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.
Best Answer
Let's record what is possible, so far. The case $n=2$, from the comments: $$\int_0^1\frac{dy}{x(x+y)}=\frac1{x^2}\int_0^1\frac{dy}{1+\frac{y}x}=\int_0^1\sum_{k\geq0}\frac{(-1)^ky^k}{x^{k+2}}dy=\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac1{x^{k+2}}.$$ Hence, $$\int_1^{\infty}\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac{dx}{x^{k+2}}=\sum_{k\geq0}\frac{(-1)^k}{(k+1)^2}=\frac{\zeta(2)}2.$$
The case $n=3$: $$\begin{aligned}\int_0^1\frac{dz}{x(x+y)(x+y+z)}&=\frac1{x(x+y)^2}\int_0^1\frac{dz}{1+\frac{z}{x+y}}\\&=\frac1x\sum_{k\geq0}\int_0^1\frac{(-1)^kz^kdz}{(x+y)^{k+2}}\\&=\frac1x\sum_k\frac{(-1)^k}{(k+1)(x+y)^{k+2}}.\end{aligned}$$ Hence, $$\frac1x\int_0^1\sum_k\frac{(-1)^kdy}{(k+1)(x+y)^{k+2}}=\frac1x\sum_k\frac{(-1)^k}{(k+1)^2}\left[\frac1{x^{k+1}}-\frac1{(x+1)^{k+1}}\right].$$ Now, integrate with respect to $x$ (standard): $$\int_1^{\infty}\frac{dx}{x^{k+2}}=\frac1{k+1} \qquad \text{and} \qquad \int_1^{\infty}\frac{dx}{x(x+1)^{k+1}}=\sum_{j=k+1}^{\infty}\frac1{j\cdot 2^j}.$$ Therefore, we compute the two series: $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}=\frac34\zeta(3) \qquad \text{and} \qquad \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{j=k}^{\infty}\frac1{j\cdot 2^j}=\frac{13}{24}\zeta(3).$$ We arrived at the valued predicted by Peter Mueller, $$\int_1^{\infty}\int_0^1\int_0^1\frac{dz\,dy\,dx}{x(x+y)(x+y+z)}=\frac5{24}\zeta(3).$$
Caveat. One may anticipate higher values of $n$ to scale up the challenge.
UPDATE. Regarding Fedor's question, one contention is a follows: the sum in question is a weight 3 polylog, so it is a rational combination of $\zeta(3), \zeta(2)\log 2$ and $\log^3(2)$. Since a large numerical agreement verifies equality with only $\frac{13}{24}\zeta(3)$, it must be the exact evaluation.
UPDATE. I like to address the request from GH from MO directed to Agno. Then, a direct answer to Fedor's question.
This time, we start integrating with respect to $x$: $$\int_1^{\infty}\frac{dx}{x(x+y)(x+y+z)} =\frac{y\,\log(1+y)+z\,\log(1+y)-y\,\log(1+y+z)}{yz(y+z)}.$$ Next, integrate in the variable $y$: $$\int_0^1\frac{y\,\log(1+y)+z\,\log(1+y)-y\,\log(1+y+z)}{yz(y+z)}\,dy =\frac{\text{Li}_2(z+2)-\text{Li}_2(z+1)-\text{Li}_2(2)}z;$$ where $\text{Li}_2(z)$ is the dilogarithm function $$\text{Li}_2(z)=\int_1^z\frac{\log t}{1-t}\,dt.$$ Finally, we integrate in the last variable $z$: $$\begin{align} \int_0^1\frac{\text{Li}_2(z+2)-\text{Li}_2(z+1)-\text{Li}_2(2)}z\,dz &=\int_0^1\left(\int_1^{z+2}\frac{\log t}{1-t}\,dt-\int_1^{z+1}\frac{\log t}{1-t}\,dt-\int_1^2\frac{\log t}{1-t}\,dt \right)\frac{dz}z \\ &=\int_0^1\left(\int_{z+1}^{z+2}\frac{\log t}{1-t}\,dt-\int_1^2\frac{\log t}{1-t}\,dt \right)\frac{dz}z \\ &=\int_0^1\left(\int_2^{z+2}\frac{\log t}{1-t}\,dt-\int_1^{z+1}\frac{\log t}{1-t}\,dt\right)\frac{dz}z \\ &=\int_2^3\frac{\log t}{1-t}\left(\int_{t-2}^1\frac{dz}z\right)dt-\int_1^2\frac{\log t}{1-t}\left(\int_{t-1}^1\frac{dz}z\right)dt \\ &=\int_2^3\frac{\log t\,\log(t-2)}{t-1}\,dt-\int_1^2\frac{\log t\,\log(t-1)}{t-1}\,dt \\ &=\int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt-\int_0^1\frac{\log(t+1)\,\log t}t\,dt \\ &=\int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt+\frac34\zeta(3). \end{align}$$ For the first integral in the last equality, write $\log(t+2)=\log 2+\log(1+\frac{t}2)$ and apply Taylor series: $$\begin{align} \int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt &=\log 2\int_0^1\frac{\log t}{t+1}\,dt+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\int_0^1\frac{t^n\log t}{t+1}\,dt \\ &=-\frac12\zeta(2)\,\log2+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\int_0^1\frac{t^n\log t}{t+1}\,dt \\ &=-\frac12\zeta(2)\,\log2+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\left[\frac{(-1)^{n-1}}2\zeta(2)+(-1)^{n-1}\sum_{k=1}^n\frac{(-1)^k}{k^2}\right] \\ &=-\frac12\zeta(2)\,\log2+\frac12\zeta(2)\sum_{n\geq1}\frac1{2^nn}+\sum_{n\geq1}\frac1{2^nn}\sum_{k=1}^n\frac{(-1)^k}{k^2} \\ &=-\frac12\,\zeta(2)\,\log2+\frac12\,\zeta(2)\,\log2+\sum_{n\geq1}\frac1{2^nn}\sum_{k=1}^n\frac{(-1)^k}{k^2} \\ &=\sum_{k\geq1}\frac{(-1)^k}{k^2}\sum_{n=k}^{\infty}\frac1{2^nn}. \end{align}$$ The above derivations indicate we do not need Agno's $\log(e^x\pm1)$ integrals, instead we got Robert Z's $\log$-integral which sends us to his useful link evaluating as $\frac{13}{24}\zeta(3)$.