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Let $G$ be a connected, reductive group over a $p$-adic field $k$, $A_0$ a maximal split torus of $G$, and $P = MU$ a parabolic subgroup with $M$ containing $A_0$. Let $\overline{P} = M \overline{U}$ be the opposite parabolic, and let $K$ be a maximal compact subgroup in good position relative to $P$ and $A_0$. Then we have $G = PK$, and for $g \in G$, we can write $g = u_P(g)m_P(g)k_P(g)$ with $u_p(g) \in U, m_P(g) \in M$, and $k_P(g) \in K$ (nonuniquely).

Let $\delta_P = q^{\langle 2 \rho, H_M(-) \rangle}$ be the modulus character for $P$. The map $\overline{U} \rightarrow \mathbb C^{\ast}$ given by $\bar{u} \mapsto \delta_P(m_P(\bar{u}))$ is well defined and continuous. In Waldspurger's writeup on Harish Chandra's unpublished notes on the Plancherel formula, he considers the integral

$$\gamma(P) = \int\limits_{\overline{U}}\delta_P(m_P(\bar{u})) \space d \bar{u}$$

Is it obvious that this integral converges? I considered the following example:

Example: $G = \operatorname{SL}_2(\mathbb Q_p)$, $P = MU$ the usual Borel, and $\overline{U}$ the usual opposite, and $K = \operatorname{SL}_2(\mathbb Z_p)$. From the Iwasawa decomposition, we can write (nonuniquely)

$$\bar{u} = \begin{pmatrix} 1 & \\ x & 1 \end{pmatrix} = \begin{pmatrix}1 & x^{-1} \\ & 1 \end{pmatrix} \begin{pmatrix} x^{-1} \\ & x \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & x^{-1} \end{pmatrix}$$

whenever $x \in \mathbb Q_p - \mathbb Z_p$. Thus $\delta_P(m_P(\bar{u})) = |x|^{-2}$ when $x \not\in \mathbb Z_p$, and $1$ for $x \in \mathbb Z_p$, giving us

$$\gamma(P) = \operatorname{meas}(\mathbb Z_p) + \sum\limits_{l=1}^{\infty} p^{-2l} \operatorname{meas}(p^{-l}\mathbb Z_p - p^{-l+1} \mathbb Z_p)$$

$$ = 1 + \sum\limits_{l=1}^{\infty}p^{-2l}(p^l-p^{l-1}) < 1 + \sum\limits_{l=1}^{\infty} p^{-2l}(p^l) = 1 + \sum\limits_{l=1}^{\infty} p^{-l} < \infty$$

if we normalize $\operatorname{meas}(\mathbb Z_p) = 1$. Maybe there is some way to reduce to the case of $\operatorname{Res} \operatorname{SL}_2$ or $\operatorname{SU}(3)$ where calculations will go similarly?

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