Lang $SL_2$ two formulas for Harish transform Let $G = SL_2$ and give it the standard Iwasawa decomposition $G = ANK$. Let: $$D(a) = \alpha(a)^{1/2} - \alpha(a)^{-1/2} := \rho(a) - \rho(a)^{-1}.$$ Now, Lang defines ($SL_2$, p.69) the Harish transform of a function $f \in C_c(G,K)$ to be $$Hf(a) := \rho(a)\int_Nf(an)dn = |D(a)|\int_{A\setminus G} f(x^{-1}ax)d\dot x$$
My trouble is in understanding why the two definitions agree for $\rho (a)≠1$. In the second integral, we're are integrating over $A\setminus G$, so we can write $x = nk$, whence $$f(x^{-1}ax) = f((nk)^{-1}ank) = f(n^{-1}an) $$ since $f \in C_c(G,K)$, i.e. it is invariant w.r.t. conjugation by elements in $K$. But now, I don't know who to get rid of the remaining $n^{-1}$ and get the factor before the integral.

This equality is not at all obvious. Just before that section, it was proven that
$$
\int_{A\backslash G} f(x^{-1}ax)\;dx\;=\; {\alpha(a)\over |D(\alpha)}
\int_K\int_N f(kank^{-1})\;dn\;dk
$$
for arbitrary $f\in C_c(G)$. For $f$ left and right $K$-invariant, the outer integral goes away, leaving just the integral over $N$.
The key point is the identity proven another page or two earlier, something like
$$
\int_N f(a^{-1}nan^{-1})\,dn\;=\; {1\over |\alpha(a)^{-1}-1|}\int_N f(n)\,dn
$$
which follows from multiplying out.
