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Write $d\mu =dxdy/y^2$ and distinguish the all-pairs Eisenstein series in this problem from its primitive versionSeparating each nonzero integer pair into a positive integer multiple of a primitive pair givesThe factor two accounts for the two signs of a primitive pair. It must not be omitted in the Rankin–Selberg unfolding.

Because $f\overline{g}{y}^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\le x<1$, $y>0$:The $x$ integral of the product Fourier expansions is ${\sum }_{n\ge 1}a(n)\overline{b(n)}{e}^{-4\pi ny}$. The Gamma function integral then givesThese steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and modular cusp decay show the square-coefficient series converges for $\mathrm{Re}s>k$; the Cauchy-Schwarz inequality gives the same absolute-convergence range for the mixed series.

Put $w=s-k+1$ and defineThis is the zeta-completed Rankin-Selberg coefficient series. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.

The exponential decay of cusp forms at infinity dominates the polynomial growth in $y$ of the Eisenstein series and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb{C}$, with possible simple poles only at $s=k-1,k$, and its functional equation isIndeed $w\mapsto 1-w$ corresponds to $s\mapsto 2k-1-s$. If $⟨f,g⟩=0$, both pole residues vanish and the completion is entire.

The normalization of the poles can also be made explicit. The constant term of the given Eisenstein series isThe second term has residue one at $w=1$, since the pole of $\zeta (2w-1)$ has residue $1/2$; the functional equation gives residue minus one at $w=0$. Hence the residues of $\mathcal{R}$ at $k$ and $k-1$ are respectively $⟨f,g⟩$ and $-⟨f,g⟩$, the Petersson inner product.

It is important to separate this completed statement from the raw $F$. Inverting the completion gives meromorphic continuation of $F$, but zeros of $\zeta (2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow 2k-1-s$. At the two displayed completion poles, for a nontrivial modular cusp modular weight,The latter is finite because the pole of $\Gamma (w)$ cancels that of the integral at $w=0$, using $\zeta (0)=-1/2$. In particular, taking $f=g\ne 0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal{R}$, not the raw series with all completion factors suppressed.

Finally suppose the two $L$-series have Euler products. With first coefficients normalized to one, their coefficients are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Thereforeinitially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized coefficients inside it.

For the usual degree-two Hecke products of Question 3, take roots ${\alpha }_p,{\beta }_p$ with sum $a(p)$ and product $p^{k-1}$, and ${\gamma }_p,{\delta }_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the Euler factor of a coefficientwise product of Hecke eigenforms:Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta (2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.