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,


$$ \int_{\Delta(n;\ S)} \exp\left(-\sum_{k=1}^n x_k\right)\cdot dx_1\ldots dx_n\,\ = \,\ 1 -\ \sum_{k=0}^{n-1} \frac{S^k}{k!} \cdot e^{-S} $$

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,


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

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


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.

  • 3
    $\begingroup$ This is related to adding independent exponential random variables, and is well-known. See, for example, Problem 12 in Chapter I of Feller's book (according to these notes). $\endgroup$ Commented Jun 12, 2018 at 7:36
  • $\begingroup$ @MateuszKwaśnicki, cześć. Thank you for information and references. Can you write the n-dimensional version of the theorem 2 (see above)? $\endgroup$
    – Wlod AA
    Commented Jun 12, 2018 at 8:32
  • 1
    $\begingroup$ Done. (Only now have I read your profile note; cześć!) $\endgroup$ Commented Jun 12, 2018 at 8:59

1 Answer 1


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 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 $$\begin{aligned} I & = \int_0^1 f_{\mathbf{A}}(s) ds = \sum_{j = 1}^n \left(\prod_{k \in \{1, \ldots, n\} \setminus \{j\}} \frac{A_k}{A_k - A_j}\right) (1 - e^{-A_j}) . \end{aligned}$$

  • $\begingroup$ Thank you, it looks perfect. (It's good I've saved my time asking an expert). I'll get some sleep and will check things exactly after I open my eyes anew. Do you know about number theoretical applications of this material? N.Th. was my motivation. $\endgroup$
    – Wlod AA
    Commented Jun 12, 2018 at 9:12
  • $\begingroup$ The pseudo-singularities due to some equalities $A_k=A_m,\ $ are a bit messy. I wonder if nonstandard analysis would make things smooth? $\endgroup$
    – Wlod AA
    Commented Jun 12, 2018 at 9:20
  • 2
    $\begingroup$ @WlodAA: You're welcome. I am not a number theorist, I have no idea if this was ever applied there. I think the general case is treated in [Amari, Misra, Closed-Form Expressions for Distribution of Sum of Exponential Random Variables]. $\endgroup$ Commented Jun 12, 2018 at 9:46
  • $\begingroup$ Note that the final result can be written as $$I=(-1)^n\big( \prod_{i=1}^n A_i\big)\, \Delta^n(\exp(-x);\,0,A_1,\ldots,A_n)$$ where $\Delta^n(f;\,x_0,\ldots,x_n)$ denotes the divided difference of $f$ corresponding to $x_0,\ldots,x_n$. So you can use the calculus of finite differences in your considerations $\endgroup$
    – esg
    Commented Jun 12, 2018 at 18:30

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