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I have a real function $q(t)$ which belongs to $L^1_2$, i.e., $\int^{\infty}_{-\infty}|q(x)|(1+x^2)dx<\infty$.

There is a claim that $\int^\infty_x\int^{x_n}_x...\int^{x_2}_{x}|q(x_1)||q(x_2)|...|q(x_n)|dx_1dx_2...dx_n=\frac{(\int^{\infty}_x|q(t)|dt)^n}{n!}$.

Again, I try this for two weeks...

My advisor showed me an easy case, but I've not caught the key to prove this claim...

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  • $\begingroup$ What is $\int_a^b\int_a^b f(x)f(y)\,dx\,dy$? $\endgroup$ Commented Dec 4, 2021 at 17:48
  • $\begingroup$ $F'(x)=f(x)$, $\int^b_a\int^b_af(x)f(y)dxdy=\int^b_a[F(b)-F(a)]f(y)dy=[F(b)-F(a)]^2$ $\endgroup$
    – xfireskyx
    Commented Dec 4, 2021 at 17:51
  • $\begingroup$ But you do not need an antiderivative to establish this. The same computation shows that it equals $\left(\int_a^b f(x)\,dx\right)^2$. $\endgroup$ Commented Dec 4, 2021 at 18:24
  • $\begingroup$ But in the case of the OP not all Integration boundaries are the same. $\endgroup$
    – PhoemueX
    Commented Dec 4, 2021 at 18:33
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    $\begingroup$ The OP has changed the problem since our discussion, without any warning that he changed it. That is totally crummy. Now the OP should change the order of integration. Write a new $2$-dimensional question to understand what's going on, but I'm not going to do it. $\endgroup$ Commented Dec 4, 2021 at 18:49

1 Answer 1

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Thanks for those comments.

Suddenly, the thought struck me that how to fix it and I thought it's great to give you guys the solution.

I consider that $m_j=\int^{x_j}_xq(t)dt$ (My advisor reminded me that $\int^x_xq(t)dt=0$) and notice $t$ is a dummy variable. $$ \frac{dm_j}{dx_1}=q(x_1)\ \Rightarrow\ dm_j=q(x_1)dx_1\\ \int^{x}_{x}q(x_1)dx_1=0\ \Rightarrow\ \int^{x_2}_xq(x_1)dx_1=\int^{m_2}_{0}dm_1\\ \int^{x_3}_x[\int^{m_2}_{0}dm_1]q(x_2)dx_2=\int^{m_3}_0\int^{m_2}_0dm_1dm_2 $$

And the remaining calculation is intuitive.

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