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I'm trying to calculate the following integral $\int\limits_S \exp\left\{\sum_{i=1}^n \lambda _ix_i\right\} \, d\sigma$ where the $\lambda_i$ are constant real parameters, $S$ is a surface in $\mathbb{R}^n$ determined by the conditions $\sum _{i=1}^n x_i=1$ and $\forall _i0\leq x_i\leq 1,$ and $d\sigma$ is the element of area on this surface.

I have the feeling that a relatively simple expression can be found. Thanks.

  • 0
    Looks like your feeling was right :-)2011-08-09

2 Answers 2

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It is easy to remap the simplex into unit hypercube by changing variables: $x_1 = u_1$, $x_2 = (1-u_1)u_2$, $x_{n-1} = (1-u_1)(1-u_2)\cdots (1-u_{n-2}) u_{n-1}$, $x_{n} = (1-u_1)(1-u_2)\cdots (1-u_{n-2})(1- u_{n-1})$. The Jacobian will be $(1-u_1)^{n-1} (1-u_2)^{n-2} \cdots (1-u_{n-2})$. The integral thus becomes

$ \int_0^1 du_1 \cdots \int_0^1 du_{n-1} (1-u_1)^{n-1} (1-u_2)^{n-2} \cdots (1-u_{n-2}) \; \mathrm{e}^{ \lambda_1 u_1 + \lambda_2 (1-u_1)u_2 + \cdots + \lambda_{n} (1-u_1)\cdots (1-u_{n-1}) } $

Now carry out integration with respect to $u_{n-1}$. The part of exponential that depends on $u_{n-1}$ is $ (1-u_1)\cdots (1-u_{n-2})(\lambda_{n-1} u_{n-1} + \lambda_{n} (1-u_{n-1}))$, hence integration over $u_{n-1}$ gives

$ \int_0^1 du_1 \cdots \int_0^1 du_{n-2} (1-u_1)^{n-2} (1-u_2)^{n-3} \cdots (1-u_{n-3}) \; \mathrm{e}^{ \lambda_1 u_1 + \lambda_2 (1-u_1)u_2 + \cdots + \lambda_{n-2} (1-u_1)\cdots (1-u_{n-3}) } f $

where $f = \frac{1}{\lambda_{n} -\lambda_{n-1}} ( e^{(1-u_1)(1-u_2)\cdots (1-u_{n-2}) \lambda_{n}} - e^{(1-u_1)(1-u_2)\cdots (1-u_{n-2}) \lambda_{n-1}}) $.

Iterating over gives the answer:

$ \sum_{k=1}^{n} \frac{e^{\lambda_k}}{\prod_{k\not= m} (\lambda_k - \lambda_m)} $

I ran numerical simulations, which confirm the answer above:

Screen-shot of simulation performed

  • 0
    See my comment in the answer below. @robjohn2011-08-11
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At least for parameters $\lambda_i$ that are all different, the value of the integral is

$\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\displaystyle \sum_i\mathrm{e}^{\lambda_i}\prod_{j\ne i}\frac1{\lambda_i-\lambda_j}. $

To prove this formula, one can denote by $J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})$ the integral of interest when there are $n+1$ parameters, hence $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\int_{[0,1]^n}\mathrm{e}^{\lambda_1x_1+\cdots+\lambda_nx_n+\lambda_{n+1}(1-x_1-\cdots-x_n)}\mathbf{1}_{0\le x_1+\cdots+x_n\le1}\text{d}x_1\cdots\text{d}x_n. $ Equivalently, $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\mathrm{e}^{\lambda_{n+1}}K_n(\mu_1,\ldots,\mu_n), $ with $\mu_i=\lambda_i-\lambda_{n+1}$ for every $i\le n$ and $ K_n(\mu_1,\ldots,\mu_n)=\int_{[0,1]^n}\mathrm{e}^{\mu_1x_1+\cdots+\mu_nx_n}\mathbf{1}_{0\le x_1+\cdots+x_n\le1}\text{d}x_1\cdots\text{d}x_n. $ Now, perform the integral along the last coordinate $x_{n}$. The domain of integration is $0\le x_{n}\le 1-x_{1}-\cdots-x_{n-1}$ and $ \int_0^{1-s}\mathrm{e}^{\mu_{n}x_{n}}\mathrm{d}x_{n}=\frac1{\mu_{n}}(\mathrm{e}^{\mu_{n}(1-s)}-1), $ hence, using the shorthand \mu'_i=\mu_i-\mu_n=\lambda_i-\lambda_{n} for every $i\le n-1$, K_{n}(\mu_1,\ldots,\mu_{n})=\frac1{\mu_n}(\mathrm{e}^{\mu_n}K_{n-1}(\mu'_1,\ldots,\mu'_{n-1})-K_{n-1}(\mu_1,\ldots,\mu_{n-1})). This translates back in terms of $J_{n+1}$ and $J_n$ as $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\frac1{\mu_n}(J_{n}(\lambda_1,\ldots,\lambda_{n})-J_{n}(\lambda_1,\ldots,\lambda_{n-1},\lambda_{n+1})), $ Starting from $ J_2(\lambda_1,\lambda_2)=\mathrm{e}^{\lambda_1}\frac1{\lambda_1-\lambda_2}+\mathrm{e}^{\lambda_2}\frac1{\lambda_2-\lambda_1}, $ this yield the desired formula through a recursion over $n$.

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    If you want, I'll edit the question to make this point clearer.2011-08-11