Calculus of an integral 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.
 A: 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:

A: 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 $n$.
