TITLE: An interesting integral expression for $\pi^n$?
QUESTION [54 upvotes]: I came on the following multiple integral while renormalizing elliptic multiple zeta values:
$$\int_0^1\cdots \int_0^1\int_1^\infty {{1}\over{t_n(t_{n-1}+t_n)\cdots (t_1+\cdots+t_n)}} dt_n\cdots dt_1.$$
Only the variable $t_n$ goes from $1$ to $\infty$, the others all go from $0$ to $1$.  Numerically, I seem to be getting a rational multiple of $\pi^n$. I would like to prove this. Has anyone ever seen an integral like this before?
Edit: After reworking out why I thought it would be a power of $\pi$, I now have a different integral, which numerically really does seem to give a rational multiple of $\zeta(n)$ for each $n$ (though I have only been able to go up to n=4 numerically).  I want to integrate $1/(z_1\cdots z_n)$ over the part of the simplex $0\le z_n\le \cdots \le z_1\le 1$ in which $|z_i -z_{i+1}|>\varepsilon$, then let $\varepsilon\rightarrow 0$.   The integral is $$\int_{n\epsilon}^{1-\varepsilon}\int_{(n-1)\varepsilon}^{z_1-\epsilon}\cdots \int_{\varepsilon}^{z_{n-1}-\varepsilon} {{1}\over{z_1\cdots z_n}} dz_n\cdots dz_1.$$ It actually diverges when $\varepsilon\rightarrow 0$, but it can be regularised like ordinary multizeta values by computing the integral as a power series in $ln(\varepsilon)$ and $\varepsilon$ and then taking its constant term.  It's related to the previous integral by the variable change $z_1=\varepsilon(t_1+\cdots+t_n),\ldots,z_n=\varepsilon t_n$, but the new bounds of the integral form an $(n+1)$-angled polyhedron, not a cube.

REPLY [4 votes]: As suggested by მამუკა ჯიბლაძე, for n=2,3 there is indeed an interesting connection of
$$I_n=\int\limits_0^1\cdots \int\limits_0^1 dt_1\cdots dt_{n-1}\int\limits_1^\infty dt_n \frac{1}{t_n(t_n+t_{n-1})\cdots(t_n+\cdots +t_1)}$$
with the Beukers-type integrals. Using the Feynman parametrization
$$\frac{1}{A_1A_2\cdots A_n}=(n-1)!\int\limits_0^1\cdots \int\limits_0^1\frac{\delta\left(1-\sum\limits_{i=1}^n x_i\right)dx_1\cdots dx_n}{(x_1A_1+x_2A_2+\cdots +x_nA_n)^n},$$
we get after the parametrization and subsequent trivial integration over $t_n$:
$$I_n=\frac{(n-1)!}{n-1}\int\limits_0^1\cdots \int\limits_0^1 \frac{dx_1\cdots dx_n dt_1\cdots dt_{n-1}\;\delta(1-x_1-\cdots-x_n)}{[1+t_{n-1}(x_{n-1}+\cdots x_1)+\cdots +t_1x_1]^{n-1}}.$$
In particular, for n=2 we get immediately 
$$I_2=\int\limits_0^1\int\limits_0^1 \frac{dx dt}{1+xt}=\frac{1}{2}\zeta(2),$$
and for n=3 we get
$$I_3=\int\limits_0^1\cdots\int\limits_0^1 \frac{dx_1 dx_2 dx_3dt_1 dt_2\;\delta(1-x_1-x_2-x_3)}{[1+t_2(x_1+x_2)+t_1x_1]^2}=$$ $$\iint\limits_0^1\frac{dx_1dx_2dx_3\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(x_1+x_2)}\left [\ln{(1+x_1+x_2)}+\ln{(1+x_1)}-\ln{(1+2x_1+x_2)}\right]=$$ $$\iiint\limits_0^1\frac{dx_1dx_2dx_3\;\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(1-x_3)}\left [\ln{(2-x_3)}+\ln{(1+x_1)}-\ln{(2+x_1-x_3)}\right].$$
Making $x=x_1,y=1-x_3,z=x_2$ change of variables in the integral, and
performing integration over $z$ thanks to the $\delta$-function, we get
$$I_3=\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right]\theta(y-x)=$$ $$
\frac{1}{2}\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right], \tag{1}$$
where in the last step we used the symmetry of the integrand and $\theta(y-x)+\theta(x-y)=1$. This result can be represented in the form
$$I_3=\frac{1}{2}\iiint\limits_0^1\frac{dx dy dz}{1+x+y+xyz}. \tag{2}$$
Note the resemblance with Beukers integrals
$$ \iiint\limits_0^1\frac{dx dy dz}{1-z+xyz}=2\zeta(3),\;\;
\iiint\limits_0^1\frac{dx dy dz}{1+xyz}=\frac{3}{4}\iiint\limits_0^1\frac{dx dy dz}{1-xyz}=\frac{3}{4}\zeta(3).$$
Analogously, for $I_4$ we get
$$I_4=\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+
\ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right]\theta(z-x)\theta(y-z).$$
$\theta(z-x)\theta(y-z)$ term is a consequence of integration with the help of the $\delta$-function and reflects $x_2=z-x>0$ and $x_3=y-z>0$ conditions. After symmetrization, we finally get
$$I_4=\frac{1}{6}\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+
\ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right],$$
and the analogy with Beukers-type integrals is less obvious, because it doesn't seem this can be represented in the form analogous to (2).
I don't know whether there are some further deeper connections to Beukers-type integrals, or to its generalizations (for example Vasilenko integrals - see section 8 in https://eudml.org/doc/249095).
P.S. From (1) we have 
$$I_3=\frac{1}{2}\int\limits_0^1f(x)\;d\ln{x}=-\frac{1}{2}\int\limits_0^1 \ln{x}\,\frac{df(x)}{dx}\;dx,$$
where
$$f(x)=\int\limits_0^1\frac{dy}{y}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right],$$
and
$$\frac{df(x)}{dx}=\int\limits_0^1 \frac{dy}{y}\left[\frac{1}{1+x}-\frac{1}{1+x+y}\right]=\frac{\ln{(2+x)}-\ln{(1+x)}}{1+x}.$$
Therefore
$$I_3=-\frac{1}{2}\left[\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}-\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}\right]dx=-\frac{1}{2}\left[-\frac{13}{24}+\frac{1}{8}\right]\zeta(3)=\frac{5}{24}\zeta(3),$$
because $$\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}dx=-\frac{13}{24}\zeta(3),$$ and $$\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\frac{1}{8}\zeta(3).$$
In the case of $I_4$, we analogously get
$$I_4=\frac{1}{6}\int\limits_0^1\int\limits_0^1\frac{\ln{x}\ln{y}}{(1+x+y)^2}\left[\ln{(2+x+y)}-\ln{(1+x+y)}+\frac{1}{2+x+y}\right],$$
however further progress doesn't seem feasible.