discuss integrability of $(x_1, \cdots, x_n) \mapsto \frac{1}{(1+x_1+x_2+ \cdots + x_n)^{\alpha}}$ on $[0,+ \infty[^n$

Question

let's first discuss integrability of $(x,y) \mapsto \frac{1}{(1+x+y)^{\alpha}}$ on $[0,+ \infty[^2$

the integrand being positive and measurable we can compute the integral in however order we want according to Tonelli's theorem.

for a fixed $ y\in [0,+ \infty[ $

$$ \int_0^{+\infty} \frac{1}{(1+x+y)^{\alpha}}dx = \int_{1+y}^{+\infty} \frac{1}{x^{\alpha}}dx = \frac{1}{(\alpha-1)(1+y)^{\alpha -1}} $$

provided that $\alpha > 1$ (our first condition) otherwise the integral is not finite.

$$ \int_0^{+\infty} \frac{1}{(\alpha-1)(1+y)^{\alpha -1}}dy = \int_0^{+\infty} \frac{1}{(\alpha-1)(1+y)^{\alpha -1}}dy = \int_1^{+\infty} \frac{1}{(\alpha-1)y^{\alpha -1}}dy = \frac{1}{(\alpha - 1)(\alpha-2)}$$

provided that $\alpha > 2$ otherwise the integral is not finite.

so intuitively in the n-th dimension case the condition of integrability should be $\alpha > n$ and its value should be $((\alpha - 1)(\alpha -2)\cdots(\alpha-n))^{-1}$

but how to formally prove that (if it's correct of course) ?

Answer 1

Let $t=x_1+\cdots+ x_n$ be one of the variables in a change-of-variables formula. The slices $t=$constant through the region of integration are all $n-1$ dimensional simplices, with volume $C_n t^{n-1}$, where the value of $C_n$ does not concern us. So an application of Fubini's theorem gives us $C_n\int_0^\infty t^{n-1-\alpha}\,dt$ which is finite just when $\alpha>n$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}\cdots\int_{0}^{\infty}{\dd x_{1}\ldots\dd x_{n} \over \pars{1 + x_{1} + \cdots + x_{n}}^{\alpha}}} \\[5mm] = &\ \int_{0}^{\infty}\cdots\int_{0}^{\infty}\bracks{ {1 \over \Gamma\pars{\alpha}}\int_{0}^{\infty}t^{\alpha - 1} \expo{-\pars{1 + x_{1} + \cdots + x_{n}}t}\dd t}\,\dd x_{1}\ldots\dd x_{n} \\[5mm] = &\ {1 \over \Gamma\pars{\alpha}}\int_{0}^{\infty}t^{\alpha - 1}\expo{-t} \pars{\int_{0}^{\infty}\expo{-tx}\dd x}^{n}\,\dd t = {1 \over \Gamma\pars{\alpha}}\int_{0}^{\infty}t^{\alpha - 1 - n}\expo{-t} \dd t \\[5mm] = &\ \bbx{{\Gamma\pars{\alpha - n} \over \Gamma\pars{\alpha}}\,,\qquad\color{#f88}{\Re\pars{\alpha} > n}} \end{align}

Note that

$$ {\Gamma\pars{\alpha - n} \over \Gamma\pars{\alpha}} = {1 \over \Gamma\pars{\alpha}/\Gamma\pars{\alpha - n}} = {1 \over \pars{\alpha - n}^{\overline{n}}} = {1 \over \pars{\alpha - n}\pars{\alpha - n + 1}\cdots \pars{\alpha - 1}} $$