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

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  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Jul 11 at 16:58
  • $\begingroup$ 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 $$\int_{a}^{x^{1/2^n}} k(t,x,a)f(t)\,dt.$$ $\endgroup$ – Thomas Andrews Jul 11 at 17:07
  • $\begingroup$ 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). $\endgroup$ – N00ber Jul 11 at 17:10
  • $\begingroup$ Then you will definitely want $k(t,x,0)=0$ for $t>x^{1/2^n}.$ $\endgroup$ – Thomas Andrews Jul 11 at 17:13
  • $\begingroup$ I added a condition on the variables to make it clearer... $\endgroup$ – AIM_BLB Jul 11 at 17:24
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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 $k_n(m,i)=\binom{m-i+n-1}{n-1}.$

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    $\begingroup$ This is very interesting. Is this your result or is this part of a larger body of literature? $\endgroup$ – AIM_BLB Jul 12 at 0:56
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    $\begingroup$ I just worked it out right here. @AIM_BLB $\endgroup$ – Thomas Andrews Jul 12 at 1:16
  • $\begingroup$ This is a neat computation! $\endgroup$ – AIM_BLB Jul 12 at 1:17
  • $\begingroup$ @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 $\endgroup$ – N00ber Jul 12 at 3:46
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    $\begingroup$ No, because you need $k(x,t)=0$ when $h^n(x)=t.$ @N00ber $\endgroup$ – Thomas Andrews Jul 12 at 13:54

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