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Let $$E = \{(x_1,\ldots,x_n) \mid x_1,\dots,x_n>0, x_1+\cdots+x_n<1 \}$$ and $f:[0,1]\to \mathbb{R}$ be continuously differentiable. Prove $$\int \cdots \int_E f(x_1+ \cdots +x_n) \, dx_1\cdots dx_n=\frac{1}{(n-1)!}\int_0^1 f(s)s^{n-1} \, ds$$

I think induction is the way to go here, the base was rather trivial but now I'm stuck. Any help?

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    $\begingroup$ hint: set $\sum_{i=1}^n x_i =s$ and $x_1=x_1,....., x_{n-1}=x_{n-1}$ $\endgroup$ – tired Aug 1 '16 at 17:49
  • $\begingroup$ @tired : Maybe you should have posted that as an answer. I noticed your comment after I posted essentially the same thing. $\qquad$ $\endgroup$ – Michael Hardy Aug 1 '16 at 17:56
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Let $E_n$ be the $n$-simplex and consider $g(s):=\int_0^{1-s} f(s+t) \, dt$. Assuming the claim is true for $n$, we have

\begin{align} & \int_{E_{n+1}} f(x_1+\dots+x_n+x_{n+1}) \, dx_1\dots dx_n \, dx_{n+1}\\ = {} & \int_{E_n}\int_0^{1-({x_1+\dots+x_n})} f(x_1+\dots+x_n+x_{n+1})\,dx_{n+1} \, dx_1\dots dx_n \\ = {} & \int_{E_n} g(x_1+\dots+x_n)\,dx_1\dots dx_n\\ = {} & \frac{1}{(n-1)!} \int_0^1 g(s)s^{n-1}\,ds \end{align} and the rest just follows from integration by parts: \begin{align} \int_0^1 g(s) s^{n-1} \, ds= \frac{1}{n}g(s)s^n \bigg|_0^1 - \frac{1}{n} \int_0^1 g'(s)s^n ds \end{align} The first term is zero because $g(1)=0$. The second term is $$\frac{1}{n}\int_0^1 f(s)s^n \, ds$$ as desired.

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  • $\begingroup$ Beautiful solution! I just don't really get the last part with the second term equality, where does the negative sign disappear? $\endgroup$ – lfc Aug 1 '16 at 18:18
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    $\begingroup$ If you take the derivative of $g$, you need to multiply with the derivative of the upper endpoint $(1-s)'=-1$ $\endgroup$ – Dimitris Aug 1 '16 at 18:21
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Do a substitution, find a Jacobian, and then simplify: $$ \begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ x_n \end{bmatrix} \mapsto \begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ x_1+\cdots + x_n \end{bmatrix} = \begin{bmatrix} x_1 \\ \vdots \\ x_{n-1} \\ s \end{bmatrix} $$

You will need the identity that says $$ \int_A \left( \int_B f(s)\cdot 1\,dx_1\cdots dx_{n-1} \right) \,ds = \int_A\left( f(s) \int_B 1\, dx_1\cdots dx_{n-1} \right) \, ds $$ (But we cannot just multiply the integral over $A$ by the integral over $B$, because the integral over $B$ depends on $s$.)

You will need to find to deal with this: $$ \int_0^s \cdots \,dx_1. \tag 1 $$ Inside of $(1)$ you have another integral: $$ \int_0^s \left( \int_0^{x_1} \cdots\,dx_2 \right) \, dx_1 $$ and another inside of that: $$ \int_0^s \left( \int_0^{s-x_1} \left( \int_0^{s-x_1-x_2} \cdots \, dx_3 \right)\,dx_2 \right) \, dx_1 $$ and so on. This is the part for which induction can be used. You ought to get $\dfrac{s^{n-1}}{(n-1)!}$.

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  • $\begingroup$ Exactly,... (+1) $\endgroup$ – tired Aug 1 '16 at 17:56
  • $\begingroup$ Everything is fine, i'm too busy at the moment so i'm very glad that you posted the answer $\endgroup$ – tired Aug 1 '16 at 17:57

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