An interesting integral expression for $\pi^n$? 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.
 A: You can easily reduce to a double integral and in a few seconds you
obtain
0.0817241169677268267134627567961242656233303062217211895...
(all digits correct, I checked at higher accuracy), but still
cannot recognize it.
A: The following is a somewhat simplified version of the proof by T. Amdeberhan, Agno, and Robert Z that
$$\int_0^1 \int_0^1 \int_1^\infty \frac{dx\,dy\,dz}{x(x+y)(x+y+z)}=\frac{5}{24}\zeta(3).$$
We integrate with respect to $x$ first, then with respect to $y$:
\begin{align*}\int_0^1\int_1^\infty \frac{dx\,dy}{x(x+y)(x+y+z)}&=\int_0^1\left(\frac{\log(1+y)}{yz}-\frac{\log(1+y+z)}{(y+z)z}\right)dy\\
&=\int_0^1\frac{\log(1+t)}{tz}\,dt-\int_z^{1+z}\frac{\log(1+t)}{tz}\,dt\\
&=\int_0^z\frac{\log(1+t)}{tz}\,dt-\int_1^{1+z}\frac{\log(1+t)}{tz}\,dt.\\
\end{align*}
The integral of the right hand side with respect to $z$ equals, by Fubini,
\begin{align*}&=\int_0^1\int_0^z\frac{\log(1+t)}{tz}\,dt\,dz-\int_0^1\int_1^{1+z}\frac{\log(1+t)}{tz}\,dt\,dz\\
&=\int_0^1\int_t^1\frac{\log(1+t)}{tz}\,dz\,dt-\int_1^2\int_{t-1}^1\frac{\log(1+t)}{tz}\,dz\,dt\\
&=-\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt+\int_1^2\frac{\log(t-1)\log(t+1)}{t}\,dt\\
&=-\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt+\int_0^1\frac{\log(t)\log(t+2)}{t+1}\,dt.
\end{align*}
On the right hand side, the second integral equals $-\frac{13}{24}\zeta(3)$ by this MSE post, while the first integral equals
\begin{align*}\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt
&=\int_0^1\frac{\log(t)}{t}\sum_{n=1}^\infty\frac{(-1)^{n-1}t^n}{n}\,dt\\
&=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\int_0^1 t^{n-1}\log(t)\,dt=\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}=-\frac{3}{4}\zeta(3).
\end{align*}
To summarize,
$$\int_0^1 \int_0^1 \int_1^\infty \frac{dx\,dy\,dz}{x(x+y)(x+y+z)}=\frac{3}{4}\zeta(3)-\frac{13}{24}\zeta(3)=\frac{5}{24}\zeta(3).$$
