The Cauchy formula for repeated integration states that for $f$ a continuous function on $\mathbb{R}$, we have: $$\int_a^x\int_a^{x_1}\cdots\int_a^{x_{n-1}}f(x_n)\,dx_n\cdots\,dx_{2}\,dx_1=\frac{1}{(n-1)!}\int_a^x(x-t)^{n-1}f(t)\,dt.$$ A natural generalization is to consider "partial convolutions" of $f$ with a bounded, continuous function $\tau$: $$S_\tau f(x)=\int_a^xf(s)\tau(x-s)\,ds.$$ In the Cauchy formula above we can take $\tau\equiv1$. Suppose we iteratively apply $S_\tau$ to a function $f$ to obtain a sequence of integrals: $$S_\tau^nf(x)=\int_a^x\tau(x-x_1)\int_a^{x_1}\tau(x_1-x_2)\cdots\int_a^{x_{n-1}}\tau(x_{n-1}-x_n)f(x_n)\,dx_n\cdots dx_2\,dx_1.$$ Can we obtain a generalized Cauchy formula for such repeated integrals?
One guess for what a repeated integration formula might look like is: $$S_\tau^nf(x)\stackrel{?}{=}\int_a^x\tau^{(1-n)}(x-t)f(t)\,dt,$$ where $\tau^{(-n)}$ is the $n^\text{th}$ antiderivative of $\tau$. This agrees with the usual Cauchy formula in the case $\tau\equiv1$; however, it cannot be correct. Take $f\equiv1$, $\tau(x)=x$, and $a=0$. Then: $$S_\tau^2f(x)=\int_0^x(x-x_1)\int_0^{x_1}(x-x_2)\,dx_2\,dx_1=\frac{x^4}{24},$$ but: $$\int_0^x\tau^{(-1)}(x-t)\,dt=\frac{1}{2}\int_0^x(x-t)\,^2dt=\frac{x^3}{6}.$$ Thus we would need a more refined guess for what such a formula might look like.
The substitution $(x_1,\dots,x_{n-1},x_n)\mapsto(t_1,\dots,t_{n-1},s)$ in the integral for $S_\tau^n f(x)$, where $t_k=x_k-x_{k+1}$ for $1\leqslant k\leqslant n-1$, and $s=x_n$, gives $S_\tau^n f(x)=\int_a^x f(s)\tau_n(x-s)\,ds$, where $$\tau_n(t)=\idotsint\limits_{\substack{t_1,\dots,t_{n-1}\geqslant 0\\t_1+\dots+t_{n-1}\leqslant t}}\tau(t_1)\cdots\tau(t_{n-1})\tau(t-t_1-\dots-t_{n-1})\,dt_1\cdots dt_{n-1}$$ (I'm assuming $x\geqslant a$; the Jacobian of the substitution is easily seen to be $1$).
The case $\tau\equiv1$ of course corresponds to $\tau_n(t)=t^{n-1}/(n-1)!$; more generally, for $\tau(t)=t^{\alpha-1}$ with $\alpha>0$, one obtains $\tau_n(t)=t^{n\alpha-1}\Gamma^n(\alpha)/\Gamma(n\alpha)$; see e.g. [$1$], [$2$], [$3$].