An Easier Way to Show ζ(3) ∉ Q

Stephen D. Miller^{∗} Department of Mathematics Yale University P.O. Box 208283 New Haven, CT 06520-8283

                      stephen.miller@yale.edu

October 19, 1998

In 1978 R. Apery proved that

ζ(3) = ∑_{n=1}^{∞} \frac{1}{n^{3}}

is an irrational number. His proof was shortened by F. Beukers, who translated it into equivalent statements about integrals of Legendre polynomials. Nevertheless, there is some “magic” in Beukers’ proof involving a complicated change of variables. In this note we will use a different integral than Beukers which gives the same approximation as Beukers’ (and Apery’s), but which is much easier to maniuplate.

1 Some Formulas

For the sake of reference, let us state some key integral formulas

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{a}}{1 − st}dsdt = ∑_{n=1}^{∞} \frac{1}{(n + a)^{2}}  (1)

∗The author was supported by an NSF Postdoctoral Research Fellowship during this work.

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{a} ln(st)}{1 − st}dsdt = −2 ∑_{n=1}^{∞} \frac{1}{(n + a)^{3}}  (2)

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{a} ln(t)}{1 − st}dsdt = − ∑_{n=1}^{∞} \frac{1}{(n + a)^{3}} .  (3)

If a ≠ b,

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{b}}{1 − st}dsdt = \frac{1}{b − a} ∑_{n=1}^{∞} (\frac{1}{n + a} − \frac{1}{n + b})  (4)

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{b} ln(s)}{1 − st}dsdt = \frac{1}{a − b} ∑_{n=1}^{∞} \frac{1}{(n + a)^{2}} + \frac{1}{(a − b)^{2}} ∑_{n=1}^{∞} (\frac{1}{n + a} − \frac{1}{n + b})

(5)

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{b} ln(t)}{1 − st}dsdt = \frac{1}{b − a} ∑_{n=1}^{∞} \frac{1}{(n + b)^{2}} + \frac{1}{(a − b)^{2}} ∑_{n=1}^{∞} (\frac{1}{n + b} − \frac{1}{n + a})

(6)

∫_{0}^{1} ∫_{0}^{1} \frac{s^{a}t^{b} ln(st)}{1 − st}dsdt = \frac{1}{a − b} ∑_{n=1}^{∞} (\frac{1}{(n + a)^{2}} − \frac{1}{(n + b)^{2}} ) .  (7)

Formulas (1) and (4) are proven directly by expanding

\frac{1}{1 − st} = ∑_{n=0}^{∞} s^{n}t^{n}

and integrating. The remaining formulas follow by differentiating (1) and (4) with respect to a and b.

If p(s, t) ∈ Z[s, t] is a polynomial of degree n with integral coefficients, then it follows from (2) and (7) that

∫_{0}^{1} ∫_{0}^{1} \frac{p(s, t) log(st)}{1 − st}dsdt = \frac{a_{n} + b_{n}ζ(3)}{d_{n}^{3}},  (8)

where a_{n}, b_{n}, d_{n} ∈ Z and d_{n} is the least common multiple of the integers 1, 2, …, n. It is not hard to see (using the Prime Number Theorem) that for any fixed ϵ > 0, d_{n} ≤ e^{(1+ϵ)n} for large n.

An Easier Way to Show ζ(3) ∉ Q 3

Legendre Polynomials

Consider P_{n}(s) = \frac{1}{n!} \frac{d^{n}}{ds^{n}} (s − s^{2})^{n}, a polynomial with integral coefficients. We will use the integrals

∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t) log(st)}{1 − st}dsdt = ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{1 − (1 − st)u}dsdtdu. (9)

Since

∫_{0}^{1} \frac{1}{1 − (1 − x)t}dx = −\frac{log(x)}{1 − x}.  (10)

and P_{n}(1 − s) = (−1)^{n}P_{n}(s), this equals

(−1)^{n} ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{1 − (1 − (1 − s)t)u}dsdtdu.

Lemma 1.1 For s, t ∈ (0, 1) fixed,

∫_{0}^{1} \frac{1}{1 − (1 − (1 − s)t)u}du = ∫_{0}^{1} \frac{1}{(1 − (1 − u)s)(1 − (1 − t)u)}du. (11)

Proof: Partial fractions give that

\frac{1}{(1 − (1 − u)s)(1 − (1 − t)u)} = \frac{1}{1 − (1 − s)t} (\frac{s}{1 − (1 − u)s} − \frac{1 − t}{1 − (1 − t)u}) ,

(12)

so

∫_{0}^{1} \frac{1}{(1 − (1 − u)s)(1 − (1 − t)u)}du  (13)
= \frac{1}{1 − (1 − s)t} (−s\frac{log(1 − s)}{s} + (1 − t)\frac{log(t)}{t − 1}) = −\frac{log(t(1 − s))}{1 − (1 − s)t}. (14)

Using (10) with x = 1 − (1 − s)t, we see that the two integrals are equal. □

Beukers used the integral (9) and, by using integration by parts and changing variables, deduced from it that ζ(3) is irrational. We will do some easy integrations-by-parts, but avoid his change of variables.

The lemma implies that

(−1)^{n} ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{1 − (1 − (1 − s)t)u}dsdtdu

= (−1)^{n} ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{(1 − (1 − u)s)(1 − (1 − t)u)}dsdtdu  (15)

= ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{(1 − (1 − u)s)(1 − tu)}dsdtdu.

We already know that this is of the form \frac{a_{n}+b_{n}ζ(3)}{d_{n}^{3}}. We will show it is very

small, in fact o (\frac{1}{d_{n}^{3}} ), which shows that ζ(3) ∉ Q.

By integrating by parts in the variables s and t, each n times, we get that

∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{P_{n}(s)P_{n}(t)}{(1 − (1 − u)s)(1 − tu)}dsdtdu  (16)
= ∫_{0}^{1} ∫_{0}^{1} ∫_{0}^{1} \frac{(s − s^{2})^{n}(t − t^{2})^{n}(u − u^{2})^{n}}{((1 − (1 − u)s)(1 − tu))^{n+1}} dsdtdu.  (17)

The function

f(s, t, u) = \frac{s(1 − s)t(1 − t)u(1 − u)}{(1 − (1 − u)s)(1 − tu)}  (18)

vanishes on the boundary of [0, 1] × [0, 1] × [0, 1] and has its maximum at

(s, t, u) = (2 − \sqrt{2}, \sqrt{2} − 1, \frac{1}{2}),

where f(2 − \sqrt{2}, \sqrt{2} − 1, \frac{1}{2}) = 17 − 12\sqrt{2} ≈ .029437.  (19)

Thus

\frac{a_{n} + b_{n}ζ(3)}{d_{n}^{3}} = O((.029437)^{n}).  (20)

However, d_{n}^{3} = O(e^{3.01n}), and since e^{3.01}(17−12\sqrt{2}) ≈ .597205, we conclude that

|\frac{a_{n} + b_{n}ζ(3)}{d_{n}^{3}} | = o (\frac{1}{d_{n}^{3}} ) ,  (21)

proving indeed that ζ(3) ≠ Q.