How are the following integrals equivalent?

Question

Several textbooks I've read make the claim that for any $f: \mathbb{R} \to \mathbb{R}$

$$ \frac{1}{k!}\int_{0}^{\infty} x^k f(x) dx =\int_{[0,\infty)^{k+1}} f(x_1+x_2+\dots + x_{k+1}) \ dx_1 \dots dx_{k+1} $$

where we assume both sides of the equality are finite (e.g $f$ is continuous and compactly supported). I would preferably like to start with the left hand side (LHS) and obtain the right hand side (RHS) because the LHS is the "natural" integral to look at in the context of these textbooks.

However, given that this equality is true, it seems easier to start with the right hand side. (If you have a motivated proof that starts with the LHS and gets to the RHS I would love to see that).

Starting with the RHS, we can substitute $ x = x_1 + \dots x_k + x_{k+1}$ and have $dx = dx_k $. Therefore,

$$\int_{[0,\infty)^{k+1}} f(x_1+x_2+\dots + x_{k+1}) \ dx_1 \dots dx_{k+1} = \int_{[0,\infty)^k} \int_{x_1+\dots + x_k}^{\infty} f(x) \ dx \ dx_1 \dots dx_k$$

We are integrating over the set $\{ x \in \mathbb{R} \ \text{and}\ x_i \in \mathbb{R}| 0 \leq x_i \leq \infty \ \text{and} \ (x_1 + \dots + x_k) \leq x \leq \infty \} $

I was thinking of changing the order of integration because that worked in the case $k=1$, but I'm running into trouble with doing that for general $k$. Any ideas?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$This is a very pretty result, thank you for sharing it.

Theorem:

Let $f:\Bbb R\to\Bbb R$ be a measurable function that is integrable over $[0,\infty)$ and for which the below $n$-dimensional integral exists, for $n\in\Bbb N$. Then: $$\int_{[0,\infty)^n}f(x_1+x_2+\cdots+x_n)\d x_1\d x_2\cdots\d x_n=\color{red}{\frac{1}{(n-1)!}}\int_0^\infty x^{n-1}f(x)\d x$$

Notes: we don’t need compactly supported and smooth (a.k.a “bump”). However, every such function will satisfy the above hypotheses.

Proof:

By induction. Clearly the case $n=1$ is correct. Now take $n>1$ and inductively assume the result for $n\mapsto n-1$. Furthermore, the function will satisfy the Fubini theorem hypotheses, so we can break the integrals down however we wish. From your own work: $$J:=\int_{[0,\infty)^n}f(x_1+\cdots+x_n)=\int_{[0,\infty)^{n-1}}\int_{x_1+\cdots+x_{n-1}}^\infty f(x)\d x\d x_1\cdots\d x_{n-1}$$

Define $F:[0,\infty)\to\Bbb R$ by $F(x)=\int_x^\infty f(y)\d y$. We have: $$J=\int_{[0,\infty)^{n-1}}F(x_1+\cdots+x_{n-1})\d x_1\cdots\d x_{n-1}$$And $F$ satisfies the hypotheses for the case $(n-1)$. By induction and Fubini: $$J=\frac{1}{(n-2)!}\int_0^\infty x^{n-2}F(x)\d x=\frac{1}{(n-2)!}\int_0^\infty x^{n-2}\int_x^\infty f(y)\d y\d x$$Being careful with the regions of integration, we interchange: $$J=\frac{1}{(n-2)!}\int_0^\infty f(y)\int_0^y x^{n-2}\d x\d y=\frac{1}{(n-1)!}\int_0^\infty y^{n-1}f(y)\d y$$As desired!

Answer 2

The answer I upvoted helped me come up with an answer in the direction I wanted which also clarifies the need for a factorial mentioned in the comments. I post the answer below which is also done by induction.

The base case $k =0$ is clear. Suppose the result is true up to $k-1$.

\begin{align} \int_0^{\infty} \frac{x^kf(x)}{k!} \ dx & = \int_0^{\infty} f(x) \bigg(\int_0^{x} \frac{x_1^{k-1}}{(k-1)!} \ dx_1 \bigg) \ dx \\ & = \int_0^{\infty} \int_0^{x} \frac{x_1^{k-1}}{(k-1)!} f(x) \ dx_1 \ dx \\ \text{(By changing the order of integration)} & = \int_0^{\infty} \int_{x_1}^{\infty} \frac{x_1^{k-1}}{(k-1)!} f(x) \ dx \ dx_1 \\ & = \int_0^{\infty}\frac{x_1^{k-1}}{(k-1)!} \int_{x_1}^{\infty} f(x) \ dx \ dx_1 \\ \text{(By simply shifting $x$ by $x_1$)} & = \int_0^{\infty} \frac{x_1^{k-1}}{(k-1)!} \int_0^{\infty} f(x +x_1) \ dx \ dx_1 \\ & = \int_0^{\infty}\frac{x_1^{k-1}}{(k-1)!} F(x_1)\ dx_1 \end{align}

where $F(x_1) = \displaystyle{\int_0^{\infty} f(x+x_1) \ dx }$. By induction, this last integral is given by

\begin{align} \int_{[0,\infty)^k} F(x_2 + x_2 + \dots x_{k+1}) \ dx_2 \dots \ dx_{k+1} & = \int_{[0,\infty)^{k+1}} f(x+x_2+ \dots x_{k+1}) \ dx\ dx_2 \dots \ dx_{k+1}\\ \text{Relabeling variables} \ & = \int_{[0,\infty)^{k+1}} f(x_1+ \dots x_k+x_{k+1}) \ dx_1 \dots \ dx_{k+1}\end{align}