Given $k$ points $\{p_1,\cdots, p_k\}$ in $\mathbb{R}^n$ and positive constants $r_1, ..., r_k$ and another positive constant $\alpha>0$. Is there a way to compute/approximate the following (Riemann) integral in terms of $\{p_1, ..., p_k\}$, $\{r_1, ..., r_k\}$ and $\alpha$? $$ \int_{\mathbb{R}^n} \frac{dx}{1+\alpha(e^{r_1\lVert x - p_1\rVert^2}+\cdots +e^{r_k\lVert x-p_k\rVert^2})} $$ where $\lVert v\rVert$ denotes the standard $2$-norm in $\mathbb{R}^n$. For example, in $\mathbb{R}^2$, I am considering integrals like $$ \int_{\mathbb{R}^2}\frac{dxdy}{1+2(e^{2\lVert (x,y)-(3,4)\rVert^2}+e^{5\lVert(x,y)-(6,7)\rVert^2})}. $$ By "approximating the integral", I mean a sequence of expressions $\{f_i(p_1, ..., p_k, r_1, ..., r_k, \alpha)\}_{i = 1}^\infty$ in terms of $\{p_1, ..., p_k\}$, $\{r_1, ..., r_k\}$ and $\alpha$, such that $f_i$ converges to the value of this integral as $i\to\infty$. Any idea or useful reference to related topics is welcome. Thanks
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$\begingroup$ Do you really want to use $k$ points and let $k\rightarrow \infty$ here, or did you inadvertently reuse "$k$" for two separate quantities? $\endgroup$– Michael EngelhardtCommented Oct 16, 2019 at 3:06
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$\begingroup$ @MichaelEngelhardt Confusion fixed. Thanks for pointing out. I made a mistake by using the same notation $k$. $\endgroup$– Min WuCommented Oct 16, 2019 at 3:48
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$\begingroup$ I believe (though do not know how to prove) there is no analytical way to evaluate this integral. Does anyone have any idea for how to approximate it? $\endgroup$– Min WuCommented Oct 18, 2019 at 3:05
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$\begingroup$ Well ... probably this is not the answer you want to hear, but in principle, any valid numerical method implemented on a computer to calculate your integral generates just such a sequence as you describe. Of course, the sequence probably can't be written down easily, and will be only for one fixed set of $p$, $r$ and $\alpha $ each time you use it ... $\endgroup$– Michael EngelhardtCommented Oct 18, 2019 at 3:17
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