I feel that there is a good chance to apply certain integrals of
 $\ \exp(-(\sum\_{k=1}^n x\_k))\ $ over corners (see below) to the analytic number theory. I have obtained two formulas to start with but am asking about their joint generalization. If all this is well known then let me be in the known too (and I apologize).


A real sequence $\ \mathbf A:=(A\_1\ \ldots\ A\_n)\in(0;\infty)^n\ $ defines a corner


$$ \Delta(\mathbf A)\ :=\ \{ (x\_1\ \ldots\ x\_n)\in[0;\infty)^n\ :
 \ \sum\_{k=1}^n\frac{x\_k}{A\_k}\ \le\ 1 \} $$


The special $n$-dimensional case is $\ \mathbf S\ :=\ (S\ \dots\ S),\ $ where $\ A\_1=\ldots=A\_n=S>0.\ $ I'll write
$\ \Delta(n;\ S):=\Delta(\mathbf S)\ $ to show
the dimension $n$ explicitly. Then,


**THEOREM 1**


$$ \int\_{\Delta(n;\ S)} 
 \exp\left(-\sum\_{k=1}^n x\_k\right)\cdot dx\_1\ldots dx\_n\,\ =
 \,\ 1 -\ \sum\_{k=0}^{n-1} \frac{S^k}{k!} \cdot e^{-S} $$


Now let's consider the general but only 2-dimensional case, where
$\ \mathbf A\ :=\ (A\ B),\ $ so that we can write
$\ \Delta(A\ B):= \Delta(\mathbf A)\ $. Then,


**THEOREM 2**


$$ \int\_{\Delta(A\ B)} \exp(-(x+y))\cdot dxdy\,\ = 
 \,\ 1\ +\ \frac B{A-B}\cdot e^{-A}\ +\ \frac A{B-A}\cdot e^{-B} $$


*See the comment below about the case* $\ A=B>0$.



> 
> **Q U E S T I O N**
> 
> 
> 


I'd welcome the n-dimensional formula for the integral over the n-dimensional corner, especially in the spirit of Theorem 2 above.



> 
> **A comment:** Singularity (removable;)
> 
> 
> 


Theorem 2 doesn't look kosher when $\ A=B>0.\ $ But it is a $2$-dimensional generalization of Theorem 1. To see it, let $\ D:=B-A.\ $ Then


$$ \int\_{\Delta(A\ B)} \exp(-(x+y))\cdot dxdy\,\ = 
 \,\ 1\ - \left(1+\frac{1-e^{-D}}D\cdot A\right)\cdot e^{-A} $$


Thus, for $\ D\rightarrow 0,\ $ the above expression approaches the 2-dimensional case of Theorem 1.



Corner integrals of $\exp$
By a substitution $x\_k = A\_k y\_k$,
$$\begin{aligned}
I & := \idotsint\limits\_{\;\;\Delta(\mathbf{A})} \exp\left(-\sum\_{k=1}^n x\_k\right) dx\_1 \ldots dx\_n \\
& = \idotsint\limits\_{\;\;\Delta(\mathbf{1})} \exp\left(-\sum\_{k=1}^n A\_k y\_k\right) A\_1 \ldots A\_n dy\_1 \ldots dy\_n \\
& = \idotsint\limits\_{\;\;\Delta(\mathbf{1})} \prod\_{k = 1}^n \bigl(A\_k \exp(-A\_k y\_k)\bigr) dy\_1 \ldots dy\_n .
\end{aligned}$$
It follows that $I$ is the CDF of the sum of independent exponential random variables with parameters $A\_1, \ldots, A\_n$, evaluated at $1$. According to [this solution](https://people.maths.bris.ac.uk/~mb13434/sumexp.pdf) of Problem 12 in Chapter I of Feller's book, if all $\lambda\_k$ are distinct, the density function of the distribution of this sum is given by
$$
 f\_{\mathbf{A}}(s) = \sum\_{j = 1}^n \left(\prod\_{k \in \{1, \ldots, n\} \setminus \{j\}} \frac{A\_k}{A\_k - A\_j}\right) A\_j e^{-A\_j s}
$$
for $s > 0$. (This also follows directly from the partial fraction expansion of the Laplace transform of $f\_{\mathbf{A}}$). We conclude that
$$\begin{aligned}
I & = \int\_0^1 f\_{\mathbf{A}}(s) ds = \sum\_{j = 1}^n \left(\prod\_{k \in \{1, \ldots, n\} \setminus \{j\}} \frac{A\_k}{A\_k - A\_j}\right) (1 - e^{-A\_j}) .
\end{aligned}$$


