$\int\limits_{\mathcal O_K^r} |f(x_1, ... , x_r)|^s dx_1 \cdots dx_r$ is holomorphic on $\textrm{Re}(s) > 0$

Question

Let $K$ be a nonarchimedean local field, $\mathcal O_K$ its ring of integers, $\varpi$ a uniformizer, and $f \in \mathcal O_K[X_1, ... , X_n]$. Is there is an easy way to see that the expression

$$\phi(s) = \int\limits_{\mathcal O_K^n} |f(x_1, ... , x_n)|^sdx_1 \cdots dx_r$$

converges and is holomorphic on $\textrm{Re}(s) > 0$? Convergence is not difficult. One can use the triangle inequality and the convergence of the given integral when $f$ is a monomial.

As far as I can tell, one can take a given polynomial $f$ and eventually explicitly calculate the given integral, and the result will be a rational function in $|\varpi|^s$, hence holomorphic.

For example, let $n = 1$, $\mathcal O_K = \mathbb{Z}_p$ for a prime $p$, and $f(x) = x$. Then

$$\int\limits_{\mathbb{Z}_p} |x|^s dx = \sum\limits_{k=0}^{\infty} \int\limits_{p^k \mathbb{Z}_p \setminus p^{k+1}\mathbb{Z}_p} p^{-ks}$$

Taking the measure of $\mathbb{Z}_p$ to be $1$, the measure of $p^k \mathbb{Z}_p \setminus p^{k+1}\mathbb{Z}_p$ will be $p^{-k} - p^{-k-1}$, and so we have

$$\sum\limits_{k=0}^{\infty} (p^{-k} - p^{-k-1})p^{-ks} = (1 - \frac{1}{p}) \sum\limits_{k=0}^{\infty} p^{(-1-s)k} = (1 - \frac{1}{p}) \frac{1}{1-p^{-1-s}}$$

When $f$ is a sum of monomials, the procedure is more complicated, because we must take into account the possible absolute values of an expression like $x+y$ when $x, y$ are of the same value. But one can still compute something like

$$\int\limits_{\mathbb{Z}_p \times \mathbb{Z}_p} |xy(x+y)|^s dxdy$$

to be holomorphic.

But I see no immediate general principle by which functions like this must be holomorphic.

Answer 1

I didn't know that I know how to do this but I do.

Let $\mu$ be the measure on $\mathcal O_K^n$, normalized so that $\mathcal O_K^n$ has measure $1$. Let

$$E_k = \{ (x_1, ... , x_n) \in \mathcal O_K^n : |f(x_1, ... , x_n)| = q^{-k} \}$$

for $k \geq 0$, and $q = |\omega|^{-1} < 1$. Then each $E_k$ is measurable, has measure less than $1$, and the given integral is equal to

$$\phi(s) = \sum\limits_{k=0}^{\infty} \mu(E_k) q^{-ks}$$

Each expression $f_k(s) = \mu(E_k)q^{-ks}$ is holomorphic on $\textrm{Re}(s) > 0$.

Fix $0 < \rho$, and let $M_k = q^{-k\rho}$. Then for $\textrm{Re}(s) \geq \rho$,

$$|f_k(s)| \leq |q^{-ks}| = q^{-k \textrm{Re}(s)} \leq M_k$$

with $$M_0 + M_1 + \cdots = \frac{1}{1 - q^{-\rho}} < \infty$$

By the Weierstrass M-test, the given sum converges absolutely and uniformly on $\textrm{Re}(s) \geq \rho$. In particular, the given sum converges uniformly on compact sets, which implies that $\phi(s)$ is holomorphic.