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I don't expect to actually compute this integral. Just an upper bound would be sufficient for my purposes, but even that is being too hard to obtain. I hope someone here knows how to bound this integral. The more sharper is the bound the better it is.

Consider the vectors $a_i =( a_{i1},a_{i2}, \ldots, a_{in} ) \in \mathbb{R}^n$ for $i = 1 \ldots m$. These vectors are composed by non-negative coordinates (not sure if this us helpful, but this is the form of my original formulation). For any $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$, denote $a_i \cdot x = a_{i1}x_1 + \ldots + a_{in} x_n$. Finally, fix two coordinates index $1 \leq j,k \leq n$. The integral I want to bound is the following:

$$\int_{-\infty}^\infty \ldots \int_{-\infty}^\infty \frac{1}{ e^{2(a_1\cdot x - a_{jk}x_k)} + \ldots + e^{2(a_m\cdot x - a_{jk}x_k)}} \ dx_1 \ldots dx_n.$$

Any ideas are very welcome. Thank you!

EDIT: the vectors $a_i$ are all distinct to each other. EDIT 2: the intervals of integration may be altered if necessary. Finite intervals may be used if this leads to some nice bound.

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  • $\begingroup$ I don't know what nice bound means in your question. $\endgroup$ Commented Mar 27, 2018 at 2:32

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Consider the case $n=2$ and $a_1 = (1, 1), a_2 = (1, 1)$. Fix $j=1, k=1$ you get \begin{align} \int^\infty_{-\infty} \int^\infty_{-\infty} \frac{1}{2e^{2y}}\ dxdy \end{align} which clearly doesn't converge.

The upper bound is trivially infinity.

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  • $\begingroup$ I thought it wasn't necessary to include this condition, but all $a_i$'s are distinct. Sorry for that. $\endgroup$
    – Integral
    Commented Mar 27, 2018 at 2:13
  • $\begingroup$ If the same issue remains, you may change the interval of integration to finite intervals, say $[0,1]$. $\endgroup$
    – Integral
    Commented Mar 27, 2018 at 2:28
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    $\begingroup$ Don't bother trying to improve your answer. It is good enough. I just realize I did a big mistake. This integral was supposed to be a expected value of a random vector with normal distribution, but I forgot to include the density function inside the integral.... I'm going to create another question. $\endgroup$
    – Integral
    Commented Mar 27, 2018 at 2:41

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