Cauchy Repeated Integral Formula with Root upper-bounds? Cauchy's formula for repeated integration states that for any continuous function on $[0,1]$ we have that the $n$-fold integral can be represented by a single integral as follows
$$
\int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1
=
\frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.
$$
Following this question, I'm wondering if there is a "known" analogue of the formula for the following variant
$$
\int_a^{\sqrt{x}} \int_a^{\sqrt{\sigma_1}} \cdots \int_a^{\sqrt{\sigma_{n-1}}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1
= \int_a^x k(t,x,a) f(t)dt,
$$
for some locally-integrable function $k(t,x,a)\in L_{loc}^1(\mathbb{R}^3)$?, where $\sigma_1\leq ...\leq \sigma_{n-1}\leq x$.
 A: Assuming throughout that $a=0.$
In $[0,x]^{n-1},$ let $$S_{x,t}=\{(x_1,x_2,\dots,x_{n-1})\mid x\geq x_1^2\geq x_2^4\geq\dots \geq x_{n-1}^{2^{n-1}}\geq t^{2^n}\}.$$
Take $k(x,t)=\mu\left(S_{x,t}\right),$ the hyper-volume of $S_{x,t}.$ Then $k(x,t)$ works.
In particular, if $t^{2^n}\geq x,$ then $k(x,t)=0.$
I'm not sure what $k(x,t)$ is, in general. When $n=1,$ $k(x,t)=1$ when $t^2<x$ and $0$ otherwise.
When $n=2,$ then $S_{x,t}=\{x_1\mid x\geq x_1^2\geq t^4\}=[t^2,\sqrt{x}].$ So then $$k(x,t)=\begin{cases}\sqrt{x}-t^2&t^2<\sqrt{x}\\0&\text{otherwise}\end{cases}$$ 
When $n=3,$ I get $$k(x,t)=\frac{2}{3}x^{3/4}-x^{1/2}t^2+\frac{t^6}{3}=\frac{1}{3}\left(x^{1/4}-t^2\right)^2(2x^{1/4}+t^2),$$ but I'm not sure that is correct. It does have the necessary condition $h\left(t^8,t\right)=0.$ 
It might be generally true that $k(x,t)$ is divisible by $(x^{1/2^{n-1}}-t^2)^{n-1}.$
Note, these won't work when $x<1,$ since then $x^{1/2^n}>x,$ so the left side of will depend on values of $f$ outside $[0,x].$ You really do need to just change the right side to $$\int_{0}^{x^{1/2^n}}k(x,t)f(t)\,dt.$$ This formulation will work in that all cases given our definition of $k(x,t).$

More generally, if $h:[a,\infty)\to[a,\infty)$ is a continuous bijection, then we define $h^{1}(x)=h(x)$ and $h^{k+1}(x)=h(h^k(x)).$ Then we can define for any $x,t\geq a$ the set:
$$S_{x,t}=\{(x_1,\cdots,x_{n-1})\mid h^n(x)\geq h^{n-1}(x_1)\geq\cdots\geq h^{1}(x_{n-1})\geq t\}$$ then define $k_h(x,t)=\mu(S_{x,t}).$ Then:
$$\int_{a}^{h(x)}\int_{a}^{h(\sigma_1)}\cdots \int_{a}^{h(\sigma_{n-1})} f(\sigma_n)\,d\sigma_n\dots d\sigma_1=\int_{0}^{h^n(x)} k_h(x,t)f(t)\,dt.$$
This is basically done by switching the order of the integrals, letting $t=\sigma_n$ then the left side is equal to:
$$\int_{a}^{h^n(x)}f(t)\left(\int_{h^{-1}(t)}^{h^{n-1}(x)}\int_{h^{-1}(\sigma_{n-1})}^{h^{n-2}(x)}\cdots \int_{h^{-1}(\sigma_2)}^{h(x)}1\,d\sigma_1\,d\sigma_{2}\cdots d\sigma_{n-1}\right)\,dt$$ where the inside integral is computing the hyper-volume of $S_{x,t}.$
Indeed, the inner integral was how I computed $k(x,t)$ in the case when $h(x)=\sqrt{x}.$
You do get a recursion based on $n$:
$$k_{n+1}(x,t)=\int_{h^{-1}(t)}^{h^n(x)}k_n(x,s)\,ds.$$

When $h(x)=x$ for all $x,$ then we have that $$S_{x,t}=\{(x_1,\dots,x_{n-1})\mid x\geq x_1\geq x_2\cdots \geq x_{n-1}\geq t\}.$$ Probabilistically, given a random element of $[t,x]^{n-1},$ the probability that a random element is sorted in descending order is $\frac{1}{(n-1)!}$ so we get $$\mu(S_{x,t})=\frac{1}{(n-1)}\mu\left([t,x]^{n-1}\right)=\frac{(x-t)^{n-1}}{(n-1)!}.$$
This retrieves Cauchy's original result.

Technically, I don't think you need $h:[a,+\infty)\to[a,+\infty)$ to be a bijection, just strictly increasing, perhaps with $h(a)=a.$ 

There is a discrete form of this. 
Assume $h:\mathbb N\to\mathbb N$ such that $h(0)=0$ and is (not necessarily strictly) monotonically increasing) then there is a function $k:\mathbb N^2\to\mathbb N$ so that: $$\sum_{i_1=0}^{h(m)}\sum_{i_2=0}^{h(i_1)}\cdots\sum_{i_n=0}^{h(i_{n-1}} f(i_{n}) = \sum_{i=0}^{h^n(m)}f(i)k(m,i)$$
And $k_n(m,i)$ can be expression in terms of counting the number of $n-1$-tuples $(x_1,x_2,\cdots,x_{n-1})$ of natural numbers such that $h^n(x)\geq h^{n-1}(x_1)\cdots \geq h(x_{n-1})\geq i.$
When $h(m)=m,$ you get that $k_n(m,i)=\binom{m-i+n-1}{n-1}.$
