The Cauchy reduction formula is
$$\int_{0}^{x}{\cdots{\int_{0}^{x_{3}}{\int_{0}^{x_{2}}{F(x_1)\,dx_1\cdots dx_n}}}}
=\frac{1}{(n-1)!}{\int_{0}^{x}{(x-t)^{n-1}F(t)\,dt}}$$
Consider a generalisation
$$\int_{0}^{x}{\cdots\alpha (x_3)~{\int_{0}^{x_{3}}\alpha (x_2)~{\int_{0}^{x_{2}}{\alpha (x_1)~F(x_1)\,dx_1\cdots dx_n}}}}$$
where $\alpha(x)= {\sqrt g}$ is a measure.
I wonder if the corresponding formula has been discussed in the literature?
Best Answer
Assume that $\sigma(x)=\int_0^x \alpha(x')\,dx'$ is invertible on $\mathbb{R}^+$, with inverse $\sigma^{-1}(x)$, then the following formula reduces the $n$-fold integration to a single integration: \begin{align} & \int_0^x \cdots\alpha (x_3)~\int_0^{x_3} \alpha (x_2)~\int_0^{x_2} \alpha (x_1)~F(x_1)\,dx_1\cdots dx_n \\[8pt] = {} & \frac{1}{(n-1)!} \int_0^{\sigma(x)} (\sigma(x)-t)^{n-1} F\bigl(\sigma^{-1}(t)\bigl)\,dt. \end{align} For $\alpha(x)=1$, $\sigma(x)=x=\sigma^{-1}(x)$ you recover Cauchy's formula.
The OP requests a test of the formula, for two examples:
example 1: $$F(x)=x,\;\;\alpha(x)=x^2,\;\;\sigma(x)=x^3/3,\;\;\sigma^{-1}(x)=(3x)^{1/3},$$ $$\int_0^x \alpha(x_3)dx_3\int_0^{x_3}\alpha(x_2)dx_2\int_0^{x_2}\alpha(x_1)dx_1\,F(x_1)=x^{10}/280,$$ which agrees with $$\frac{1}{(n-1)!} \int_0^{\sigma(x)} (\sigma(x)-t)^{n-1} F\bigl(\sigma^{-1}(t)\bigl)\,dt=x^{10}/280,\;\;\text{for}\;\;n=3.$$
example 2: $$F(x)=x,\;\;\alpha(x)=1/(1-x),\;\;\sigma(x)=-\ln(1-x),\;\;\sigma^{-1}(x)=1-e^{-x},$$ $$\int_0^x \alpha(x_3)dx_3\int_0^{x_3}\alpha(x_2)dx_2\int_0^{x_2}\alpha(x_1)dx_1\,F(x_1)=-x-\tfrac{1}{6}\ln(1-x)\bigl(6+3\ln(1-x)+\ln^2(1-x)\bigr)$$ $$\qquad=\frac{1}{(n-1)!} \int_0^{\sigma(x)} (\sigma(x)-t)^{n-1} F\bigl(\sigma^{-1}(t)\bigl)\,dt,\;\;\text{for}\;\;n=3,\;\;0<x<1.$$