Integrating dependent variables If $f(p_1,...,p_n)=c$ only if $0\le p_i\le 1$ for all $i$ and $p_1+\cdots +p_n=1$ and equals $0$ otherwise, what must $c$ be in order for $f$ to be a probability distribution? 

Naturally, I need to make sure the integral over the whole domain equals $1$ so $c$ must be the reciprocal of the integral. However, given the constraints, one variable will depend on all others, so how should the integration go?

$$\int\cdots \int _{0\le p_i \le 1 \ , \ i=1,...,n \ \sum_1^n p_i=1} f(p_1,...,p_n) \text{d} p_1 \cdots \text{d} p_n$$
I know the above transforms into
$$\int\cdots \int _{0\le p_i \le 1 \ , \ i=2,...,n \ 0\le \sum_2^n p_i\le 1} f(1-\sum_2^n p_i,...,p_n) \text{d} p_1 \cdots \text{d} p_n$$
However, what should happen to the innermost integral? Should it be disregarded or integrated within the previous bounds? I.e. should the above transform into this
$$\int\cdots \int _{0\le p_i \le 1 \ , \ i=2,...,n \ 0\le \sum_2^n p_i\le 1} f(1-\sum_2^n p_i,...,p_n)\left (\int_0^1 \text{d} p_1\right)\text{d}p_2 \cdots \text{d} p_n$$
Or this
$$\int\cdots \int _{0\le p_i \le 1 \ , \ i=2,...,n \ 0\le \sum_2^n p_i\le 1} f(1-\sum_2^n p_i,...,p_n) \text{d} p_2 \cdots \text{d} p_n$$
And why? How does one go about these situations?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{\substack{\large\bracks{0,1}^{n}
                 \\[1mm]
                 p_{1} + \cdots + p_{n} = 1}}\mrm{f}\pars{p_{1},\ldots,p_{n}}
\,\dd p_{1}\ldots\dd p_{n} =
\int_{\large\bracks{0,1}^{n}}c\bracks{z^{1}}z^{p_{1} + \cdots + p_{n}}
\,\dd p_{1}\ldots\dd p_{n}
\\[5mm] = &\
c\bracks{z^{1}}\pars{\int_{0}^{1}z^{p}\,\dd p}^{n} =
c\bracks{z^{1}}\pars{{1 \over \ln\pars{z}}\int_{0}^{1}\totald{z^{p}}{p}
\,\dd p}^{n} = 
c\bracks{z^{1}}\bracks{z - 1 \over \ln\pars{z}}^{n}
\\[5mm] = &\ c\,\,\,
\underbrace{\lim_{z \to 0}\partiald{}{z}\bracks{z - 1 \over \ln\pars{z}}^{n}}
_{\ds{\color{#f00}{\large ?}}}\qquad\qquad
\underline{What\ is\ your\ conclusion}\ ?.
\end{align}

Lets assume a different restriction. Namely,
  $\ds{p_{1} + \cdots + p_{n} \color{#f00}{>} 1}$.

\begin{align}
&\int_{\substack{\large\bracks{0,1}^{n}
                 \\[1mm]
                 p_{1} + \cdots + p_{n} \color{#f00}{>} 1}}\mrm{f}\pars{p_{1},\ldots,p_{n}}
\,\dd p_{1}\ldots\dd p_{n} =
\int_{\large\bracks{0,1}^{n}}c\bracks{p_{1} + \cdots + p_{n} > 1}
\,\dd p_{1}\ldots\dd p_{n}
\\[5mm] = &\
c\int_{\large\bracks{0,1}^{n}}
\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}
{\exp\pars{\bracks{p_{1} + \cdots + p_{n} - 1}s} \over s}\,{\dd s \over 2\pi\ic}
\,\dd p_{1}\ldots\dd p_{n}
\\[5mm] = &\
c\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}{\expo{s} \over s}
\pars{\int_{0}^{1}\expo{sp}\dd p}^{n}\,{\dd s \over 2\pi\ic} =
c\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}{\expo{s} \over s}
\pars{\expo{s} - 1 \over s}^{n}\,{\dd s \over 2\pi\ic}
\\[5mm] = &\
c\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}
{\expo{s} \over s^{n + 1}}
\pars{\expo{s} - 1}^{n}\,{\dd s \over 2\pi\ic} =
c\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}
{\expo{s} \over s^{n + 1}}
\pars{n!\sum_{k = 0}^{\infty}{k \brace n}{s^{k} \over k!}}
\,{\dd s \over 2\pi\ic}
\end{align}
$\ds{p \brace q}$ is a
Stirling Number of the Second Kind. Then,
\begin{align}
&\int_{\substack{\large\bracks{0,1}^{n}
                 \\[1mm]
                 p_{1} + \cdots + p_{n} \color{#f00}{>} 1}}\mrm{f}\pars{p_{1},\ldots,p_{n}}
\,\dd p_{1}\ldots\dd p_{n}
\\[5mm] = &\
c\,n!\sum_{k = 0}^{\infty}{{k \brace n} \over k!}\
\underbrace{\int_{0^{+} -\infty\ic}^{0^{+} + \infty\ic}
{\expo{s} \over s^{n + 1 - k}}\,{\dd s \over 2\pi\ic}}
_{\ds{\delta_{kn} \over \pars{n - k}!}} = c \implies \bbx{c = 1}
\end{align}
