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Let $\delta>0$ be a small real number and consider the $k$-dimensional region consisting of points for which $$\delta\leq x_1\leq x_2\leq\ldots \leq x_k$$ and $$x_1+\ldots+x_k\leq 1.$$ I am interested in the integral of $\frac{1}{x_1x_2\ldots x_k}$ over this region. I am also interested in generalising this slightly by replacing the second constraint with $$x_1+\ldots+nx_k\leq 1,$$ for some $n>0$.

I would like to evaluate these integrals, or give an upper bound which is of the correct order of magnitude, for $k\approx 100$. Currently, I can get an upper bound of $\delta^{-k}$ times the volume of the region, the volume can be computed by changing variables to $(x_1-\delta,x_2-x_1,\ldots,x_k-x_{k-1})$. However, I suspect that this is a considerable overestimate.

I would be interested to know if these integrals can be evaluated exactly. I believe they could be expressed in terms of polylogarithms, but even for $k=3$ the expressions appear to be very messy. If not are there any suitable numerical methods I could use?

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  • $\begingroup$ As $\frac{dx}{x}=d\log x$, the logarithmic change of variable $t_i=\log x_i$ transforms your integral into the calculation of a volume of a domain $$\log \delta\le t_1\le \dots \le t_k, \, e^{t_1}+\dots+e^{t_k} \le 1.$$ In particular, as $t_j\le 0$, the new domain is thus a subset of a cube $[\log \delta, 0]^k$, and you get an upper estimate by $(\log \delta^{-1})^k$; taking into account the order of $t_i$'s, you can further reduce it $k!$ times (as it is a subset of the corresponding simplex). $\endgroup$ – Victor Kleptsyn Nov 6 '14 at 20:51
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Let's just deal with the constraint $x_1+\ldots + x_k \le 1$. You can give a pretty good bound as follows: For any $\alpha >0$ the integral you want is at most $$ \frac{1}{k!} \int_{x_1,\ldots,x_k \ge \delta} \exp(\alpha(1-x_1-\ldots-x_k)) \frac{dx_1}{x_1}\cdots \frac{dx_k}{x_k}= \frac{e^{\alpha}}{k!} \Big( \int_{\delta}^{\infty} e^{-\alpha x} \frac{dx}{x}\Big)^{k}. $$ Now choose $\alpha$ to minimize this quantity. For example, if $\delta \le 1/(ke^2)$, say, then choosing $\alpha$ to be $k/\log (1/(k\delta))$, one gets an upper bound of $$ \frac{1}{k!} \Big( \log \frac{1}{k\delta} +\log \log \frac{1}{k\delta} + O(1)\Big)^k. $$ Clearly by just focussing on the region $x_j \le 1/k$ we also have the lower bound $$ \frac{1}{k!} \Big( \log \frac{1}{k\delta}\Big)^k. $$

This idea crops up in many places -- e.g. as Rankin's trick in number theory, or in the Chernov bounds. The upper bound is usually pretty accurate. Also, one can in many situations refine the idea to get an asymptotic. In this problem, you can try to use Laplace inversion now to estimate the integral you need. The idea would be to integrate along a vertical line with real part $\alpha$, and then the method of stationary phase would apply.

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Use symmetry to remove the ordering of the variables, so that the integral in question is

$$\frac{1}{k!}\displaystyle\idotsint\limits_{\substack{x_1,\ldots, x_k\ge\delta\\ x_1+\cdots+x_k\le 1}}\frac{dx_1}{x_1}\cdots\frac{dx_k}{x_k}.$$

I prefer to make the change of variables $t_i=x_i/\delta$ and to let $\delta=1/u$, so that the integrals are

$$ K_k(u):=\frac{1}{k!}\displaystyle\idotsint\limits_{\substack{t_1,\ldots, t_k\ge 1\\ t_1+\cdots+t_k\le u}}\frac{dt_1}{t_1}\cdots\frac{dt_k}{t_k}.$$

Soundararajan showed that,

$$ K_k(u)=\sum_{r=0}^{k}\frac{(-1)^r}{(k-r)!}C_r\log^{k-r}u+O_{k}\left(\frac{\log^k u}{u}\right),$$

where the constants, $C_r$, are given by the generating function

$$ \sum_{r=0}^{\infty}C_r z^r=\frac{e^{\gamma z}}{\Gamma(1-z)}.$$

With a little more work, found here, you find that for any $N\ge0$,

$$ K_k(u)=\sum_{j=0}^{N}\sum_{m=0}^{k}\sum_{r=0}^{k-m}\frac{(-1)^{r}}{m!(k-m-r)!}E_{j,m}C_r\left(\log^{k-m-r} u\right)^{(j)}+O_{k,N}\left(\frac{\log^k eu}{u^{N+1}}\right),$$

where the constants, $E_{j,m}$, are given by the generating function

$$\sum_{j=0}^{\infty}E_{j,m}z^j=\left(\int_{0}^{z}\frac{1-e^{-t}}{t}dt\right)^m.$$

The convergence is pretty quick, and avoids the complicated polylog expressions you mentioned.

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