# Applying the Beukers-like irrationality proof for $\zeta(2)$/$\zeta(3)$ to Catalan's Constant: Where does it fail?

Many people have tried and failed to extend Apery's Irrationality proof of $$\zeta(3)$$ to Catalan's constant, by looking for a fast converging series for Catalan's constant analogous to the one for $$\zeta(3)$$ that Apery utilized:

{\begin{aligned}\zeta (3)&={\frac {5}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{{\binom {2k}{k}}k^{3}}}\end{aligned}}}.

My question is different and specifically relates to Beukers-like irrationality proof for $$\zeta(2)$$/$$\zeta(3)$$ as most clearly articulated recently by F. M. S. Lima in Beukers-like irrationality proofs for $$\zeta(2)$$ and $$\zeta(3)$$

For analogous Beukers-like irrationality proof applied in the case Catalan's Constant, does a proof fail in the initial lemma's or in evaluating and applying the unit square integral?

Lets rewrite the Lemma's applying to $$\zeta(2)$$ given by F. M. S. Lima and apply them to an analogous unit square integral for Catalan's Constant, $$G$$, that is "Lemma N" in the paper becomes "Lemma N_G" here:

Lemma 1G A unit square integral for Catalan's Constant $$\int_0^1 \int_0^1 \frac{1}{1+(x y)^2} \,dx\,dy= G$$

Lemma 2G ($$I_{2r,2r}$$) For all odd integers $$r>0$$ $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2r}}{1+(x y)^2} \,dx\,dy= G-\sum_{m=1}^{2r}\frac{(-1)^{m-1}}{(2m-1)^2}$$

Lemma 3G ($$I_{2r,2s}$$) Let r and s be non-negative odd integers, $$r\ne s$$. Then $$\int_0^1 \int_0^1 \frac{x^{2r}y^{2s}}{1+(x y)^2} \,dx\,dy=\frac{\widetilde{h_s}-\widetilde{h_r}}{2(r-s)}$$ where $$\widetilde{h_n}=\sum_{m=1}^n \frac{(-1)^{m-1}}{(2m-1)}$$, an alternating analog of the Harmonic Number.

Lemma 4G ($$I_{2r,2r}$$ as a linear form). For all odd integers $$r>0$$ $$I_{2r,2r}=G-\frac{z_{2r}}{(d_{2r})^2}$$ for some $$z_{2r} \in {\mathbb{N}}^*$$. Where $$d_{r}=lcm(1^2,3^2,5^2,...,r^2)$$

Lemma 5G ($$I_{2r,2s}$$ is a positive rational). For all odd $$r,s \in {\mathbb{N}},\, r \ne s,$$ $$I_{2r,2s}=\frac{z_{2r,2s}}{(d_{2r})^2}$$ for some $$z_{2r,2s}\in {\mathbb{N}^*}$$

Lemma 6G and Lemma 7G These are written in terms of $$n$$ and therefore both Lemma's can be written in terms of $$2n$$

If the analogous lemma's can be all proved correct for odd integer $$r,s>0$$ then presumably a proof in regards to $$G$$ must fail in the main part of the proof, what Lima labels Theorem 1, i.e. in the process of evaluating the new unit square integral and applying the result; which for me is the hardest part of this proof.

Added One difficulty is defining the even powered polynomials to multiply together - the resultant terms will have more than one factor of 2 so the lemmas above need to be modified.