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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.

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This is integral is equal to zero. Maybe, you meant related surface integral? –  Ilya Aug 9 '11 at 17:26
    
@Gortaur: Yes, that is correct. I'm fixing it now. –  becko Aug 9 '11 at 17:28
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there may be more elegant methods like use of Stokes theorem, but one of the methods is to use an induction since if you will calculate it using sections, the sections again will be simplices. –  Ilya Aug 9 '11 at 17:30
    
I'm not sure how I can invoke Stokes theorem, since $S$ is not a closed surface. The other method you mention, can you elaborate? Thanks. –  becko Aug 9 '11 at 17:37
    
The lower bound $0$ is enough; together with the normalization condition, it implies the upper bound $1$. –  joriki Aug 9 '11 at 17:37
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2 Answers 2

up vote 5 down vote accepted

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

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it's a nice solution. I'm just curious why you and Didier used $n+1$ variables instead of $n$. Do you think I should further edit OP's question to match your notation? ) –  Ilya Aug 9 '11 at 20:16
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@Sasha: I don't think changing the question to say $n+1$ is a good idea. $n+1$ is a natural aspect of the derivation, not of the question. One would usually want to know what happens with $n$ variables, not with $n+1$ variables. –  joriki Aug 9 '11 at 20:37
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@joriki I have updated my post to use $n$ variable to conform to OP's question. –  Sasha Aug 9 '11 at 20:47
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I think that the $d\sigma$ in the original question was meant to be the "surface area" on the simplex, but your Jacobian and integral were done integrating against $dx_1\wedge dx_2\wedge\cdots\wedge dx_{n-1}$. Luckily, on the simplex, $d\sigma=\sqrt{n}dx_1\wedge dx_2\wedge\cdots\wedge dx_{n-1}$, if I'm not mistaken. –  robjohn Aug 9 '11 at 23:22
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@becko But if $d\sigma$ is meant to be measure induced from ambient $\mathbb{R}^n$ on a simplex $x_1+x_2+\ldots+x_n=1$ then the measure I used it correct: $\int_{\mathbb{R}^n} f(x_1,\ldots,x_n) \delta(x_1+x_2+\ldots+x_n -1) dV \int_0^1 dx_1 \int_0^{1-x_1} dx_2 \int_0^{1-x_1-x_2} dx_3 \ldots \int_0^{1-x_1-x_2-\ldots-x_{n-2}} dx_{n-1} f $ –  Sasha Aug 11 '11 at 1:31
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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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Can you explain how you got this result? –  becko Aug 9 '11 at 17:57
    
See above. Please whistle if something is not clear. –  Did Aug 9 '11 at 18:41
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@Didier: If you integrate over $n+1$ variables, there's a delta function in the integrand; you can't just integrate the integrand over $\mathrm dx_1\dotsc\mathrm dx_{n+1}$ instead of $\mathrm d\sigma$. Your domain of integration ensures that the sum is $\le1$, not $=1$. –  joriki Aug 9 '11 at 19:50
    
@joriki: Thanks, you spotted the problem all right. Revised version. –  Did Aug 10 '11 at 0:30
    
See @robjohn comment in the answer above. –  becko Aug 11 '11 at 1:25
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