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.

  • $\begingroup$ Explain how it might not be positive. $\endgroup$ – GEdgar Dec 19 '17 at 13:46

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.