$\DeclareMathOperator\GL{GL}$The classical Godement–Jacquet zeta integral is of this form:
$f$ is a matrix coefficient of a cuspidal automorphic representation of $\GL_n(\mathbb{A}_\mathbb{Q})$, and $\Phi$ is a Bruhat–Schwartz function on $M_n(\mathbb{A}_\mathbb{Q})$. We define:
$$Z(f,\Phi,s)=\int_{\GL_n(\mathbb{A}_\mathbb{Q})}f(g)\Phi(g)\lvert\det(g)\rvert^s\mathrm{d}g.$$
I am recently considering this integral:
For $i=1,2,\dotsc,k$, $f_i$ is a matrix coefficient of a cuspidal automorphic representation of $\GL_{n_i}(\mathbb{A}_\mathbb{Q})$, and $\Phi_i$ is a Bruhat–Schwartz function on $M_{n_i}(\mathbb{A}_\mathbb{Q})$. $\GL_{n_i}^\mathbb{1}$ is defined to be the subgroup $\{\lvert\det(g)\rvert=1: g\in \GL_{n_i}(\mathbb{A}_\mathbb{Q})\}$. For $X\in\mathbb{R}$, $e^X$ is defined to be $(e^X,1,\dotsc,1)\in\mathbb{A}^\times$. We consider the following:
$$\int_{\prod_{i=1}^k \GL_{n_i}^\mathbb{1}}\prod_{i=1}^kf_i(g_i)\left(\int_\mathbb{R} e^{Xs} \prod_{i=1}^k\Phi_i(e^Xg_i)\mathrm{d}X\right) \mathrm{d}g_1\dotsm\mathrm{d}g_k.$$
Notice that we ignore the compact $\mathbb{Q}^\times\backslash\mathbb{A}^\mathbb{1}$.
This integral comes from some constant term computation in a regularization problem. This term appears when the dimension of a linear space is not enough (we have only one $\mathbb{R}$ in the integral). So I wish to show that this integral VANISHES (and then in my problem only the term with enough dimension is left).
I tried to compute the Archimedean term directly. For simplicity, we assume all $(\Phi_i)_\infty$ are $e^{\operatorname{tr} (^t(g_i)_\infty(g_i)_\infty)}$. Then the Archimedean term is:
$$\left|\sum_{i=1}^k\operatorname{tr} (^t(g_i)_\infty(g_i)_\infty)\right|^{-\frac{s}{2}}\frac{1}{2}\Gamma(\frac{s}{2}).$$
We find that it becomes a summation. It is very difficult to separate variables.
Please give me some help if there is any good idea.
