For $\epsilon>0$ and $y\gg\epsilon$ I have an integral on the region $$P(n,y)=\left\{(x_1,\ldots,x_n)\in\mathbb{R}^n\;:\; \sum_k^nx_k\geq y, \;x_i\geq \epsilon, \forall i \right\}.$$ I want to explicitly write down the limits of integration when the integral is broken down. For example, for $n=1,2$ $$\int_{P(1,y)}f(x)d(x)=\int_y^{\infty}f(x)dx$$ $$\int_{P(2,y)}f(x)d(x) = \int_{\epsilon}^{y-\epsilon}\int_{y-x_2}^{\infty}f(x_1,x_2)dx_1dx_2+ \int_{y-\epsilon}^{\infty}\int_{\epsilon}^{\infty}f(x_1,x_2)dx_1dx_2$$

Up to $n=3$ it is easy to visualize. The region $P(3,y)$ is just getting everything except a small ``triangle" near the origin. In the lower level, each slice is simply the region $P(2,\cdot)$, but after the plane $x_1+x_2+x_3=\epsilon$ intersects the $x_3$-axis you just get a giant cube. With this, I have written the regions for $P(3,y)$ as $$\int_{P_3}dx_1dx_2dx_3=\int_{\epsilon}^{y-2\epsilon}\int_{P(2,\epsilon-u_3)}dx_1dx_2dx_3+\int_{y-2\epsilon}\int_{\epsilon}^{\infty}\int_{\epsilon}^{\infty}dx_1dx_2dx_3.$$ My goal is to write out the integral without any $P_n$'s. For $n=3$, the reasoning above leads to the explicit limits $$\int_{P_3}=\int_{\epsilon}^{y-2\epsilon}\int_{\epsilon}^{y-x_3-\epsilon}\int_{y-x_1-x_2}^\infty+\int_{\epsilon}^{y-2\epsilon}\int^{\infty}_{y-x_3-\epsilon}\int_{\epsilon}^\infty+\int^{\infty}_{y-2\epsilon}\int_{\epsilon}^{\infty}\int_{\epsilon}^\infty$$

For general $n\geq 2$, I am looking for something like $$\int_{P_n}dx_1\ldots dx_n=\int_{y-(n-1\epsilon)}^\infty\underbrace{\int_\epsilon^\infty \cdots\int_\epsilon^\infty}_{n\text{ times}}dx_1\ldots dx_n + \int_{\epsilon}^{y-(n-1)\epsilon}``(n-1)\text{ integrals"}dx_1\ldots dx_n$$

Any help finding the $n-1$ ``integrals?" Thanks,

UPDATE: I think I found something much simpler. You can write $$\int_{x_1,\ldots,x_n\geq\epsilon}dx_1\ldots dx_n=\int_{P(n,y)}dx_1\ldots dx_n+\int_{D(n,y)}dx_1\ldots dx_n,$$ with $D(n,y)=\left\{(x_1,\ldots,x_n)\in\mathbb{R}^n\;:\; \sum_k^nx_k< y, \;x_i\geq \epsilon, \forall i \right\}$. Note the region $D$ is the ``triangle" part I was talking about. The limits over that region for general $n$ are $$\int_{D(n,y)}dx_1\ldots dx_n=\int_{\epsilon}^{y-(n-1)\epsilon}\int_{\epsilon}^{y-(n-2)\epsilon-x_{n-1}}\int_{\epsilon}^{y-(n-3)\epsilon-x_{n-1}-x_{n-2}}\cdots\int_{\epsilon}^{y-\sum_{i=1}^nx_i}dx_1\ldots dx_n$$ Use these two results to rewrite the integral over $P(n,y)$.


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

$\ds{y \gg \epsilon > 0}$.

\begin{align} &\int_{\ds{\mathbb{R}^{n}}}\mrm{f}\pars{x_{1},\ldots,x_{n}} \bracks{\sum_{k = 1}^{n}x_{k} > y}\prod_{j = 1}^{n}\bracks{x_{j} > \epsilon}\, \dd x_{1}\ldots\, x_{n} \\ = &\ \int_{\ds{\mathbb{R}^{n}}}\mrm{f}\pars{x_{1},\ldots,x_{n}} \prod_{j = 1}^{n}\bracks{x_{j} > \epsilon}\ \overbrace{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic}{\exp\pars{ \braces{\sum_{k = 1}^{n}x_{k} - y}s} \over s}\,{\dd s \over 2\pi\ic}} ^{\ds{=\ \bracks{\sum_{k = 1}^{n}x_{k} > y}}}\ \,\dd x_{1}\ldots\, x_{n} \\[5mm] = &\ \int_{0^{+} - \infty\ic}^{0^{+} + \infty}{\expo{-ys} \over s} \bracks{\int_{\epsilon}^{\infty}\cdots\int_{\epsilon}^{\infty} \mrm{f}\pars{x_{1},\ldots,x_{n}}\exp\pars{\bracks{x_{1} + \cdots + x_{n}}s} \,\dd x_{1}\ldots\dd x_{n}}\,{\dd s \over 2\pi\ic} \end{align}

$\ds{\large Indeed}$, this expression is useful whenever we know an explicit form of$\ds{\,\mrm{f}\pars{x_{1},\ldots,x_{n}}}$. Namely, it avoids 'to worry' for the integration limits.

  • $\begingroup$ Thanks Felix. Your responses are always helpful. I was satisfieed with the update I added to the post but your reply is very curious; I would never have thought of that. Two questions: (1) the brackets on the first integral are Iverson brackets, is that also true of the brackets in the argument of the exponential function? (2) the rewriting of $[\sum_k x_k>y]$ into an integral is an application of the Fourier transform? $\endgroup$ – strawberryBeef Mar 4 '17 at 19:31
  • $\begingroup$ @strawberryBeef Thanks. $\large\left(1\right)$: Right. It's an$\texttt{Iverson Bracket}$. In the exponential function it encloses an expression. I just changed to $\left\{\right\}$ to avoid the confusion. $\large\left(2\right)$: It's convenient to use the Laplace Transform because the $x_{k}$-arguments are always positive $> \epsilon > 0$. $\endgroup$ – Felix Marin Mar 4 '17 at 21:26
  • $\begingroup$ Great. Can you help me out with the derivation. Here is my attempt: for a vector $x=(x_1,..x_n)$ define $h(x)=\sum_ix_i$. The function $\left[h(x) \geq y\right]$ can be written as $g(y,x)=\mathbb{1}_{L(h,y)}(x),$ with $\mathbb{1}$ an indicator function and $L(h,y)$ the ``super-level set" defined by $L(h,y)=\{(x_1,..x_n):h(x)\geq y\}.$ With this notation, it is easier to see that $\mathbb{1}_{L(h,y)}(x)=\mathbb{1}_{[0,h(x)]}(y)$. The Laplace transform of $g$ w.r.t. $y$ is then given by $G(s)=\int_0^{\infty}g(t,x)e^{-st}\;dt=\int_0^{\infty}\mathbb{1}_{[0,h(x)]}(t)e^{-st}\;dt=(1-e^{-sh(x)})/s$. $\endgroup$ – strawberryBeef Mar 4 '17 at 23:47
  • $\begingroup$ Plugging $G(s)$ into the inversion formula gives us different answers. What am I overlooking? Thanks, $\endgroup$ – strawberryBeef Mar 4 '17 at 23:50

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