= Solution
Set $r=k-l$. Since $k$ and $l$ are even and $k\geq l+6$, one has $r\geq6$. The holomorphic <Eisenstein series> in the question is absolutely convergent and decomposes as
$$
G_r(\tau)
=2\zeta(r)\sum_{\gamma\in\Gamma_\infty\backslash\Gamma(1)}j(\gamma,\tau)^{-r},
$$
because every nonzero integer pair is a positive multiple of a primitive pair and the two signs contribute the factor two.
Absolute convergence, including that established in part (c) at $s=k-1$, permits <Rankin–Selberg unfolding>. Unfolding the <Petersson inner product> from the fundamental domain to the strip $0\leq x<1$, $y>0$, gives
$$
\langle f,gG_r\rangle
=2\zeta(r)\int_0^\infty\int_0^1
f(\tau)\overline{g(\tau)}y^{k-2}\,dx\,dy.
$$
The $x$-integral uses <orthogonality of complex exponentials> to retain equal Fourier indices:
$$
\int_0^1f(\tau)\overline{g(\tau)}\,dx
=\sum_{n\geq1}a_n\overline{b_n}e^{-4\pi ny}.
$$
Finally,
$$
\int_0^\infty e^{-4\pi ny}y^{k-2}\,dy
=\frac{\Gamma(k-1)}{(4\pi n)^{k-1}}.
$$
Substitution gives the <Rankin–Selberg unfolding identity for a holomorphic Eisenstein series>
$$
\langle f,gG_{k-l}\rangle
=\frac{2\zeta(k-l)\Gamma(k-1)}{(4\pi)^{k-1}}L(f,g,k-1).
$$
Back to article page
