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Let $\Delta^n=\{x\in[0,1]^n\colon \sum_{i=1}^nx_i\le 1\}$ be a simplex.

I want to compute $\int_{\Delta^n}\exp\left (\sum_{i=1}^nx_i\right )\,\mathrm{d}\lambda(x)$.

As I have already determined the volume of the simplex, I tried a similar approach for the integral.

$\int_0^1...\int_0^{1-x_1-x_2...-x_{n-1}}\exp\left ( \sum_{i=1}^nx_i\right ) \mathrm{d}x_n...\mathrm{d}x_1$

But if I compute this integral it might not be positive (depending on $n$), so this cannot be correct.

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  • $\begingroup$ Explain how it might not be positive. $\endgroup$ – GEdgar Dec 19 '17 at 13:46
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Let $x_{n+1} = 1-\sum_{i=1}^n x_i$, then $e^{x_{n+1}} = \frac{e}{e^{\sum_{i=1}^n x_i}}$. The integral can be rewritten as:

$$I = \int_{\Delta^n} e^{\sum_{i=1}^nx_i}e^{x_{n+1}}\cdot\frac{1}{e^{x_{n+1}}}\, \,dx_1\cdots\,dx_n\\ = \frac{1}{e}\int_{\Delta^n} \prod_{i=1}^n e^{2x_i}e^{x_{n+1}}\, \,dx_1\cdots\,dx_n \triangleq \frac{1}{e}\int_{\Delta^n} \prod_{i=1}^{n+1} h_i(x_i)\,dx_1\cdots\,dx_n.$$

The integral on the RHS is a product of functions integrated on the probability simplex, and to calculate that, refer to the following and the references therein for general procedure:

Integration of product of functions on a probability simplex.

Basically, you'll need

  • (Inverse) Laplace Transform
  • Laplace Convolution Theorem
  • Partial Fraction Expansion

In particular, by change of variables, you can again rewrite the integral as convolutions:

$$I=\frac{1}{e}\left(\otimes_{i=1}^{n+1} h_i(x_i)\right)(\tau)|_{\tau=1},$$

and by applying the above bullet points, you get:

$$I = \mathcal{L}^{-1}\big[\mathcal{L}[I](s)\big](\tau)\rvert_{\tau=1} =\mathcal{L}^{-1}\Bigg[\frac{1}{e}\prod_{i=1}^{n+1}\mathcal{L}[h_i(x_i)](s)\Bigg](\tau)\rvert_{\tau=1} =\frac{1}{e}\mathcal{L}^{-1}\Bigg[\frac{1}{(s-2)^n(s-1)}\Bigg](\tau)\rvert_{\tau=1}\\ =\frac{1}{e}\mathcal{L}^{-1}\Bigg[\frac{(-1)^{n-1}}{s-2}+\frac{(-1)^{n-2}}{(s-2)^2}+\cdots+\frac{(-1)^0}{(s-2)^n}+\frac{(-1)^n}{s-1}\Bigg](\tau)\rvert_{\tau=1} =\sum_{k=0}^{n-1}\frac{(-1)^k\cdot e}{(n-1-k)!} + (-1)^n.$$

Check this out:

verify the integral in closed form with Monte Carlo integration.

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