Proving $ \idotsint_{A}e^{t\left ( x_1+\ldots +x_n\right )}\,dx_1 \ldots dx_n=\left ( \frac{e^t-1}{t} \right )^n$

Question

Let $$A=\left \{ \left ( x_1,\ldots ,x_n \right )\in \mathbb{R}^n \,\Bigg|\sum\limits_{1\leq k\leq n}\frac{x_k}{k}\leq 1,x_1,\ldots,x_n\geq 0 \right \}$$ Prove that for any $t\in \mathbb{R}$ the equality holds $$ \idotsint\limits_{A}e^{t\left ( x_1+\ldots +x_n\right )}\,dx_1 \ldots dx_n=\left ( \frac{e^t-1}{t} \right )^n$$ I tried to solve this problem using the method of mathematical induction, but nothing happened ... Are there any other ways ?

Answer 1

When $n=1$, the integral is $$ \int_0^1 e^{t x_1} dx_1 = \frac{1}{t} \left. e^{t x_1} \right|_{x=0}^1 = \frac{e^t-1}{t}. $$ Suppose we have $$ \iint_A e^{t(x_1+\dotsb+x_n)} dV = \left( \frac{e^t-1}{t} \right)^n $$ as claimed, for some particular $n$. Consider the integral in dimension $n+1$. Let's use $A$ to denote the region in $n$ dimensions, and $B$ for the region in dimension $n+1$: $$ A = \left\{ (x_1,\dotsc,x_n) \colon \sum \frac{x_k}{k} \leq 1, x_1,\dotsc,x_n \geq 0 \right\}, \\ B = \left\{ (x_1,\dotsc,x_n,x_{n+1}) \colon \sum \frac{x_k}{k} \leq 1, x_1,\dotsc,x_{n+1} \geq 0 \right\}. $$ For each $0 \leq u \leq n+1$, let $B_u = B \cap \{x_{n+1}=u\}$. Then $$ B_u = \left\{ (x_1,\dotsc,x_n,u) \colon \sum_{1 \leq k \leq n} \frac{x_k}{k} \leq 1-\frac{u}{n+1}, x_1,\dotsc,x_n \geq 0 \right\}, $$ which can be obtained from $A$ by dilating by a factor of $1 - \frac{u}{n+1}$ (and translating by $u$ in the $x_{n+1}$ direction). Let $u' = 1 - \frac{u}{n+1}$, so $$ \sum_{1 \leq k \leq n} \frac{x_k}{k} \leq u' \iff \sum_{1 \leq k \leq n} \frac{(x_k/u')}{k} \leq 1, $$ which means $(x_1,\dotsc,x_n,u) \in B_u$ if and only if $(x_1/u',\dotsc,x_n/u') \in A$. Substitute $(x'_1,\dotsc,x'_n) = (x_1/u',\dotsc,x_n/u')$ to get $$ \begin{split} \iint_B e^{t(x_1+\dotsb+x_{n+1})} dV_{n+1} &= \int_0^{n+1} \iint_{B_u} e^{t(x_1+\dotsb+x_n)} \, dV_n \, e^{tu} \, du \\ &= \int_0^{n+1} \iint_{B_u} e^{(tu')(x_1/u'+\dotsb+x_n/u')} \, dV_n \, e^{tu} \, du \\ &= \int_0^{n+1} \iint_A e^{(tu')(x'_1+\dotsb+x'_n)} \, u'^n dV_n \, e^{tu} \, du \\ &= \int_0^{n+1} \left(\frac{e^{tu'}-1}{tu'}\right)^n u'^n \ e^{tu} \, du \\ &= \int_0^{n+1} \left(\frac{e^{tu'}-1}{t}\right)^n \ e^{tu} \, du \end{split} $$ by induction. To simplify, first substitute $u' = 1 - \frac{u}{n+1}$, so $du' = - \frac{1}{n+1} du$. Then $u = (n+1)(1-u')$. When $u=0$ then $u'=1$, and when $u=n+1$, then $u'=0$. Substituting and reversing the direction of the integral, that last integral is equal to $$ \begin{split} \int_0^1 \left(\frac{e^{tu'}-1}{t}\right)^n e^{t(n+1)(1-u')} (n+1) du' &= \int_0^1 \left(\frac{e^{tu'}-1}{t}\right)^n e^{tn(1-u')} e^{t(1-u')} (n+1) du' \\ &= \int_0^1 \left(\frac{e^{tu'}-1}{t} e^{t(1-u')}\right)^n e^{t(1-u')} (n+1) du' \\ &= \int_0^1 \left(\frac{e^t-e^{t(1-u')}}{t}\right)^n e^{t(1-u')} (n+1) du' \\ &= \int_0^{(e^t-1)/t} v^n (n+1) \, dv, \end{split} $$ where $v = \frac{e^t-e^{t(1-u')}}{t}$, $dv = e^{t(1-u')} du'$. And this integral is equal to $\left( \frac{e^t-1}{t} \right)^{n+1}$ as desired.