Evaluating Integral of $x_1^{b_1-1}\cdots x_k^{b_k-1}$ over $x_1+\dots+x_k\leq 1$ 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}
 A: 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 $B(a,b) = \Gamma(a)\Gamma(b)/\Gamma(a+b)$.
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\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
$$
{\cal C} \equiv
\braces{\pars{x_{1},\ldots,x_{k}}\ \mid\ x_{1} + \cdots + x_{k} \leq 1\,\,\, \wedge\,\,\, x_{i} > 0\,\,\,\forall i
\in \braces{1,\ldots,k} }
$$

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
$\ds{1 \over
\Gamma\pars{1 + b_{1} + \cdots + b_{k}}}$.
