# 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 $$\sigma_1\leq ...\leq \sigma_{n-1}\leq x$$.

• It's a little odd, since if $\sigma_1\in [a,\sqrt{x}]$ then it is quite possible to have $\sqrt{\sigma_1}<a.$ That doesn't mean that it is not possible, just that it necessarily be done in a simple way. – Thomas Andrews Jul 11 at 16:58
• In particular, the value of the left side will depend on values of $f(t)$ outside $[a,x],$ so we can't find such a $k(t,x,a).$ But we might find a way to write it as $$\int_{a}^{x^{1/2^n}} k(t,x,a)f(t)\,dt.$$ – Thomas Andrews Jul 11 at 17:07
• Well, we can set $a=0$ (to be honest this is the only case I really care about)... in that case issue cannot arise (I believe). – N00ber Jul 11 at 17:10
• Then you will definitely want $k(t,x,0)=0$ for $t>x^{1/2^n}.$ – Thomas Andrews Jul 11 at 17:13
• I added a condition on the variables to make it clearer... – AIM_BLB Jul 11 at 17:24

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 $$t^2 and $$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 $$k_n(m,i)=\binom{m-i+n-1}{n-1}.$$

• This is very interesting. Is this your result or is this part of a larger body of literature? – AIM_BLB Jul 12 at 0:56
• I just worked it out right here. @AIM_BLB – Thomas Andrews Jul 12 at 1:16
• This is a neat computation! – AIM_BLB Jul 12 at 1:17
• @ThomasAndrews In your notation, if $h(x)=x^2$, is $\mu(S_{x,t})$ is equal to $(x-t)^{n-1}$ up to a factor of a constant? It seems like it should be in my intuition – N00ber Jul 12 at 3:46
• No, because you need $k(x,t)=0$ when $h^n(x)=t.$ @N00ber – Thomas Andrews Jul 12 at 13:54