How to calculate the following integral: $$I = \idotsint_V dx_1 \, dx_2 \cdots dx_k$$ where $$V = \big\langle 0\leq x_1 \leq 1, \space 0\leq x_2 \leq 1-x_1, \space \ldots, \space 0\leq x_k \leq 1-x_1-x_2-\cdots-x_{k-1} \big\rangle$$ I'm pretty sure that this is a well-known integral, and it is equal to $\dfrac{1}{k!}$, but can't prove it.
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5This integral is the volume of the $n$-simplex. See this answer for one way of computing it. For another way see this answer. – Winther Jan 21 '16 at 19:51
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@Winther Oh, thank you very much, that solved my problem. Well, should this question now be closed or marked as dublicate or smth? – Chan Kha Vu Jan 21 '16 at 20:13
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There are many clever ways to derive this, see for example the answers in the two questions [1] and [2]. I just wanted to quickly mention a more direct route which you can use if we don't know or see the simpler approach: we perform the first few integrations, try to notice a pattern and prove this pattern using induction.
Let $s_j = \sum_{i=1}^j x_i$. After performing the $x_k$ integral $\int_0^{1-x_1-\ldots-x_{k-1}}{\rm d}x_k$ the integrand is simply $1-x_1-\ldots-x_{k-1} = 1-s_{k-1}$. The next integral in line is
$$\int_0^{1-s_{k-2}}[1-s_{k-1}]{\rm d}x_{k-1} = \left[-\frac{(1-s_{k-1})^2}{2}\right]_{x_{k-1}=0}^{x_{k-1}=1-s_{k-2}} = \frac{(1-s_{k-2})^2}{2}$$
Likewise for the third integral we find
$$\int_0^{1-s_{k-3}}\frac{(1-s_{k-2})^2}{2}{\rm d}x_{k-2} = \frac{(1-s_{k-3})^3}{6}$$
We now see a pattern. After performing $i$ of the integrals the integrand looks like it's simply
$$P_i = \frac{(1-s_{k-i})^i}{i!}$$
To prove that this holds we use induction. It holds for the base case $i=1$ and
$$\int_0^{1-s_{k-i-1}}P_i\,{\rm d}x_{k-i} = \left[-\frac{(1-s_{k-i})^{i+1}}{(i+1)!}\right]_{x_{k-i}=0}^{x_{k-i}=1-s_{k-i-1}} = \frac{(1-s_{k-i-1})^{i+1}}{(i+1)!}$$
which is nothing but $P_{i+1}$ confirming our induction guess. Finally taking $i=k$, which corresponds to the result after evaluating all the $k$ integrals, we get $\frac{1}{k!}$.
