Irrationality of ζ(2) and ζ(3): Part 2

In the last post we proved that $\zeta(2)$ is irrational. Now we shall prove in a similar manner that $\zeta(3)$ is irrational. Note that this proof is based on Beukers' paper "A Note on the Irrationality of $\zeta(2)$ and $\zeta(3)$."

Irrationality of $\zeta(3)$

Like the case of $\zeta(2)$ we first establish certain formulas concerning some double integrals which are related to $\zeta(3)$. The derivation of these formulas is based on the integral formulas established in last post.

Preliminary Results

Let $r, s$ be non-negative integers with $r > s$. Then we have
$$\begin{align} \int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}x^{r}y^{r}\,dx\,dy &= 2\sum_{n = 1}^{\infty}\frac{1}{(n + r)^{3}}\tag{1}\\ \int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}x^{r}y^{s}\,dx\,dy &= \frac{1}{r - s}\left\{\frac{1}{(s + 1)^{2}} + \frac{1}{(s + 2)^{2}} + \cdots + \frac{1}{r^{2}}\right\}\tag{2} \end{align}$$ Using equation $(2)$ from the last post we get $$\int_{0}^{1}\int_{0}^{1}\frac{x^{r + \alpha}y^{r + \alpha}}{1 - xy}\,dx\,dy = \sum_{n = 1}^{\infty}\frac{1}{(n + r + \alpha)^{2}}$$ Differentiating the above relation with respect to $\alpha$ we get $$\int_{0}^{1}\int_{0}^{1}\frac{x^{r + \alpha}y^{r + \alpha}\log xy}{1 - xy}\,dx\,dy = \sum_{n = 1}^{\infty}\frac{-2}{(n + r + \alpha)^{3}}$$
Now putting $\alpha = 0$ the first result is established. This means that $$\int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}\,dx\,dy = 2\zeta(3)$$ and if $r$ is a positive integer then $$\int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}x^{r}y^{r}\,dx\,dy = 2\left\{\zeta(3) - \left(\frac{1}{1^{3}} + \frac{1}{2^{3}} + \cdots + \frac{1}{r^{3}}\right)\right\}$$
Next from equation $(3)$ of last post we have $$\int_{0}^{1}\int_{0}^{1}\frac{x^{r + \alpha}y^{s + \alpha}}{1 - xy}\,dx\,dy = \frac{1}{r - s}\left\{\frac{1}{s + \alpha + 1} + \frac{1}{s + \alpha + 2} + \cdots + \frac{1}{r + \alpha}\right\}$$ Differentiating the above relation with respect to $\alpha$ we get $$\begin{align} &\int_{0}^{1}\int_{0}^{1}\frac{x^{r + \alpha}y^{s + \alpha}\log xy}{1 - xy}\,dx\,dy\notag\\ &\,\,\,\,\,\,\,\,= \frac{1}{r - s}\left\{\frac{-1}{(s + \alpha + 1)^{2}} + \frac{-1}{(s + \alpha + 2)^{2}} + \cdots + \frac{-1}{(r + \alpha)^{2}}\right\}\notag \end{align}$$ Putting $\alpha = 0$ in the above equation we obtain equation $(2)$.

From the above results it is now clear that if $P(x), Q(x)$ are polynomials of degree $n$ with integer coefficients then $$\int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}P(x)Q(y)\,dx\,dy = \frac{a\zeta(3) + b}{d_{n}^{3}}$$ where $a, b$ are some integers dependent on polynomials $P(x), Q(x)$ and $d_{n}$ denotes the LCM of numbers $1, 2, \ldots, n$ (aslo for completeness we can assume $d_{0} = 1$).

Strategy of the Proof

Now we choose a specific polynomial $P_{n}(x)$ defined by $$P_{n}(x) = \frac{1}{n!}\frac{d^{n}}{dx^{n}}\{x^{n}(1 - x)^{n}\}$$ Since $P_{n}(x)$ is a polynomial of degree $n$ with integer coefficients it follows that the integral defined by $$I_{n} = \int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}P_{n}(x)P_{n}(y)\,dx\,dy$$ can be expressed in the form $$I_{n} = \frac{a_{n}\zeta(3) + b_{n}}{d_{n}^{3}}$$ where $a_{n}, b_{n}$ are integers dependent on $n$.

We will establish that

1. $I_{n} \neq 0$ for all positive integers $n$.
2. $d_{n}^{3}I_{n} \to 0$ as $n \to \infty$.

This will imply that the expression $a_{n}\zeta(3) + b_{n} \to 0$ as $n \to \infty$ and is never zero for any value of $n$. If $\zeta(3)$ were rational, say $p/q$, then we would have $|a_{n}\zeta(3) + b_{n}|\geq 1/q$ and hence $a_{n}\zeta(3) + b_{n}$ would not tend to zero. This contradiction proves that $\zeta(3)$ is irrational.

Estimation of $I_{n}$

Now we come to the proof of the two claims mentioned above which are vital to obtain a contradiction needed to prove the irrationality of $\zeta(3)$.

First we need to observe that $$\begin{align} \int_{0}^{1}\frac{dz}{1 - az} &= \left[\frac{-1}{a}\log(1 - az)\right]_{z = 0}^{z = 1}\notag\\ &= -\frac{\log(1 - a)}{a}\notag \end{align}$$ hence on putting $a = 1 - xy$ we get $$\frac{-\log xy}{1 - xy} = \int_{0}^{1}\frac{dz}{1 - (1 - xy)z}$$ Using the above equation we can write the integral $I_{n}$ as a triple integral $$I_{n} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{P_{n}(x)P_{n}(y)}{1 - (1 - xy)z}\,dx\,dy\,dz\tag{3}$$ Using integration by parts $n$ times with respect to $x$ we get $$I_{n} =  \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{(xyz)^{n}(1 - x)^{n}P_{n}(y)}{\{1 - (1 - xy)z\}^{n + 1}}\,dx\,dy\,dz$$ Following Beukers, we apply the substitution $$w = \frac{1 - z}{1 - (1 - xy)z}$$ so that $$z = \frac{1 - w}{1 - (1 - xy)w},\, 1 - z = \frac{xyw}{1 - (1 - xy)w}$$ Hence $$dz = \frac{-xy}{\{1 - (1 - xy)w\}^{2}}dw$$ and $$\begin{align} \frac{z^{n}}{\{1 - (1 - xy)z\}^{n + 1}} &= \frac{(1 - w)^{n}}{\{1 - (1 - xy)w\}^{n}}\frac{w^{n + 1}}{(1 - z)^{n + 1}}\notag\\ &= \frac{(1 - w)^{n}(1 - (1 - xy)w)}{(xy)^{n + 1}}\notag \end{align}$$ Also note that as $z$ moves from $0$ to $1$, $w$ moves from $1$ to $0$.

After substituting these expressions we get $$I_{n} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{(1 - x)^{n}(1 - w)^{n}P_{n}(y)}{1 - (1 - xy)w}\,dx\,dy\,dw$$ Using integration by parts $n$ times with respect to $y$ we get $$I_{n} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{x^{n}(1 - x)^{n}y^{n}(1 - y)^{n}w^{n}(1 - w)^{n}}{\{1 - (1 - xy)w\}^{n + 1}}\,dx\,dy\,dw\tag{4}$$ and from this expression it is clear that $I_{n} > 0$ as the integrand is positive for all $x, y, w \in (0, 1)$.

We next need to find an estimate for the function $f(x, y, w)$ defined by $$f(x, y, w) = \frac{x(1 - x)y(1 - y)w(1 - w)}{1 - (1 - xy)w}$$ for $x, y, w \in (0, 1)$. Finding the maximum value using first and second partial derivatives seems a bit complicated hence it is better to go for a simpler approach based on inequalities.

The denominator of $f(x, y, w)$ is $1 - w + xyw$ and clearly we have $$1 - w + xyw \geq 2\sqrt{(1 - w)xyw}$$ and hence we have $$f(x, y, w) \leq \frac{1}{2}\sqrt{x}(1 - x)\sqrt{y}(1 - y)\sqrt{w(1 - w)}$$ If we put $x = t^{2}$ then $\sqrt{x}(1 - x) = t(1 - t^{2}) = t - t^{3}$ which is maximum when $t = 1/\sqrt{3}$ and the maximum value is $2/(3\sqrt{3})$. Similar is the case for $\sqrt{y}(1 - y)$. The maximum value of $\sqrt{w(1 - w)}$ is clearly $(w + 1 - w)/2 = 1/2$. Hence we have $$f(x, y, w) \leq \frac{1}{2}\frac{2}{3\sqrt{3}}\frac{2}{3\sqrt{3}}\frac{1}{2} = \frac{1}{27}$$ Therefore by equation $(4)$ we get $$\begin{align} I_{n} &= \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{\{f(x, y, w)\}^{n}}{1 - (1 - xy)w}\,dx\,dy\,dw\notag\\ &\leq \left(\frac{1}{27}\right)^{n} \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1 - (1 - xy)w}\,dx\,dy\,dw\notag\\ &= \left(\frac{1}{27}\right)^{n} \int_{0}^{1}\int_{0}^{1}\frac{-\log xy}{1 - xy}\,dx\,dy\notag\\ &= 2\zeta(3)\left(\frac{1}{27}\right)^{n}\notag \end{align}$$ From the last post we know that if $K > e$ is a fixed number then $d_{n} < K^{n}$ for all sufficiently large values of $n$. Hence it follows that $$d_{n}^{3}I_{n} < 2\zeta(3)\left(\frac{K^{3}}{27}\right)^{n}\tag{5}$$ for all sufficiently large values of $n$. If we choose $K$ such that $e < K < 3$ then we can see that the right hand side of equation $(5)$ above tends to zero as $n \to \infty$. Therefore $d_{n}^{3}I_{n} \to 0$ as $n \to \infty$. We have thus completed the proof of irrationality of $\zeta(3)$.

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