Proving an integral of products of a function is equal to a regular integral.

Question

Given $f \in C([a,b])$, we also define $V= \{x\in \mathbb{R}^n | a\leq x_1 \leq x_2 \leq \cdots \ x_n \leq b \} .$

Now, I want to solve the following equation:

$$ \int_V \prod_{1\leq j \leq n} f(x_j) dx= \frac{1}{n!}\left( \int_a^b f(t)dt)\right). $$

I have shown this to be true for $n=1$ and $n=2$, and so I thought that perhaps I should use induction to prove the whole equation to be true. However, I wasn't able to figure out how to apply the inductive step to this integral. Any help in solving this would be greatly appreciated.

Answer 1

I'm afraid its just not true. Pick $f=1$ and $b=1$ and $a=-1$, then $$\int_a^bf(t)\, dt= 2 \mbox{ and } \int_V1dx=(b-a)^n=2^n.$$ Hence $$2^n=\frac{2}{n!},$$ which is not true. What you may want to look into is Fubini's Theorem. It yields $$\int_V\Pi_{j}f(x_j)dx=\Pi_j\int_a^b f(t)\, dt= (\int_a^bf(t)\, dt)^n$$

Answer 2

If I understand your notation, it is not a true identity, for $n=2$, $f(x)=x$, $a=0$, $b=1$ $$ \int_0^1 \int_0^x xy\,dy \,dx=\frac1{2!}\int_0^1 x^3\,dx=\frac1{2!4}\neq \frac 1{2!2} = \frac 1{2!}\int_0^1 x\,dx. $$