Here's a nice little Mathematica routine for evaluating Tito's continued fraction with precision prec:
prec = 10^4;
y = N[4, prec];
c = y; d = 0; k = 1;
u = 1; v = y;
While[True,
c = 1 + u/c; d = 1/(1 + u d);
h = c*d; y *= h;
v += 96 k^2 + 8;
c = v + u/c; d = 1/(v + u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
u += 3 k (k + 1) + 1;
k++];
6/y
where I used the Lentz-Thompson-Barnett method for the evaluation.
For prec = 10^4, the thing evaluates in 120 seconds (via AbsoluteTiming[]), giving a result that agrees with $\zeta(3)$ to 10,000 digits.
One can consider the even part of Tito's CF, which converges at twice the rate of the original:
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
Here's Mathematica code corresponding to this CF:
prec = 10^4;
y = N[5, prec];
c = y; d = 0; k = 1;
While[True,
u = k^6;
v = (2 k + 1) ((17 k + 17) k + 5);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
k++];
6/y
For prec = 10^4, the thing evaluates in 70 seconds (via AbsoluteTiming[]). There may be further ways to accelerate the convergence of the CF, but I have yet to look into them.
Added, quite a bit later:
As it turns out, the even part I derived is precisely Apéry's CF for $\zeta(3)$ (thanks Américo!). Conversely put, Tito's CF is an extension of Apéry's CF. Here's how to derive Apéry's CF from Tito's CF (while proving convergence along the way).
We start from an equivalence transformation of Tito's CF. A general equivalence transformation of a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$
with some sequence $\mu_k, k>0$ looks like this:
$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$
Now, given a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cdots}}$$
one can transform this into a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
where $w_1=\dfrac{a_1}{b_1}$ and $w_k=\dfrac{a_k}{b_k b_{k-1}}$ for $k > 1$, where we used $\mu_k=\dfrac1{b_k}$. Applying this transformation to Tito's CF yields the CF
$$\cfrac{\frac32}{1+\cfrac{w_2}{1+\cfrac{w_3}{1+\cdots}}}$$
where $w_{2k}=\dfrac{k^3}{4(2k-1)^3}$ and $w_{2k+1}=\dfrac{k^3}{4(2k+1)^3}$. (You can easily demonstrate that this transformed CF and Tito's CF have identical convergents.)
At this point, we find that since the $w_k \leq\dfrac14$, we have convergence of the CF by Worpitzky's theorem.
Now, we move on to extracting the even part of this transformed CF. Recall that if a CF has the sequence of convergents
$$u_0=b_0,u_1=b_0+\cfrac{a_1}{b_1},u_2=b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2}},\dots$$
then the even part is the CF whose convergents are $u_0,u_2,u_4,\dots$ (Analogously, there is the odd part with the sequence of convergents $u_1,u_3,u_5,\dots$)
Now, given a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
its even part is the CF
$$b_0+\cfrac{w_1}{1+w_2-\cfrac{w_2 w_3}{1+w_3+w_4-\cfrac{w_4 w_5}{1+w_5+w_6-\cdots}}}$$
Thus, the even part of the previously transformed CF is given by
$$\cfrac{\frac32}{\frac54-\cfrac{\beta_1}{\delta_1-\cfrac{\beta_2}{\delta_2-\cdots}}}$$
where
$$\begin{align*}
\beta_k&=\frac{k^3}{4(2k-1)^3}\frac{k^3}{4(2k+1)^3}=\frac{k^6}{16(2k-1)^3(2k+1)^3}\\
\delta_k&=1+\frac{k^3}{4(2k+1)^3}+\frac{(k+1)^3}{4(2k+1)^3}=\frac{17k^2+17k+5}{4(2k+1)^2}
\end{align*}$$
We're almost there! We only need to perform another equivalence transformation, which I'll split into two steps to ease understanding. First, the easy one with $\mu_k=4$, which yields the CF
$$\cfrac{6}{5-\cfrac{16\beta_1}{4\delta_1-\cfrac{16\beta_2}{4\delta_2-\cdots}}}$$
The last step is to cancel out the odd integer denominators of the $\beta_k$ and $\delta_k$; to do this, we take $\mu_k=(2k+1)^3$; this finally yields the CF
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
and this is Apéry's CF.
For completeness, I present a formula for the odd part of Tito's CF, after some post-processing with a few equivalence transformations:
$$\zeta(3)=\frac32-\cfrac{81}{\lambda_1-\cfrac{\eta_1}{\lambda_2-\cfrac{\eta_2}{\lambda_3-\ddots}}}$$
where
$$\begin{align*}
\eta_k&=4\times(4k^4+8k^3+k^2-3k)^3=4\times10^3,\,4\times126^3,\dots\\
\lambda_k&=8\times(68k^6-45k^4+12k^2-1)=8\times34,\,8\times3679,\dots
\end{align*}$$
The formula is somewhat more complicated, and converges at the same rate as the even part.
Article $[1]$ by Alfred van der Poorten is a good starting point, the best I know of.
Convergence speed
I mean "brutal force" by summing handly or in a computer, the
original formula for $\zeta(3)$.
It is known that the given series converges very slowly. Apéry's rational approximation $a_n/b_n$ below improves the speed of convergence to $\zeta(3)$, while controlling the increase of the size of $b_n$.
Apéry stated the following equality:
$$\zeta (3)\overset{\mathrm{def}}{=}\sum_{n=1}^{\infty }\frac{1}{n^{3}}=\frac{5}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}.\tag{1}$$
In section 3 van der Poorten shows that
$$\sum_{n=1}^{N}\frac{1}{n^{3}}+\sum_{k=1}^{N}\frac{(-1)^{k-1}}{2k^{3}\binom{N}{k}\binom{N+k}{k}}=\frac{5}{2}\sum_{k=1}^{N }\frac{(-1)^{n-1}}{k^{3}\binom{2k}{k}},\tag{2}$$
from which $(1)$ follows by letting $N$ tend to infinity (see this answer of mine). The formula $(2)$ is not very intuitive though.
The second series in $(1)$ is a fast convergent series, faster by far than the defining series for the Apéry's constant $\zeta(3)$. There are even faster convergent series, obtained by techniques of convergence acceleration.
Proof of the irrationality
As far as the irrationality concerns, Apéry constructed two sequences $(a_n),(b_n)$ $^1$ whose ratio $a_n/b_n\to\zeta(3)$ and
- $2(b_{n}\zeta (3)-a_{n})$ satisfies $\lim\sup \left\vert 2(b_{n}\zeta (3)-a_{n})\right\vert^{1/n}\le(\sqrt{2}-1)^4 $.
- $b_{n}\in \mathbb{Z},2(\operatorname{lcm}(1,2,\ldots ,n))^{3}a_{n}\in
\mathbb{Z}$.
- $\left\vert b_{n}\zeta (3)-a_{n}\right\vert >0$.
This is enough to prove the irrationality of $\zeta (3)$ by contradiction. $[2]$.
Intuition
Given that in van der Poorten's words
Those who listened casually, or who were afflicted with being
non-Francophone, appeared to hear only a sequence of unlikely
assertion.
it is natural to ask: Where did these ideas come from? My tentative explanation is based on the following fact. A few years later after his proof $[3]$, Roger Apéry derived the rational approximation $a_n/b_n$ to $\zeta(3)$ by transforming the defining series for $\zeta(3)$ into a continued fraction and applying iterated transformations to it, which improved the speed of convergence, and obtained the recurrence relations satisfied by $a_{n},b_{n}$ $[4]$.
--
$^1$ The sequences are $$\begin{equation*}
a_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}c_{n,k},
\qquad b_{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2},\end{equation*}$$
where
$$\begin{equation*}
c_{n,k}=\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\frac{\left( -1\right)
^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\quad k\leq n.
\end{equation*}$$
References.
$[1]$ Poorten, Alf., A Proof that Euler Missed…, Apéry’s proof of the irrationality of $\zeta(3)$. An informal report, Math. Intelligencer 1, nº 4, 1978/79, pp. 195-203.
$[2]$ Fischler, Stéfane, Irrationalité de valeurs de zêta (d’ après Apéry, Rivoal, …), Séminaire Bourbaki 2002-2003, exposé nº 910 (nov. 2002), Astérisque 294 (2004), 27-62
$[3]$ Apéry, Roger (1979), Irrationalité de $\zeta2$ et $\zeta3$, Astérisque 61: 11–13
$[4]$ Apéry, Roger (1981) Interpolations de Fractions Continues et Irrationalité de certaines Constantes, Bull. section des sciences du C.T.H.S., n.º3, p.37-53
Best Answer
I think your problem lies in the interchanging of the product and the integral. Even if it wasn't an infinite product, the interchanging would still be invalid. Here's a simple example:
$$\int_0^1x\sin x\ dx\approx 0.30116867894$$ while $$\left(\int_0^1xdx\right)\left(\int_0^1 \sin x\ dx\right)\approx 0.229848847066,$$ so those are obviously not the same.
Although $$\int \sum_{i}f_i(x)\ dx=\sum_{i}\int f_i(x)dx$$ holds under special circumstances, very rarely (if ever) does $$\int\prod_{i}f_i(x)\ dx=\prod_i \int f_i(x)dx$$ hold.