Let $V$ be an $n$-dimensional simplex, let $f(\boldsymbol{x}) = f(x_1,\cdots,x_n)\in \mathbb{C}[x_1,\cdots,x_n]$ be a product of linear polynomials that is non-zero in interior of $V$. Also let $E(\boldsymbol{x})$ be an entire function.
Consider the integral $$I_{f,E}(s) = \int_V (f(\boldsymbol{x}))^s E(\boldsymbol{x}) dx_i$$ which is analytic on (at least) $\Re(s)\geq 0$.
Is it true that $I_{f,E}(s)$ admits a meromorphic continuation to $s\in \mathbb{C}$, and order of poles at negative integer $s$ is $\leq n$?
If $f$ is non-zero on whole $V$ (not only its interior), then the defining integral $I_{f,E}(s)$ converges for all $s\in \mathbb{C}$, and is an entire function. The claim is trivial in this case.
So more interesting case is when $f$ vanishes somewhere on boundary of $V$, for example, consider $$F_1(s) = \int_0^1 x^s E(x) dx$$ $$F_2(s) = \int_{0<x_1<x_2<x_3<1} [x_1 x_2 (x_1+x_2)(1-x_1-x_3)(x_1+x_2+x_3)]^s E(x_1,x_2,x_3) dx_i$$
the claim implies they have meromorphic continuation to $\mathbb{C}$, poles of $F_1$ at negative integer is (at most) simple, and for $F_2$, (at most) order $3$.
For $F_1(s)$, one could prove this using a special trick $$F_1(s) = (1-e^{2\pi i s})^{-1}\int_C x^s E(x)dx$$
with $C$ being contour 
.
Since $C$ doesn't pass through $0$, $\int_C x^s E(x)dx$ is entire in $s$, and $(1-e^{2\pi i s})^{-1}$ has simple pole at $s\in \mathbb{Z}$, so the statement is true for $F_1$.
One could perform ad hoc trick on $F_2$ to show claim is also true. However, I wonder if a more conceptual and uniform approach exists.
