Return to Question

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 $\leq n$?

I have a trick that can prove the statement in simple cases, but proving the full statement using this idea might require a very delicate argument. A more conceptual proof would be preferred.

The trick is as follows: consider a special case: $$I_{f,E}(s) = \int_0^1 x^s E(x) dx$$ if we let $0\leq \arg x < 2\pi$, then above equals $$(1-e^{2\pi i s})^{-1}\int_C x^s E(x)dx$$ with $C$ being contour .

Since $C$ doesn't pass through $0$, $\int_C x^s E(x)dx$ is now entire in $s$, and $(1-e^{2\pi i s})^{-1}$ has simple pole at $s\in \mathbb{Z}$, so the statement is true in this case. This can be generalized to $$I_{f,E}(s) = \int_{[0,1]^n} x_1^{s}\cdots x_n^{s} E(\boldsymbol{x}) dx_i = (1-e^{2\pi i s})^{-n} \int_{C^n} (x_1,\cdots,x_n)^s E(x)dx$$

As another example, again let $n=1$, $$I_{f,E}(s) = \int_0^1 x^s (1-x)^s E(x) dx$$ also satisfies the statement: break the integral into $\int_0^{1/2} + \int_{1/2}^1$, then apply above argument separately to these two parts.

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.

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) $\Re(s)>0$.

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 $\leq n$?

I have a trick that can prove the statement in simple cases, but proving the full statement using this idea might require a very delicate argument. So I am looking for a more conceptual proof.

The trick is as follows: consider the $n=1$ case: $$I_{f,E}(s) = \int_0^1 x^s E(x) dx$$ if we let $0\leq \arg x < 2\pi$, then above equals $$(1-e^{2\pi i s})^{-1}\int_C x^s E(x)dx$$ with $C$ being the following contour

Since $C$ doesn't pass through $0$, $\int_C x^s E(x)dx$ is now entire in $s$, and $(1-e^{2\pi i s})^{-1}$ has simple pole at $s\in \mathbb{Z}$, so the statement is true in this case. This can be generalized to $$I_{f,E}(s) = \int_{[0,1]^n} x_1^{s}\cdots x_n^{s} E(\boldsymbol{x}) dx_i = (1-e^{2\pi i s})^{-n} \int_{C^n} (x_1,\cdots,x_n)^s E(x)dx$$

As another example, again let $n=1$, $$I_{f,E}(s) = \int_0^1 x^s (1-x)^s E(x) dx$$ also satisfies the statement: break the integral into $\int_0^{1/2} + \int_{1/2}^1$, then apply above argument separately to these two parts.

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 $\leq n$?

I have a trick that can prove the statement in simple cases, but proving the full statement using this idea might require a very delicate argument. A more conceptual proof would be preferred.

The trick is as follows: consider a special case: $$I_{f,E}(s) = \int_0^1 x^s E(x) dx$$ if we let $0\leq \arg x < 2\pi$, then above equals $$(1-e^{2\pi i s})^{-1}\int_C x^s E(x)dx$$ with $C$ being contour .

Since $C$ doesn't pass through $0$, $\int_C x^s E(x)dx$ is now entire in $s$, and $(1-e^{2\pi i s})^{-1}$ has simple pole at $s\in \mathbb{Z}$, so the statement is true in this case. This can be generalized to $$I_{f,E}(s) = \int_{[0,1]^n} x_1^{s}\cdots x_n^{s} E(\boldsymbol{x}) dx_i = (1-e^{2\pi i s})^{-n} \int_{C^n} (x_1,\cdots,x_n)^s E(x)dx$$

As another example, again let $n=1$, $$I_{f,E}(s) = \int_0^1 x^s (1-x)^s E(x) dx$$ also satisfies the statement: break the integral into $\int_0^{1/2} + \int_{1/2}^1$, then apply above argument separately to these two parts.