Trying to use recursion on Iterated Integral

Question

I have the following integral: $$F_3(y)=\int_R |x_1x_2x_3|^{-(1+\alpha)}\;dx_1dx_2dx_3,$$ over the region $R=\left\{(x_1,x_2,x_3)\in\mathbb{R}:x_1+x_2+x_3\geq y, |x_1|\geq 1, |x_2|\geq 1,|x_3|\geq 1\right\}$ where $1<\alpha<2$ and $y>1.$

There is no closed form for the integral but I would like to write it in terms of $F_1$ and $F_2$, where $F_1(y)$ and $F_2(y)$ are defined in a similar fashion for $y>1$ . For example, we may first write $$F_3(y)=\int_R |x_1x_2x_3|^{-(1+\alpha)}\;dx_1dx_2dx_3=\int_R |x_1|^{-(1+\alpha)}F_2(y-x_1)\;dx_1.$$

The issue that now arises is that as $x_1$ varies the quantity $y-x_1$ will not be greater than one in some regions. While it is possible to define these $F$ integrals for all $y$, I would like to stay with positive $y$.

If there a general framework to achieve this? Also, would it be easier to evaluate the integral using complex analysis? Thanks in advance.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$ \begin{array}{l} \ds{R \equiv \braces{\pars{x_{1}, x_{2}, x_{3}} \in \mathbb{R}^{3}\ \mid\ x_{1} + x_{2} + x_{3} > y\,;\quad\verts{x_{1}} \geq 1\,, \verts{x_{2}} \geq 1\,,\verts{x_{3}} \geq 1}.} \\[2mm] \ds{\alpha \in \pars{1,2}\quad\mbox{and}\quad y > 1.} \end{array}$

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Hereafter, $\ds{\bracks{\cdots}}$ is a notation for the Iverson Bracket. Namely, $\ds{\bracks{\mrm{P}} = 1}$ whenever $\ds{\,\mrm{P}}$ $\bbox[#dfe,5px]{\texttt{is true}}$. $\bbox[#dfe,5px]{\texttt{Otherwise}}$, $\ds{\bracks{\mrm{P}} = 0}$.

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\begin{align} \mrm{F}_3\pars{y} & \equiv \int_{R}\verts{x_{1}x_{2}x_{3}}^{\,-\pars{1 + \alpha}} \,\dd x_{1}\dd x_{2}\dd x_{3} \\[5mm] & = \iint_{\verts{x_{1}}\ \geq\ 1 \atop \verts{x_{2}}\ \geq\ 1} \verts{x_{1}x_{2}}^{\,-\pars{1 + \alpha}} \int_{\verts{x_{3}}\ \geq\ 1}\verts{x_{3}}^{\,-\pars{1 + \alpha}} \bracks{x_{1} + x_{2} + x_{3} > y}\dd x_{3}\dd x_{2}\dd x_{1} \\[1cm] & = \iint_{\verts{x_{1}}\ \geq\ 1 \atop \verts{x_{2}}\ \geq\ 1} \verts{x_{1}x_{2}}^{\,-\pars{1 + \alpha}} \\[5mm] & \pars{\color{#00f}{\int_{1}^{\infty}x_{3}^{\,-\pars{1 + \alpha}} \bracks{x_{3} > y - x_{1} - x_{2}}\dd x_{3}} + \color{#f00}{\int_{1}^{\infty}x_{3}^{\,-\pars{1 + \alpha}} \bracks{x_{3} < x_{1} + x_{2} - y}\dd x_{3}}}\label{1}\tag{1} \end{align}

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\begin{align} &\color{#00f}{\int_{1}^{\infty}x_{3}^{\,-\pars{1 + \alpha}} \bracks{x_{3} > y - x_{1} - x_{2}}\dd x_{3}} \\[5mm] = &\ \bracks{y - x_{1} - x_{2} < 1}\int_{1}^{\infty}x_{3}^{-\alpha - 1}\,\dd x_{3} + \bracks{y - x_{1} - x_{2} > 1}\int_{y - x_{1} - x_{2}}^{\infty} x_{3}^{-\alpha - 1}\,\dd x_{3} \\[5mm] & = \color{#00f}{{\bracks{x_{2} > y - x_{1} - 1} \over \alpha} + {\bracks{x_{2} < y - x_{1} - 1} \over \alpha}\, \pars{y - x_{1} - x_{2}}^{-\alpha}}\label{2}\tag{2} \\[1cm] &\color{#f00}{\int_{1}^{\infty}x_{3}^{\,-\pars{1 + \alpha}} \bracks{x_{3} < x_{1} + x_{2} - y}\dd x_{3}} = \bracks{x_{1} + x_{2} - y > 1} \int_{1}^{x_{1} + x_{2} - y}x_{3}^{-\alpha - 1}\dd x_{3} \\[5mm] = &\ \color{#f00}{{\bracks{x_{2} > y - x_{1} + 1} \over \alpha}\bracks{% 1 - \pars{x_{1} + x_{2} - y}^{-\alpha}}}\label{3}\tag{3} \end{align}

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With \eqref{2} and \eqref{3}, \eqref{1} is written as $\ds{\,\mrm{F}_{3}\pars{y} = \int_{\verts{x_{1}}\ \geq\ 1}\verts{x_{1}}^{-\pars{1 + \alpha}}\,\mrm{F}_{2}\pars{y - x_{1}}\,\dd x_{1}}$ where: \begin{align} \mrm{F}_{2}\pars{z} & \equiv {1 \over \alpha}\int_{\verts{x_{2}}\ \geq\ 1}\verts{x_{2}}^{-\alpha - 1} \bracks{x_{2} > z - 1}\,\dd x_{2}\label{4}\tag{4} \\[5mm] & + {1 \over \alpha}\int_{\verts{x_{2}}\ \geq\ 1}\verts{x_{2}}^{-\alpha - 1} \pars{z - x_{2}}^{-\alpha}\bracks{x_{2} < z - 1}\,\dd x_{2}\label{5}\tag{5} \\[5mm] & + {1 \over \alpha}\int_{\verts{x_{2}}\ \geq\ 1}\verts{x_{2}}^{-\alpha - 1} \bracks{x_{2} > z + 1}\,\dd x_{2}\label{6}\tag{6} \\[5mm] & - {1 \over \alpha}\int_{\verts{x_{2}}\ \geq\ 1}\verts{x_{2}}^{-\alpha - 1} \pars{x_{2} - z}^{-\alpha}\bracks{x_{2} > z + 1}\,\dd x_{2}\label{7}\tag{7} \end{align}

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$\ds{x_{2}}$-Integration in Line \eqref{4}: \begin{align} &{1 \over \alpha}\int_{\verts{x_{2}}\ \geq\ 1}\verts{x_{2}}^{-\alpha - 1} \bracks{x_{2} > z - 1}\,\dd x_{2} \\[5mm] = &\ {1 \over \alpha}\int_{1}^{\infty}x_{2}^{-\alpha - 1} \bracks{x_{2} > z - 1}\,\dd x_{2} + {1 \over \alpha}\int_{1}^{\infty}x_{2}^{-\alpha - 1} \bracks{x_{2} < 1 - z}\,\dd x_{2} \\[5mm] & = {\bracks{z - 1 < 1}\over \alpha}\int_{1}^{\infty}x_{2}^{-\alpha - 1}\,\dd x_{2} + {\bracks{z - 1 > 1}\over \alpha}\int_{z - 1}^{\infty}x_{2}^{-\alpha - 1} \,\dd x_{2} + {\bracks{1 - z > 1}\over \alpha}\int_{1}^{1 - z}x_{2}^{-\alpha - 1} \,\dd x_{2} \\[5mm] & = {\bracks{z < 2}\over \alpha^{2}} + {\bracks{z > 2}\over \alpha^{2}}\pars{z - 1}^{-\alpha} + {\bracks{z < 0}\over \alpha^{2}}\bracks{1 - \pars{1 - z}^{-\alpha}} \\[5mm] & = \bbx{\ds{% {\bracks{z < 0}\bracks{2 - \pars{1 - z}^{-\alpha}} + \bracks{0 < z < 2} + \bracks{z > 2}\pars{z - 1}^{-\alpha} \over \alpha^{2}}}} \end{align}

$\ds{x_{2}}$-integrations in lines \eqref{5}, \eqref{6} and \eqref{7} are rather similar to the above integration. Can you take it from here ?.