# Try to convert a n-fold integral into a integral to the power of n

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...

• What is $\int_a^b\int_a^b f(x)f(y)\,dx\,dy$? Commented Dec 4, 2021 at 17:48
• $F'(x)=f(x)$, $\int^b_a\int^b_af(x)f(y)dxdy=\int^b_a[F(b)-F(a)]f(y)dy=[F(b)-F(a)]^2$ Commented Dec 4, 2021 at 17:51
• But you do not need an antiderivative to establish this. The same computation shows that it equals $\left(\int_a^b f(x)\,dx\right)^2$. Commented Dec 4, 2021 at 18:24
• But in the case of the OP not all Integration boundaries are the same. Commented Dec 4, 2021 at 18:33
• 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. Commented Dec 4, 2021 at 18:49

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. $$\frac{dm_j}{dx_1}=q(x_1)\ \Rightarrow\ dm_j=q(x_1)dx_1\\ \int^{x}_{x}q(x_1)dx_1=0\ \Rightarrow\ \int^{x_2}_xq(x_1)dx_1=\int^{m_2}_{0}dm_1\\ \int^{x_3}_x[\int^{m_2}_{0}dm_1]q(x_2)dx_2=\int^{m_3}_0\int^{m_2}_0dm_1dm_2$$