But, as a specific example, consider $K(x,t)=(1+x-t)e^{-\frac{(x-t)^2}{2}}$.
Thank you for your help.