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\begin{document}

I have tried to find the formula for $\delta_1^n\delta_2^m G_k^\HH$ using the second expansion at infinity. However I only succeded in calculating $\delta_1^n G_k^\HH$.

Recall, that (see [WW]) $\calQ_{k-1}$ has the following expansion at infinity:
\[
\begin{split}
\calQ_{k-1}(t) = \frac{\pi^{1/2} \Gamma(k)} {(2 t)^k \Gamma(k+\frac12)} F(\frac{k}2, \frac{k+1}2; k+\frac12; t^{-2}) \\
= \frac{2^{k-1} (k-1)!^2}{(2k-1)!} t^{-k} F(\frac{k}2, \frac{k+1}2; k+\frac12; t^{-2}),
\end{split}
\]
here $F$ denotes the hypergeometric series.
We are going to compute various derivatives of $G_k^\HH$. For this purpose we introduce the following function of weight $-2$ in $\tau_1$ and $0$ in $\tau_2$:
\[
Q_{\tau_2}(\tau_1) = \frac{(\tau_1 - \tau_2)(\tau_1 - \bar\tau_2)}{\tau_2-\bar\tau_2},
\]
we also assign to $t(\tau_1, \tau_2)$ weight $0$ in both variables.
This function is invariant (taking the weight into account) with respect to the simultaneous action of $SL_2(\R)$ on both arguments. We compute:
\[
\delta_1 Q_{\tau_2}(\tau_1) = \frac{\partial Q_{\tau_2}(\tau_1)}{\partial\tau_1} - \frac{2}{\tau_1-\bar\tau_1} Q_{\tau_2}(\tau_1) = t,
\]
\[
\delta_1 t = \frac{\partial t}{\partial\tau_1} = 2\frac{(\bar\tau_1-\tau_2)(\bar\tau_1-\bar\tau_2)}{(\tau_1-\bar\tau_1)^2(\tau_2-\bar\tau_2)} = \frac{t^2 - 1}2 Q_{\tau_2}(\tau_1)^{-1},
\]
\[
\delta_1^2 t = t \frac{t^2 - 1}2 Q_{\tau_2}(\tau_1)^{-2} - t \frac{t^2 - 1}2 Q_{\tau_2}(\tau_1)^{-2} = 0.
\]
Using the following formula for the derivative of the hypergeometric series:
\[
\frac{\partial F(a,b;c;x)}{\partial x} = a \frac{F(a+1,b;c;x)-F(a,b;c;x)}{x},
\]
we find, that
\begin{multline*}
\frac{\partial t^{-m} F(\frac{m}2, \frac{m+1}2; c; t^{-2})}{\partial t} = 
-m t^{-m-1} F(\frac{m}2 + 1, \frac{m+1}2; c; t^{-2}) \\= -m t^{-m-1} F(\frac{m+1}2, \frac{m+2}2;c;t^{-2}), 
\end{multline*}
so
\begin{multline*}
\delta_1^n G_k^\HH(\tau_1, \tau_2) = (-1)^{n+1} 2^k \frac{(k-1)! (k+n-1)!}{(2k-1)!} \times \\ 
t^{-k-n} F(\frac{k+n}2,\frac{k+n+1}2;k+\frac12;t^{-2}) \left(\frac{t^2-1}2\right)^n Q_{\tau_2}(\tau_1)^{-n}.
\end{multline*}
In particular, when $n=k$ we have
\begin{multline*}
\delta_1^k G_k^\HH(\tau_1, \tau_2) = (-1)^{k+1} 2^k (k-1)! \times \\ 
t^{-2k} F(k, k+\frac12;k+\frac12;t^{-2}) \left(\frac{t^2-1}2\right)^k Q_{\tau_2}(\tau_1)^{-k},
\end{multline*}
and using the identity
\[
F(a, b; b; x) = (1-x)^{-a}
\]
we obtain
\begin{equation*}
\delta_1^k G_k^\HH(\tau_1, \tau_2) = (-1)^{k-1} (k-1)! Q_{\tau_2}(\tau_1)^{-k}.
\end{equation*}

\end{document}@