Irrationality of Zeta(3) (Beuker’s Proof)

We want to prove irrationality of $\displaystyle \zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}=1.20205690\cdots$

This is the only odd positive integer which is known to be irrational. First proof of irrationality was by Apery who used constructed some very good rational approximations to zeta(3) using some recurrence relations. This proof has a lot of connections to hypergeometric functions, modular forms and many interesting things!

Beukers found another proof using approximations by multiple integrals involving Legendre polynomials.

Beukers’s proof:

The proof is quite mysterious. We use some integrals representations of ${A_n +B_n \zeta(3)}$ for some integer ${A_n, B_n}$ and show that ${0< |A_n+B_n\zeta(3)| =o(1)}$ . Therefore ${\zeta(3)}$ cannot be rational.

Consider the integrals

$\displaystyle I_{r,s } =\int_{0}^{1} \int_{0}^{1}-\frac{\log (x y)}{1-x y} x^{r} y^{s} d x d y=\frac{1}{r-s} \sum_{k=1}^{r-s} \frac{1}{(s+k)^{2}}$

$\displaystyle I_{r, r}=\int_{0}^{1} \int_{0}^{1}-\frac{\log (x y)}{1-x y} x^{r} y^{r} d x d y=2\left(\zeta(3)-\sum_{k=1}^{r} \frac{1}{k^{3}}\right)=2 \sum_{k=0}^{\infty} \frac{1}{(r+k+1)^{3}}$

So for ${r>s}$ , ${I_{r,s }}$ is a rational number with denominator dividing ${\left(\text{lcm}[1,2, \cdots, n]\right)^3.}$

Proof:

$\displaystyle \int_{0}^{1} \int_{0}^{1}-\frac{\log (x y)}{1-x y} x^{r} y^{s} d x d y=-\int_{0}^{1}\left(\sum_{k=0}^{\infty} \int_{0}^{1} \log (x y) x^{r+k} y^{s+k} d x\right) d y$

$\displaystyle =-\sum_{k=0}^{\infty} \int_{0}^{1}\left(\frac{y^{s+k} \log y}{r+k+1}-\frac{y^{s+k}}{(r+k+1)^{2}}\right) d y$

$\displaystyle =\sum_{k=0}^{\infty}\left(\frac{1}{(r+k+1)(s+k+1)^{2}}+\frac{1}{(r+k+1)^{2}(s+k+1)}\right)$

where we used the integral

$\displaystyle \int_{0}^{1}(\log x) x^{a} d x = -\frac{1}{(a+1)^2}.$

If ${r >s}$ , we get

$\displaystyle I_{r,s} = \sum_{k=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{(s+k+1)^{2}}-\frac{1}{(r+k+1)^{2}}\right)=\frac{1}{r-s} \sum_{k=1}^{r-s} \frac{1}{(s+k)^{2}}$

If ${r=s}$ , we get

$\displaystyle I=2 \sum_{k=0}^{\infty} \frac{1}{(r+k+1)^{3}}=2\left(\zeta(3)-\sum_{k=1}^{r} \frac{1}{k^{3}}\right).$

Thus we get that for any integer polynomial ${P_n(X)}$ ,

$\displaystyle I_n= \int_{0}^{1} \int_{0}^{1} \frac{P_{n}(x) P_{n}(y) \log x y}{1-x y} d x d y=\frac{A_n+B_n\zeta(3)}{[1,2,3, \cdots n]^3}$

with ${A_n, B_n}$ integers.

Choice of polynomial $P_n(X)$ : We use Legendra polynomials

$\displaystyle P_{n}(x)=\frac{1}{n !} \frac{d^{n}}{d x^{n}}\left(x^{n}(1-x)^{n}\right)$

We have

$\displaystyle P_{n}(x)=\sum_{m=0}^{n}\left(\begin{array}{c} n \\ m \end{array}\right)\left(\begin{array}{c} n+m \\ n \end{array}\right)(-x)^{m}$

and hence it’s an integer polynomial.

We want to prove that the integral ${I_n}$ is non-zero and find bounds on it.

Proof that ${I_n}$ is non-zero.

$\displaystyle \frac{-\log x y}{1-x y}=\int_{0}^{1} \frac{d z}{1-(1-x y) z}$

$\displaystyle \implies I_{n}=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{P_{n}(x) P_{n}(y)}{1-(1-x y) z} d x d y d z$

$\displaystyle \frac{1}{n !} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} P_{n}(y) \frac{\frac{d^{n}}{d x^{n}}\left(x^{n}(1-x)^{n}\right)}{1-(1-x y) z} d x d y d z$

Integrating by parts gives with respect to ${x}$ gives

$\displaystyle I_{n}=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{(x y z)^{n}(1-x)^{n} P_{n}(y)}{\{1-(1-x y) z\}^{n+1}} d x d y d z$

Change of variables

$\displaystyle w=\frac{1-z}{1-(1-x y) z}$

so that

$\displaystyle z^{n}=\frac{(1-w)^{n}}{(1-(1-u v) w)^{n}}$

gives

$\displaystyle I_{n}=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{(1-x)^{n}(1-w)^{n} P_{n}(y)}{1-(1-x y) w} d x d y d w.$

Integration by parts with respect to ${y}$ gives

$\displaystyle I_{n}=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{x^{n}(1-x)^{n} y^{n}(1-y)^{n} w^{n}(1-w)^{n}}{\{1-(1-x y) w\}^{n+1}} d x d y d w.$

This expression is clearly positive.

Next we want to prove bounds on ${I_n}$ :

The function

$\displaystyle f(x, y, w)= \frac{x(1-x) y(1-y) w(1-w)}{1-(1-x y) w}$

is bounded by ${\frac{1}{27}}$ on the region.

$\displaystyle 1-w+x y w \geq 2 \sqrt{(1-w) x y w} \implies f(x, y, w) \leq \frac{1}{2} \sqrt{x}(1-x) \sqrt{y}(1-y) \sqrt{w(1-w)} \le \frac{1}{27}$

$\displaystyle I_n \le (27)^{-n} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{d u d v d w}{1-(1-u v) w}=(27)^{-n} \int_{0}^{1} \int_{0}^{1}-\frac{\log (u v)}{1-u v} d u d v=2(27)^{-n} \zeta(3)$

So we get

$\displaystyle 0 <I_n = \frac{A_n +B_\zeta(3)}{[1, 2, \cdots, n]^3} \le 2(27)^{-n} \zeta(3)$

which implies

$\displaystyle 0 < |A_n +B_n\zeta(3)| =o(1)$

which is not possible if ${\zeta(3)}$ was rational.

We can also prove irrationality of ${\zeta(2)}$ using similar integrals.

Starting with

$\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{x^{r} y^{r}}{1-x y} d x d y=\sum_{n=1}^{\infty} \frac{1}{(n+r)^{2}}$

$\displaystyle \int_{0}^{1} \int_{0}^{1} \frac{x^{r} y^{s}}{1-x y} d x d y=\frac{1}{r-s}\left\{\frac{1}{s+1}+\frac{1}{s+2}+\cdots+\frac{1}{r}\right\}$

we get that that

$\displaystyle I_2(n) = \int_{0}^{1} \int_{0}^{1} \frac{P_n(x) (1-y)^n}{1-x y} d x d y = \frac{A_n+B_n\zeta(2)}{[1,2,3, \cdots n]^2}$

for the same Legendre polynomial ${P_n(x)}$ .

Integrating by parts we get

$\displaystyle I_2(n)=(-1)^n\int_{0}^{1} \int_{0}^{1} \frac{(x(1-x) y(1-y))^{n}}{(1-x y)^{n+1}} d x d y$

It’s easy to see that this is non-zero and from the estimate

$\displaystyle \frac{x(1-x) y(1-y)}{1-x y} \le \left(\frac{\sqrt{5}-1}{2}\right)^{5}$

we have

$\displaystyle 0<|I_2(n)| \le \left(\frac{\sqrt{5}-1}{2}\right)^{5 n} \int_{0}^{1} \int_{0}^{1} \frac{1}{1-x y} d x d y \le \left(\frac{\sqrt{5}-1}{2}\right)^{5 n}\zeta(2).$

Questions:

Where are these integrals coming from? Is there some conceptual way to think about them?
How do we find them?
How good are these approximations?
What happens if we change the Legendre polynomials or introduce more parameters?

Share this:

Related

Leave a comment Cancel reply