Irrationality of zeta constants

In this note I want to present a beautiful proof, due to Beukers, of the following result of Apéry. For comparison, the irrationality of $\zeta(2k)$ for all $k\geq 1$ has been known for centuries (at least since Euler), and the irrationality of any odd $k\geq 3$ remains an open question. Apéry's proof is, itself, quite amazing, and I recommend the account given in van der Poorten's survey [vdPo]. The alternative and ingenious proof I present here is taken entirely from the short paper of Beukers [Be].

The proof revolves around the fact that for any $z\in[0,1]$ there are two useful ways to write $\log z$: one is as the derivative of $z^\sigma$ (with respect to $\sigma$) at $\sigma=0$. The other is as \[-\int_z^{1} \frac{1}{u}\mathrm{d}u=-(1-z)\int_0^1 \frac{1}{1-(1-z)t}\mathrm{d}t.\]

Step One

The key object in this proof is the following integral, for any integral $r,s\geq 0$: \[I_{r,s} = \int_0^1\int_0^1 \frac{\log xy}{1-xy}x^ry^s\mathrm{d}x\mathrm{d}y.\] To understand this integral, we will use the first representation of $\log xy$ detailed above, and 'differentiate under the integral sign'. We thus consider the related integral, for any $\sigma>0$, \[\int_0^1\int_0^1 \frac{x^{r+\sigma}y^{s+\sigma}}{1-xy}\mathrm{d}x\mathrm{d}y.\] Writing $1-xy$ as a geometric series and integrating term by term, we see that this integral is equal to \[\sum_{k=0}^\infty \frac{1}{(k+r+\sigma+1)(k+s+\sigma+1)}.\] When $r>s$, this sum telescopes, to \[\sum_{k=0}^{\infty}\frac{1}{r-s}\left(\frac{1}{k+s+\sigma+1}-\frac{1}{k+r+\sigma+1}\right)=\frac{1}{r-s}\left( \frac{1}{s+1+\sigma}+\cdots+\frac{1}{r+\sigma}\right).\] To recover $I_{r,s}$ we differentiate with respect to $\sigma$ and then set $\sigma=0$, so that (when $r> s$) \[I_{r,s} = -\frac{1}{r-s}\left( \frac{1}{(s+1)^2}+\cdots+\frac{1}{r^2}\right).\] (We could instead just set $\sigma=0$ without differentiating, and then the resulting analogous proof would yield the irrationality of $\zeta(2)$.) In particular, and this is all we will need, $I_{r,s}$ can be written as $A/d_r^3$, where $A$ is some integer (depending on both $r$ and $s$) and where $d_r$ is the lowest common multiple of $1,\ldots,r$.

On the other hand, when $r=s$, the series does not telescope, but again by differentiating and setting $\sigma=0$, we get that \[I_{r,r}=\sum_{k=0}^\infty \frac{-2}{(k+r+1)^3}=-2\left( \zeta(3)-\sum_{k=1}^r\frac{1}{k^3}\right).\] In particular, $I_{r,r}=(A+B\zeta(3))/d_r^3$ for some integers $A,B$ (depending only on $r$). This is the way in which $\zeta(3)$ enters our proof. Note, for example, that in particular \[-2\zeta(3)=\int_0^1\int_0^1 \frac{\log xy}{1-xy}\mathrm{d}x\mathrm{d}y,\] which should be compared to the well-known integral representation of $\zeta(2)$ as $\int_0^1\int_0^1\frac{1}{1-xy}\mathrm{d}x\mathrm{d}y$.

Step Two

We now consider the integral \[J_n=-\int_0^1\int_0^1 \frac{\log xy}{1-xy}P_n(x)P_n(y)\mathrm{d}x\mathrm{d}y,\] where $P_n$ is some polynomial of degree $n$, with integer coefficients, that we will choose later. By Step One, this integral is equal to $(A+B\zeta(3))d_n^{-3}$ for some integers $A,B\in \mathbb{Z}$. On the other hand, using the second representation of $\log$ to get \[\frac{\log xy}{1-xy} = -\int_0^1 \frac{1}{1-(1-xy)z}\mathrm{d}z,\] we can write this integral as the triple integral \[\int_{0}^1\int_{0}^1\int_0^1 \frac{P_n(x)P_n(y)}{1-(1-xy)z}\mathrm{d}x\mathrm{d}y\mathrm{d}z.\] We will tackle this by partial integration in the $x$ and $y$ variables. Given the outcome of partial integration, for the simplest outcome it makes sense to choose $P_n(x)=f^{(n)}(x)$ for some polynomial $f(x)$ such that $f^{(j)}(0)=f^{(j)}(1)=0$ for all $0\leq j\leq n$. The simplest such choice is $f(x)=\frac{1}{n!}x^n(1-x)^n$ (note that we can introduce the $1/n!$ factor while ensuring that $P_n$ still has integer coefficients).

Doing $n$ partial integrations with $x$, then another $n$ with $y$, we get that \[J_n=\int_{0}^1\int_{0}^1\int_0^1 \frac{x^n(1-x)^ny^n(1-y)^nz^n(1-z)^n}{(1-(1-xy)z)^{n+1}}\mathrm{d}x\mathrm{d}y\mathrm{d}z.\] Let \[g(x,y,z)=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}.\] We want to get an upper bound for this function, which is just a straightforward calculus maximisation problem. Here is one possible approach. To simply matters, let $w=1-(1-xy)z$, so that $z=(1-w)/(1-xy)$, and \[g(x,y,z)= \frac{xy(1-x)(1-y)(1-w)(w-xy)}{(1-xy)^2w}.\] If the value of $xy$ is fixed, then calculus verifies that the maximum value of $(1-x)(1-y)=1-(x+y)+xy$ occurs when $x=y$, since the maximum of $x\mapsto x+u/x$ occurs at $x^2=u$. Therefore $g(x,y,z)$ is maximised when $x=y$, and hence \[g(x,y,z)\leq \frac{x^2(1-x)^2(1-w)(w-x^2)}{(1-x^2)^2w}=\frac{x^2}{(1+x)^2}\left(1-x^2-\frac{x^2}{w}-w\right).\] For any fixed $x\in(0,1)$, we can find the maximum of the expression in $w$ by differentiating, which again tells us that the maximum occurs when $w=x$, and so \[g(x,y,z)\leq \frac{x^2(1-x)^2}{(1+x)^2}.\] Finally, another application of calculus tells us that the maximum of $x(1-x)/(1+x)$ occurs at $x=\sqrt{2}-1$, and so finally we have \[g(x,y,z) \leq (\sqrt{2}-1)^4.\] In particular, for all $0\leq x,y,z\leq 1$, we have $g(x,y,z)\leq 2^{-5}$, and so \[J_n\leq 2^{-5n}\int_{0}^1\int_{0}^1\int_0^1 \frac{1}{1-(1-xy)z}\mathrm{d}x\mathrm{d}y\mathrm{d}z\ll 2^{-5n}.\] (In fact using Step One again it's easy to check that the implicit constant is $2\zeta(3)$, but that is irrelevant here.)

Step Three

Now we're basically done. The integral $J_n$ from the previous section is not zero (since the integrand is non-negative and there is clearly a positive contribution from some small interval around $(1/2,1/2,1/2)$, for example), and is equal to $(A_n+B_n\zeta(3))d_n^{-3}$ for some integers depending only on $n$, and is bounded above by $O(2^{-5n})$. It follows that, using the elementary bound $d_n< n^{\pi(n)}< 3^n$ for sufficiently large $n$, \[0<\lvert A_n+B_n\zeta(3)\rvert \ll 3^{3n}2^{-5n}.\] Here (and only here!) is where the supposed rationality of $\zeta(3)$ comes in. If $\zeta(3) = p/q$ then the left-hand side is at least $1/q$ (regardless of the values of $A_n$ and $B_n$, since a non-zero rational with denominator $q$ is at least $1/q$ from the origin), but since $3^3 < 2^5$ the right-hand side decays exponentially with $n$, and so we have a contradiction for large enough $n$.