How does the following change of variables occur?

Question

While reading a book, the author goes from writing that $$I=\int_0^1 \dots \int_0^1(u_1\dots u_s)^{-1+1/k}\frac{\sin 2\pi \lambda(u_1+\dots+u_s-1)}{\pi(u_1+\dots u_s-1)}\mathrm{d}u_1\dots \mathrm{d}u_s$$ to $$I=\int_0^s \phi(u)\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)} \mathrm{d}u \dots(\star)$$ where $$\phi(u)=\int_0^1 \dots \int_0^1 \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k} \mathrm{d}u_1\dots \mathrm{d}u_{s-1}$$ and the region of integration is over $(u_1,\dots, u_s)$ such that $u-1<u_1+\dots u_{s-1}<u$.

I don't really understand what theorem is being used to do this change of variables. Clearly, they are taking $u=u_1+\dots u_s$. Then $u$ can be between $0$ and $(1+\dots + 1)=s$. We have the following change of coordinates function $(u_1,\dots,u_{s-1}, u_s)\mapsto (u_1,\dots, u_{s-1}, u-u_1-\dots - u_{s-1})$. But then our Jacobian matrix is given by

\begin{pmatrix} 1 & 0 &\dots 0 & 0\\ 0 & 1 & \dots 0 & 0\\ 0 & \ddots & 1 & 0\\ -1 & -1 & \dots -1 & 0 \end{pmatrix}

After that, I am not sure what's going on here and how the author gets to $(\star)$. Also, it seems so weird that now this is a integral of one variable (sort of) and I am quite confused. Thanks, any help is appreciated!

Answer 1

The only variable changed is $u_s$, which is replaced with $u-u_1-u_2-\dots-u_{s-1}$. The transformation involved is thus $$T(u_1,u_2,\dots u_{s-1},u)=(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})$$ and: $$DT(u_1,u_2,\dots u_{s-1},u)= \begin{pmatrix} 1 & 0 &\dots 0 & 0\\ 0 & 1 & \dots 0 & 0\\ 0 & \ddots & 1 & 0\\ -1 & -1 & \dots -1 & 1 \end{pmatrix}$$ whose determinat is equal to $1$. Changing variables yields: $$\int_{(0,1)^s}f=\int_{T^{-1}((0,1)^s)}(f\circ T)\cdot |\det DT|$$ where $f(u_1,\dots ,u_s)=(u_1\dots u_s)^{-1+1/k}\frac{\sin 2\pi \lambda(u_1+\dots+u_s-1)}{\pi(u_1+\dots u_s-1)}$. Therefore, setting $E=T^{-1}((0,1)^s)$ we get: $$I=\int_E f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\textrm{d}u$$ What we need to apply now is Fubini's theorem (by the way, Fubini's theorem and also the change of variables theorem hold under suitable hypoteses, which should be checked first). Since $E\subset\{(u_1,\dots,u_{s-1},u): 0<u<s\} $, thanks to Fubini's theorem we have: $$I=\int_0^s \bigg(\int_{E^*(u)} f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\bigg)\textrm{d}u$$ where $$E^*(t)=\{(u_1,\dots,u_{s-1}) : (u_1,\dots,u_{s-1},t)\in E\}\subset \Bbb R^{s-1}$$ for ever real $t$. Now, for $0<u<s$: $$ \begin{align} &\int_{E^*(u)} f(u_1,u_2,\dots u_{s-1},u-u_1-u_2-\dots -u_{s-1})\textrm{d}u_1\dots\textrm{d}u_{s-1}\\ =&\int_{E^*(u)} \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k}\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)}\textrm{d}u_1\dots\textrm{d}u_{s-1}\\ =&\frac{\sin 2\pi \lambda (u-1)}{\pi(u-1)}\int_{E^*(u)} \Bigg(u_1\dots u_{s-1}(u-u_1-\dots-u_{s-1})\Bigg)^{-1+1/k}\textrm{d}u_1\dots\textrm{d}u_{s-1} \end{align} $$ Finally, observe that: $$E^*(u)=\{(u_1,\dots,u_{s-1})\in(0,1)^{s-1}: u-1<u_1+\dots u_{s-1}<u\} $$