Let $G = \operatorname{GL}_2$, with center $Z$, diagonal matrices $T$, and upper triangular unipotents $N$. Let $K$ be the standard maximal compact subgroup of $G(\mathbb A) = G(\mathbb A_{\mathbb Q})$. For a character $\omega$ of $\mathbb A^{\ast}/\mathbb Q^{\ast}$, trivial on $(0,\infty)$, define a character $\omega_s$ of $N(\mathbb A)Z(\mathbb A)T(\mathbb Q)T(\mathbb R)^0$ by
$$\omega_s( n \begin{pmatrix} a \alpha u & x\\ & a \beta v \end{pmatrix}) = \omega(a)|\frac{u}{v}|^{s} \tag{$x \in \mathbb A, a \in \mathbb A^{\ast}, \alpha, \beta \in \mathbb Q^{\ast}, u,v \in (0,\infty)$}$$
If we let $H(s) = \operatorname{Ind}_{N(\mathbb A)Z(\mathbb A)T(\mathbb Q)T(\mathbb R)^0}^{G(\mathbb A)} \omega_s$ (normalized Mackey induction), then this is a Hilbert space for $s \in i\mathbb R$, and the representation $\pi_s$ of $G(\mathbb A)$ by right translation is unitary.
If $\varphi$ is a smooth function on $G(\mathbb A)$, compactly supported modulo $Z(\mathbb A)$, and satisfies $\varphi(zg) = \omega(z)^{-1} \varphi(g)$, we can define a bounded linear operator $R(\varphi)$ on $H(s)$ by
$$R(\varphi)f(x) = \int\limits_{Z(\mathbb A) \backslash G(\mathbb A)} \varphi(y)f(xy)dy.$$
This operator should be a Hilbert-Schmidt operator of trace class, and I want to compute its trace. In the Corvallis article Forms of GL(2) from the analytic point of view, the trace computation is given on page 239.
This computation shows up in integrating the truncated continuous kernel in $L^2_{\textrm{cont}}(Z(\mathbb A)G(\mathbb A)\backslash G(\mathbb A),\omega)$. What I get here should cancel out with a term that is picked up in the truncated kernel on all of $L^2(Z(\mathbb A)G(\mathbb A)\backslash G(\mathbb A),\omega)$.
I haven't been able to make this formula come out. I'm guessing the process is to unwind the integral $R(\varphi)$ over, say, $N(\mathbb A)Z(\mathbb A)T(\mathbb Q)T(\mathbb R)^0$, and show that $R(\varphi)$ is like an integral operator, its trace being given by an integral over a diagonal. I haven't yet been able to obtain a quotient of $G(\mathbb A)$ whose integration is given by the above. My guess is something like $Z(\mathbb A)T(\mathbb A^{\ast 1}) \backslash G(\mathbb A)$, since
$$T(\mathbb A)/T(\mathbb A^{\ast 1})Z(\mathbb A) \cong \mathbb Q^{\ast} \times (0,\infty)$$
and from $G(\mathbb A) = T(\mathbb A)N(\mathbb A)K$ we should have an Iwasawa decomposition-like decomposition
$$\int\limits_{Z(\mathbb A)T(\mathbb A^{\ast 1}) \backslash G(\mathbb A)} = \int\limits_{T(\mathbb A)/T(\mathbb A^{\ast 1})Z(\mathbb A)} \space \space \int\limits_{N(\mathbb A)} \int\limits_K$$
I think I've got it. Let $H = N(\mathbb A)T(\mathbb R)^0 T(\mathbb Q)Z(\mathbb A)$. We can write
$$R(\varphi)f(x) = \int\limits_{Z(\mathbb A)\backslash G(\mathbb A)} \varphi(x^{-1}y) f(y)dy = \int\limits_{H \backslash G(\mathbb A)} \space \int\limits_{Z(\mathbb A) \backslash H} \varphi(x^{-1}hy) f(hy)dhdy $$
$$ = \int\limits_{H \backslash G(\mathbb A)} \space K(x,y) f(y)dy$$
where $K(x,y) = \int\limits_{Z(\mathbb A) \backslash H} \varphi(x^{-1}hy)\omega_s(y)dy$. For general reasons,
$$\operatorname{Tr} R(\varphi) = \int\limits_{H\backslash G(\mathbb A)} K(x,x)dx.$$
By some Iwasawa decomposition argument (see e.g. page 221 of the article), integration over $H \backslash G(\mathbb A)$ satisfies
$$\int\limits_{H \backslash G(\mathbb A)} K(x,x)dx = \int\limits_K \int\limits_{\mathbb A^{\ast 1}/\mathbb Q^{\ast}} K( \begin{pmatrix} a \\ & 1 \end{pmatrix}k, \begin{pmatrix} a \\ & 1 \end{pmatrix}k)d^{\ast}adk.$$
Note that
$$K( \begin{pmatrix} a \\ & 1 \end{pmatrix}k, \begin{pmatrix} a \\ & 1 \end{pmatrix}k) = \int\limits_{Z(\mathbb A) \backslash H} \varphi\Bigg[k^{-1} \begin{pmatrix} a^{-1} \\ & 1 \end{pmatrix}h \begin{pmatrix} a \\ & 1 \end{pmatrix}k\Bigg] \omega_s(h) dh$$
Also, $Z(\mathbb A) \backslash H$ is the product of $N(\mathbb A)$ with $\Big(Z(\mathbb A) \cap \big(T(\mathbb R)^0 T(\mathbb Q)\big) \Big) \backslash T(\mathbb R)^0 T(\mathbb Q)\big] \cong \mathbb Q^{\ast} \times (0,\infty)$. Therefore, the trace is
$$\int\limits_K \int\limits_{\mathbb A^{\ast 1}/\mathbb Q^{\ast}}\int\limits_{N(\mathbb A)} \sum\limits_{\alpha \in \mathbb Q^{\ast}} \int_0^{\infty} \phi\Bigg[k^{-1} \begin{pmatrix} a^{-1} \\ & 1 \end{pmatrix} \begin{pmatrix} t \alpha \\ & 1 \end{pmatrix} n\begin{pmatrix} a \\ & 1 \end{pmatrix} k\Bigg]\omega_s\begin{pmatrix} t \alpha \\ & 1 \end{pmatrix} dt dn d^{\ast}a dk.$$
We get a cancelation of $\begin{pmatrix} a \\ & 1 \end{pmatrix}$ for $a \in \mathbb A^{\ast 1}/\mathbb Q^{\ast}$, so the integral is independent of that variable. Also, the character $\omega_s$ evaluates to $|t|^{s+1/2}$. So, the trace is now
$$\operatorname{vol}(\mathbb A^{\ast 1}/\mathbb Q^{\ast}) \int\limits_K \int\limits_{N(\mathbb A)}\sum\limits_{\alpha \in \mathbb Q^{\ast}} \int_0^{\infty} \phi\Bigg[k^{-1}\begin{pmatrix} t \alpha \\ & 1 \end{pmatrix}n k\Bigg]|t|^{s+1/2} dt dn dk$$ which is exactly the required formula.