Evaluating Integral of $x_1^{b_1-1}\cdots x_k^{b_k-1}$ over $x_1+\dots+x_k\leq 1$

Question

I am struggling with this Integral: $$\idotsint\limits_{\begin{subarray}{l}x_1\ +\ \dots\ +\ x_k\ \leq\ 1 \\[1mm] x1,\,\dots\,,\,x_k\ \geq\ 0 \end {subarray}} x_1^{b_1-1}\cdots x_k^{b_k-1} dx_1 \cdots dx_k $$ for $b_1,\dots,b_k>0$.

My attempt so far is to use Fubini's theorem and get to $\frac{x_1^{b_1}\cdots x_k^{b_k}}{b_1\cdots b_k}$ but I am struggling to evaluate for $\Big|_{\begin{subarray}{l}x_1+\dots+x_k\leq 1 \\ x1,\dots,x_k \geq 0 \end {subarray}}$. I have also thought about using induction and while $k=1$ is clear, I'm having a hard time with $k\geq 2$.

Edit: The integral evaluates to $\frac{\Gamma(b_1)\cdots \Gamma(b_k)}{\Gamma(1 + b_1 + \cdots + b_k)}$. I have proven this in a way similar to @Shannon Starr's answer. I now want to evaluate the integral using a substitution of polar coordinates.
I already proved that in polar coordinates: \begin{align} x_1&=r \cos(\varphi_1)\\ x_2&= r \sin(\varphi_1)\cos(\varphi_2)\\ &\vdots\\ x_{k-1}&=r\sin(\varphi_1)\cdots \sin(\varphi_{k-2})\cos(\varphi_{k-1})\\ x_k&=r\sin(\varphi_1)\cdots \sin(\varphi_{k-2})\sin(\varphi_{k-1}) \end{align} and the calculation for $x_i^{b_k-1}$ is obvious. I was however given the hint of calculating this by substituting the $\textbf{square} $ of the polar coordinates, so I am not sure how to proceed.

Edit edit: I would just like to emphasize that I am looking for a solution using the substitution \begin{align} x_1&=(r \cos(\varphi_1))^2\\ x_2&= (r \sin(\varphi_1)\cos(\varphi_2))^2\\ &\vdots\\ x_{k-1}&=(r\sin(\varphi_1)\cdots \sin(\varphi_{k-2})\cos(\varphi_{k-1}))^2\\ x_k&=(r\sin(\varphi_1)\cdots \sin(\varphi_{k-2})\sin(\varphi_{k-1}))^2 \end{align}

Answer 1

The value of the integral in the OP is $$\begin{align} \int_\limits{\substack{x_1,\ldots,x_n>0\\ x_1+\ldots+x_n\leq1}}x^{b_1-1}_1\cdot\ldots\cdot x^{b_n-1}_n\,d\mathbf{x}&= \frac{\Gamma(b_1)\cdot\ldots\cdot\Gamma(b_n)}{\Gamma(b_1+\ldots+b_n+1)}\tag{OP}\label{op} \end{align} $$ where $\Gamma(x)=\int^\infty_0t^{x-1}e^{-t}\,dt$ is the Gamma function.

As it has been pointed out, the integral in the OP is related to the generalized beta function. There are many ways to study this integral. Here I present two methods: (1) using spherical coordinates (2) using a linear change of variables followed by introduction of the generalized beta function.

Consider the integral \begin{align} I:=\int_{\mathbb{R}^n_+}f(x_1+\ldots+x_n)x^{b_1-1}_1\cdot\ldots\cdot x^{b_n-1}_n\,d\boldsymbol{x}\tag{0}\label{zero} \end{align} where $b_1,\ldots, b_n>0$. The case $f(x)=\mathbb{1}_{[0,1]}(x)$ corresponds to the integral in the OP.

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Method 1: spherical coordinates

As the OP wishes to use spherical coordinates, we consider again integrals of the form \eqref{zero}. The change of variable $$\begin{align} X&:\mathbb{R}^n_+\rightarrow\mathbb{R}^n_+\\ \mathbf{z}&\mapsto[z^2_1,\ldots,z^2_n]^\intercal \end{align}$$ yields \begin{align} I &= \int_\limits{\mathbb{R}^n_+}f(x_1+\ldots+x_n)x^{b_1-1}_1\cdot\ldots\cdot x^{b_n-1}_n\,d\boldsymbol{x}\\ &=2^n\int_\limits{\mathbb{R}^n_+}f(|\mathbf{z}|^2)z_1^{2b_1-1}\cdot\ldots\cdot z_n^{2b_n-1}\,d\mathbf{z} \end{align} (notice that $G$ is a diffeomorphism from $(0,\infty)^n$ to itself, and that the Jacobian $J_X(\mathbf{z})=2^nz_1\cdot\ldots\cdot z_n$). Using spherical coordinates gives $$\begin{align} I &= 2^n\int^\infty_0f(r^2)r^{2b_1+\ldots+2b_n -1}\,dr \int_{\mathbf{S}_{n-1}}\mathbb{1}_{P_+}(\mathbf{u})u^{2b_1-1}_1\cdot\ldots\cdot u^{2b_n-1}_n\,\sigma(d\mathbf{u})\\ &= F_n \int^\infty_0f(r^2)r^{2b_1+\ldots+2b_n -1}\,dr \\ &=\frac12 F_n\int^\infty_0 f(s) s^{b_1+\ldots + b_n-1}\,ds \end{align}$$ where $P_+$ is the postive octant in $\mathbb{R}^n$, and $F_n=2^n\int_{\mathbb{S}_{n-1}}\mathbb{1}_{P_+}(\mathbf{u})u^{2b_1-1}_1\cdot\ldots\cdot u^{2b_n-1}_n\,\sigma(d\mathbf{u})$.

When $f(t)=e^{-t}$, we obtain $$\begin{align} \Gamma(b_1)\cdot\ldots\cdot\Gamma(b_n) =\frac12 F_n\int^\infty_0e^{-s}s^{b_1+\ldots+b_n-1}=\frac12F_n\, \Gamma(b_1+\ldots+b_n) \end{align} $$ Consequently, $$ \frac12 F_n=\frac{\Gamma(b_1)\cdot\ldots\cdot\Gamma(b_n)}{\Gamma(b_1+\ldots+b_n)}=: B(b_1,\ldots,b_n) $$ and so $$\begin{align} \int_\limits{\mathbb{R}^n_+} &f(x_1+\ldots+x_n)x^{b_1-1}_1\cdot\ldots\cdot x^{b_n-1}_n\,d\boldsymbol{x}\\ &\qquad =B(b_1,\ldots,b_n)\int^\infty_0f(s)s^{b_1+\ldots+b_n-1}\,ds\tag{1}\label{one} \end{align} $$

The estimate \eqref{op} follows by applying \eqref{one} to the case $f(t)=\mathbb{1}_{[0,1]}(t)$.

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Method 2: Linear transformation and generalized beta function:

On $\mathbb{R}^n_+$ define \begin{align} g(t_1,\ldots,t_n):=f(t_n)t^{b_1-1}_1\cdot\ldots\cdot t^{b_{n-1}-1}_{n-1}\big(t_n-(t_1+\ldots+t_{n-1})\big)^{b_n-1}, \end{align} and \begin{align} T\boldsymbol{x}:=\left(\begin{array}{llcrr} 1 & 0 & \ldots &0 &0\\ 0 & 1 & \ldots &0 &0\\ \vdots & \vdots &\ddots &\vdots &\vdots \\ 0 & 0 & \ldots & 1 & 0\\ 1 & 1& \ldots &1 & 1 \end{array} \right) \left(\begin{array}{l} x_1\\ x_2\\ \vdots\\ x_{n-1}\\ x_n \end{array}\right) \end{align} Then $I=\int_{\mathbb{R}^n_+}g(T\boldsymbol{x})|J_T(\boldsymbol{x})|\,d\boldsymbol{x}$, and $T(\mathbb{R}^n_+)=\{\boldsymbol{t}\in\mathbb{R}^n_+: t_1+\ldots+t_{n-1}<t_n\}$ and so, \begin{align} I=\int^\infty_0 f(t_n)\left(\int\limits_{\substack{t_1,\ldots,t_{n-1}>0\\ t_1+\ldots+t_{n-1}<t_n}} t^{b_1-1}_1\ldots t^{b_{n-1}-1}_{n-1}(t_n-t_1-\ldots-t_{n-1})^{b_n-1}\,dt_1\ldots dt_{n-1}\right)\,dt_n \end{align} Setting $G(\boldsymbol{t}):=(t_1/t_n,\ldots,t_{n-1}/t_n, t_n)$ we obtain that $|J_G(\boldsymbol{t})|=t^{-(n-1)}_n$, and $D_{n-1}:=G(T(\mathbb{R}^n_+))=\{\boldsymbol{v}\in\boldsymbol{R}^n_+: v_1,\ldots, v_n>0,v_1+\ldots v_{n-1}<1\}$. Hence \begin{align} I=\left(\int_\limits{D_{n-1}} v^{b_1-1}_1\ldots v^{b_{n-1}-1}_{n-1}\big(1-(v_1+\ldots + v_{n-1})\big)^{b_n-1}\,dv_1\ldots dv_{n-1}\right)\int^\infty_0f(v)v^{\alpha-1}\,dv \end{align} where $\alpha=b_1+\ldots+b_n$. The generalized Beta function is defined as \begin{align} B(b_1,\ldots, b_n):=\int\limits_{\substack{v_1,\ldots,v_{n-1}>0\\ v_1+\ldots+v_{n-1}<1}} v^{b_1-1}_1\ldots v^{b_{n-1}-1}_{n-1}\big(1-(v_1+\ldots + v_{n-1})\big)^{b_n-1}\,dv_1\ldots dv_{n-1} \end{align} Then, \begin{align} \int\limits_{\mathbb{R}^n_+}f(x_1+\ldots+x_n)x^{b_1-1}_1\ldots x^{b_n-1}_n\,d\boldsymbol{x} = B(b_1,\ldots,b_n)\int^\infty_0 f(s) s^{b_1+\ldots+b_n-1}\,ds \end{align} The case $f(t)=e^{-t}$ yields \begin{align} B(b_1,\ldots,b_n)=\frac{\Gamma(b_1)\cdot\ldots\cdot\Gamma(b_n)}{\Gamma(b_1+\ldots+b_n)} \end{align} Putting things together, we obtain formula \eqref{one} above.

As before, \eqref{op} follows by applying \eqref{one} $f(t)=\mathbb{1}_{[0,1]}(t)$.

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Answer 2

Try a change of variables to $y_1,\dots,y_n$ where $$ y_1=x_1\, ,\ x_2=(1-x_1)y_2\, ,\ x_3=(1-x_1-x_2)y_3\, , $$ up to $$ x_n\, =\, (1-(x_1+\dots+x_{n-1}))y_n\, . $$ Then your domain is $(y_1,\dots,y_n) \in [0,1]^n$. Note that $x_2=(1-y_1)y_2$ and $$ x_3\, =\, (1-y_1 - (1-y_1)y_2)y_3\, =\, (1-y_1)(1-y_2)y_3\, . $$ Then inductively, we have $x_k = (1-y_1)(1-y_2)\cdots (1-y_{k-1})y_k$ and $1-(x_1+\dots+x_k) = (1-y_1)(1-y_2)\cdots (1-y_{k-1})(1-y_k)$. Also, if $y_1,\dots,y_{k-1}$ are held fixed, then $$ dx_k\, =\, (1-y_1)(1-y_2)\cdots (1-y_{k-1})\, dy_k\, . $$ So you can turn your iterated integrals into a product of Beta-integrals which you can evaluate using $B(a,b) = \Gamma(a)\Gamma(b)/\Gamma(a+b)$.

Answer 3

I will use $n$ as the fixed dimension and $k$ as a dummy variable in the following writing.

Denote for $1\le k\le n$, $$ T_k(h):=\{(x_1,\cdots,x_k)\mid x_1+\cdots+x_k\le h; x_1,\cdots,x_k\ge 0\} $$ and $$ f_k(x_1,\cdots,x_k)=x_1^{b_1-1}\cdots x_k^{b_k-1},\quad g_k(x)=x^{b_k-1} $$

Let $$ I_k(h):=\int_{T_k(h)}f_k(x_1,\cdots,x_k)\;dx_1\cdots dx_k $$

Then the integral we want is $$ \begin{align} I_n(1)&=\int_{T_{n}(1)} f_{n}(x_1,\cdots,x_n)dx_1\cdots dx_n\\ &=\int_0^1 x_n^{b_n-1} \left( \int_{T_{n-1}(1-x_n)}f_{n-1}(x_1,\cdots,x_{n-1})\; dx_1\cdots dx_{n-1} \right) dx_n \end{align} $$ Now we do a change of variable $x_j=(1-x_n)y_j$ to get $$ \begin{align} &\int_{T_{n-1}(1-x_n)}f_{n-1}(x)\; dx_1\cdots dx_{n-1}\\ &=(1-x_n)^{n-1}(1-x_n)^{\sum_{j=1}^{n-1} (b_j-1)} \int_{T_{n-1}(1)} f_{n-1}(y_1,\cdots,y_{n-1})dy_1\cdots dy_{n-1}\\ &=(1-x_n)^{\sum_{j=1}^{n-1} b_j} \int_{T_{n-1}(1)} f_{n-1}(y_1,\cdots,y_{n-1})dy_1\cdots dy_{n-1}\\ &=(1-x_n)^{\sum_{j=1}^{n-1} b_j} I_{n-1}(1) \end{align} $$ So you have $$ I_n(1)=\left(\int_0^1 x_n^{b_n-1}(1-x_n)^{\sum_{j=1}^{n-1} b_j} dx_n \right)\cdot I_{n-1}(1) =B(b_n,1+\sum_{j=1}^{n-1} b_j) \tag{1} $$

When $n=1$, $$ I_1(1)=\int_0^1 x^{b_1-1}dx= \frac{1}{b_1}x^{b_1} \tag{2} $$

Combining (1) and (2) you have a recursive formula for the integral.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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\begin{align} &\bbox[5px,#ffd]{\int\cdots\int_{\cal C} x_{1}^{b_{1} - 1}\ldots x_{k}^{b_{k} - 1}\,\dd x_{1}\ldots\dd x_{k}} = {\large ?} \end{align} where $\ds{b_{i} > 0\,\,\, \forall i \in \braces{1, \ldots, k}}$ and $$ {\cal C} \equiv \braces{\pars{x_{1},\ldots,x_{k}}\ \mid\ x_{1} + \cdots + x_{k} \leq 1\,\,\, \wedge\,\,\, x_{i} > 0\,\,\,\forall i \in \braces{1,\ldots,k} } $$

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Then, \begin{align} & \bbox[5px,#ffd]{\int\cdots\int_{\cal C} x_{1}^{b_{1} - 1}\ldots x_{k}^{b_{k} - 1}\,\dd x_{1}\ldots\dd x_{k}} \\[2mm] & \int_{0}^{\infty}\cdots\int_{0}^{\infty} x_{1}^{b_{1} - 1}\ldots x_{k}^{b_{k} - 1}\ \times \\ & \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\exp\pars{\bracks{1 - x_{1} - \cdots -x_{k}}s} \over s}{\dd s \over 2\pi\ic} \,\dd x_{1}\ldots\dd x_{k} \end{align} The $\ds{s}$-integration is the ${Heaviside\ Step\ Function}$ evaluated at $\ds{1 - x_{1} - \cdots - x_{k}}$. So, I'll get \begin{align} & \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}{\expo{s} \over s}\pars{\int_{0}^{\infty} x_{1}^{b_{1} - 1}\expo{-sx_{1}}\dd x_{1}}\cdots \pars{\int_{0}^{\infty} x_{k}^{b_{k} - 1}\expo{-sx_{k}}\dd x_{k}} {\dd s \over 2\pi\ic} \\[5mm] = & \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{s} \over s} {\Gamma\pars{b_{1}} \over s_{1}^{b_{1}}}\cdots {\Gamma\pars{b_{k}} \over s_{k}^{b_{k}}} {\dd s \over 2\pi\ic} = \\[5mm] = & \bracks{\prod_{\ell = 1}^{k}\Gamma\pars{b_{\ell}}} \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{s} \over s^{1\ +\ b_{1}\ +\ \cdots\ +\ b_{k}}\,\,\,\,}{\dd s \over 2\pi\ic} = {\prod_{\ell = 1}^{k}\Gamma\pars{b_{\ell}} \over \pars{b_{1} + \cdots + b_{k}}!} \\[5mm] = & \ \bbx{\color{#44f}{\Gamma\pars{b_{1}}\ldots\Gamma\pars{b_{k}} \over \Gamma\pars{b_{1} + \cdots + b_{k} + 1}}} \end{align} Note that $\ds{\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic} {\expo{s} \over s^{1\ +\ b_{1}\ +\ \cdots\ +\ b_{k}}\,\,\,\,}{\dd s \over 2\pi\ic} = \oint_{\cal H} {\expo{s} \over s^{1\ +\ b_{1}\ +\ \cdots\ +\ b_{k}}\,\,\,\,}{\dd s \over 2\pi\ic}}$ where $\ds{\cal H}$ is the $\ds{Hankel\ Contour}$ and the last integral is an integral representation of $\ds{1 \over \Gamma\pars{1 + b_{1} + \cdots + b_{k}}}$.