A rather famous formula for the Reimann-Zeta function states:
$$\zeta(S) = \int_{[0,1]^S} \frac{1}{1- \prod \limits_{i=1}^{S}x_i} \ \prod_{i=1}^S \text{d}x_i$$
Now, I understand reasonably well why that is. Write $$\frac{1}{1- \prod \limits_{i=1}^{S}x_i} = \sum_{k=0}^\infty \left( \prod \limits_{i=1}^{S}x_i \right)^k$$
And then bring the integral to the front to get: $$\zeta(S)= \sum_{k=0}^\infty \prod_{i=1}^S \int_0^1x_i \ \text{d}x_i = \sum_{k=0}^\infty \left( \left. \left[ \frac{x_i^{k+1}}{k+1}\right] \right|_0^1 \right)^S = \sum_{k=0}^\infty \left(\frac{1}{k+1}\right)^S= \sum_{k=1}^\infty k^{-S} $$
The Zeta Function is obviously a complicated function but essentially we were able to take these multiple integrals and make them into a more understandable form.
So, I was considering an additive analog of this integral, and the first thing you might think of is:
$$\int_{[0,1]^n} \frac{1}{1-\sum \limits_{j=1}^{n}x_j} \ \prod_{j=1}^n \text{d}x_j$$
But then this is undefined at a whole slew of values, and doesn't much feel like an analog. What's is more analogous is if we divide our sum by $n$ so that again we only approach infinity when all our variables approach 1. So, consider:
$$S_n := \int_{[0,1]^n} \frac{1}{1- \sum \limits_{j=1}^{n} \frac{x_j}{n}} \ \prod_{j=1}^{n}x_j$$
Wolfram Alpha confirms:
$$S_1 = \infty \ \text{ (The integral Diverges)} \\ S_2 = \ln (16) \approx 2.77258872224 \\ S_3 = \ln \left( \frac{1594323 \sqrt{3}}{262144} \right) \approx 2.35461664694 \\ S_4 \approx 2.22918 $$
This is interesting, and I think it's possible $\lim \limits_{n \rightarrow \infty} S_n$ exists, but I don't know.
So, generally speaking, I'd like to know what methodologies one can use to simplify multiple integral like these. But I'm particularly interested in the example given.
For a little bit of background it may be helpful to know that I have familiarity with calculus and multivariable calculus, but almost familiarity with real or complex analysis.
I've tried some things to simplify this integral so far. The furthest I've gotten so far begins with rewriting our fraction as a geometric series.
Thus we have:
$$S_n = \int_{[0,1]^n} \frac{1}{1 - \sum \limits_{j=1}^{n} \frac{x_j}{n}} \ \prod_{j=1}^n \text{d}x_j = \int_{[0,1]^n} \sum_{k=0}^\infty \left( \sum_{j=1}^n \frac{x_j}{n} \right)^k \ \prod_{j=1}^n \text{d}x_j$$
Next we switch the order:
$$S_n = \sum_{k=0}^\infty \int_{[0,1]^n} \left( \sum_{j=1}^n \frac{x_j}{n}\right)^k \ \prod_{j=1}^n \text{d}x_j$$
Factor out the $\frac{1}{n}$ from the sum and expand via the multinomial theorem:
$$S_n = \sum_{k=0}^\infty \frac{1}{n^k}\int_{[0,1]^n} \left( \sum_{\alpha_1+\alpha_2+...+\alpha_n=k} \frac{k!}{\prod \limits_{i=1}^{n}(\alpha_i !)} \prod_{t=1}^n x_t^{\alpha_t}\right) \prod_{j=1}^n \text{d}x_j$$ $$= \sum_{k=0}^\infty \frac{1}{n^k} \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \frac{k!}{ \prod \limits_{i=1}^{n} (\alpha_i!)} \int_{[0,1]^n} \prod_{t=1}^n x_t^{\alpha_t} \ \prod_{j=1}^n\text{d}x_j$$ $$= \sum_{k=0}^\infty \frac{1}{n^k} \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \frac{k!}{ \prod \limits_{i=1}^{n} (\alpha_i!)} \prod_{t=1}^n \int_0^1 x_t^{\alpha_t} \ \text{d}x_t$$ $$= \sum_{k=0}^\infty \frac{1}{n^k} \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \frac{k!}{ \prod \limits_{i=1}^{n} (\alpha_i!)} \prod_{t=1}^n \left. \left[ \frac{x_t^{\alpha_t+1}}{\alpha_t+1} \right] \right|_0^1$$ $$ = \sum_{k=0}^\infty \frac{1}{n^k} \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \frac{k!}{ \prod \limits_{i=1}^{n} (\alpha_i!)} \prod_{t=1}^n \frac{1}{\alpha_t+1} $$ $$ = \sum_{k=0}^\infty \frac{k!}{n^k} \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \prod_{t=1}^n \frac{1}{(\alpha_t !)(\alpha_t+1)} $$
Effectively that's the best I can do. I don't think it's much better but this can be written:
$$S_n= \sum_{k=0}^\infty \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \prod_{t=1}^n \frac{\sqrt[n]{k!}}{(\alpha_t !)(\alpha_t+1)n^{k/n}} $$
Another potentially valuable idea is that we also have:
$$S_n= \sum_{k=0}^\infty \frac{1}{n^k}\sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} {k \choose \alpha_1, \alpha_2, ..., \alpha_n}\prod_{t=1}^n \frac{1}{\alpha_t+1} $$
$$ = S_n= \sum_{k=0}^\infty \frac{1}{n^k}\sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} {k \choose \alpha_1, \alpha_2, ..., \alpha_n}\left( \sum_{J \subseteq\{1,2,...n \}} \prod_{r \in J}\alpha_r\right)^{-1}$$
Note that we can from this derive convergence. We have:
$$S_n= \sum_{k=0}^\infty \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} \frac{{k \choose \alpha_1, \alpha_2, ..., \alpha_n}}{n^k}\left( \sum_{J \subseteq\{1,2,...n \}} \prod_{r \in J}r\right)^{-1}$$
Since:
$$ \sum_{\alpha_1 + \alpha_2 +...+ \alpha_n=k} {k \choose \alpha_1, \alpha_2, ..., \alpha_n} = n^k$$
We have:
$$\frac{{k \choose \alpha_1, \alpha_2, ..., \alpha_n}}{n^k}\leq 1$$
Now, there are $2^n-n$ subsets with cardinality not equal to 1, so very roughly speaking there are an exponential number of sets of $\alpha$ thus the quotient of that and a sum that should be much larger it will seemingly be at least exponentially larger since the quotients of two exponentials is an exponential. (Here we approximate taking the sum as a multiplication). So that's very hand-wavy but maybe it implies convergent.
We could prove convergence and that $\lim \limits_{n \rightarrow \infty} S_n$ exists if we could prove this is decreasing. I sort of think we could apply Liebniz's rule carefully to take the derivative to try this but I'm not sure.
If I could find a nice numeric conjecture for these values I think I could probably evaluate the successive integration using integration by parts and induction/recursion.
My Questions Are As Follows
How much more can we simplify these integrals to better understood or more elementary? (In other words can you simplify these integrals further?)
More generally what methods can be used to solve problems with multiple integrals like this?
Use the integral identity $$ \frac{1}{z}=\int_0^\infty ds\ e^{-sz}\ ,\qquad z>0 $$ to write $$ S_n := \int_{[0,1]^n} \frac{1}{1- \sum \limits_{j=1}^{n} \frac{x_j}{n}} \ \prod_{j=1}^{n}x_j=\int_0^\infty ds\ e^{-s}\int_{[0,1]^n} e^{s \sum \limits_{j=1}^{n} \frac{x_j}{n}} \ \prod_{j=1}^{n}x_j $$ $$ =\int_0^\infty ds\ e^{-s}\left[\int_0^1 dx\ e^{sx/n}\right]^n=n^n \int_0^\infty ds\ \frac{e^{-s}}{s^n}(e^{s/n}-1)^n\ . $$ Setting $s=n\tau$, we have $$ =n\int_0^\infty d\tau\ e^{n(-\tau+\log(e^\tau-1)-\log\tau)}\ , $$ which can be evaluated by a Laplace approximation. The exponent is a decreasing function, whose maximum is achieved at $\tau=0$. Expanding the exponent around $\tau=0$, the leading order is equal to $-t/2$, therefore the limit $$ S_\infty=\lim_{n\to\infty}n \int_0^\infty d\tau e^{-n\tau/2}=2\ . $$