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# On the real roots of ζ(x) = a

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Let $\mathbb N$ be the set of natural numbers and $\mathbb Q^+$ be the set of positive rational numbers. For every positive integer $n \ge 2$, there exists a unique real number $s_n$ such that $\zeta (s_n) = n$. The question is does there exists an $n \in \mathbb N$ such that $s_n \in \mathbb Q^+$ be rational? In other words, does there exist a positive rational $\frac{p}{q}$ such that $\zeta(\frac{p}{q}) \in \mathbb N$? At present, we don’t even know if $s_n$ is always rational, let alone irrational or transcendental. Investigation seems to suggest that $\zeta(\frac{p}{q}) \notin \mathbb N$ for any $\frac{p}{q} \in \mathbb Q^+$. However this has not been completely established. In this post, we shall look at some on the results in this line of research.

Expanding the Riemann zeta function about $s=1$ gives

$\displaystyle{\zeta(s) = \frac{1}{s-1} + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n (s-1)^n}$

where the constants

$\displaystyle{\gamma_n = \lim_{m \rightarrow \infty}\Bigg [\sum_{k=1}^{m}\frac{(\ln k)^n}{k} - \frac{(\ln m)^{n+1}}{n+1}\Bigg]}$

are known as Stieltjes constants. Inverting the above series we obtain:

Theorem 1Let $x > 1$ and $\zeta(x) = a$; then,

$\displaystyle{x = 1 + \frac{1}{a-\gamma _0} - \frac{\gamma _1}{1!(a-\gamma _0)^2} + \frac{\gamma _2}{2!(a-\gamma _0)^3}- \frac{\gamma _3 - 12\gamma _1}{3!(a-\gamma _0)^4} + O\Big(\frac{1}{a^5}\Big)}$.

Proof. The proof follows by inverting the Stieltjes series expansion of $\zeta (x)$$\Box$

We can rewrite the Stieltjes series expansion of the  Riemann zeta function in the following form which we shall use regularly in our analysis:

$\displaystyle{\zeta \Big (1+\frac{1}{s}\Big ) = s + \gamma_0 + \sum_{n=0}^{\infty}\frac{(-1)^n\gamma_n}{n!^n}}$.

We want to find an interval where the function $\zeta(1+\frac{1}{s})$ is strictly decreasing. Clearly $\zeta(1+\frac{1}{s})$ is continuous and strictly decreasing in the interval $(0, \infty)$. The trivial zeros of $\zeta(s)$ occur at $s=-2n$ where $n \in N$. In other words, the trivial zeros of $\zeta(1+\frac{1}{s})$ occur at $s = \frac{-1}{2n+1}$ and all these zeros lie in the interval $(-1/3, 0)$. Hence in this interval, $\zeta(1+\frac{1}{s})$ will have infinitely many points of inflexion. Excluding the interval $(-1/3, 0)$, we can divide the real line in to two parts $(-\infty, -1/3)$ and $(0,\infty)$; and in each of these two intervals, $\zeta(1+\frac{1}{s})$ in strictly decreasing. For positive rationals $\frac{p}{q}$, the scope of our current study is the interval  $(0,\infty)$.

Lemma 2For every finite real $a>1$, there exists a unique real number $c_a, 0< c_a < 1-\gamma _0$, such that

$\displaystyle{\zeta\Big(1+\frac{1}{a-1 +c_a}\Big) = a}$.

Proof. Let lemma follows form the fact that in our interval of interest $(0,\infty)$$\zeta (1+\frac{1}{s}) - s$ is strictly decreasing. $\Box$

Theorem 3. Let $c_a$ be as defined in Lemma 2. The asymptotic expansion of $c_a$ is

$\displaystyle{c_a = 1-\gamma_0 + \frac{\gamma_1}{a-1} - \frac{\gamma_2 + \gamma_1-\gamma_1\gamma_0}{(a-1)^2}}$

$\displaystyle{+ \frac{\gamma_3 + 2\gamma_2- 2\gamma_2\gamma_0 + \gamma_1 - 2\gamma_1\gamma_0 + \gamma_1\gamma_0^2 - \gamma_1^2}{(a-1)^3}+ O\Big(\frac{1}{a^4}\Big)}.$

Proof. This is obtained by inverting the above Stieltjes series expansion of $\zeta (1+\frac{1}{a-1+c_a})$$\Box$

The next result gives us a method to compute $c_a$ numerically. We have

Lemma 4. Let $k_r$ be defined recursively as

$\displaystyle{k_{r+1} = a + k_r - \zeta \Big (\frac{a+k_r}{a-1+k_r}\Big )}$

where $k_0 = 1-\gamma_0$. Then $\lim_{r \rightarrow \infty}k_r = c_a$ .

Marek Wolf, has sent me the computed values of $c_n$ for $2 \le c_n \le 1000$. Some of these numbers were used in the proof of the following theorem.

Proof. Trivial. $\Box$

Theorem 5. If $1 \le p-q \le 40$ then $\zeta (\frac{p}{q}) \notin \mathbb N$.

Proof. We have $c_2 \approx 0.37240621590$. There are exactly twenty three positive fractions $\frac{m}{n}$ with $n \le 40$ such that

$\displaystyle{c_2 \le \frac{m}{n} < 1- \gamma_0}$.

For each of these fractions, we find (by actual computation) that there exists a integer $r \le 43$ such that

$\displaystyle{c_r \le \frac{m}{n} < c_{r+1}}$.

Since $c_n$ is strictly increasing, it implies that if $c_n \in \mathbb Q^+$ then the denominator of $c_n$ has to be $\ge 41$. Hence the theorem follows. $\Box$

If we continue working with the idea given the proof of Theorem 5, we can prove $p-q > n_0$ for some large natural number $n_o$. But this method gives no hope of proving that $\zeta (\frac{p}{q}) \notin \mathbb N$. Therefore we must look for some other ideas.

Theorem 6. Let $\zeta (1+\frac{1}{x_i}) \in \mathbb N$, $x_i \in \mathbb R^+ , i=1, 2, \ldots , k$ and $a_j \in \mathbb N$ such that

$\sum a_j = \{1, 2, 3, 4, 5, 7, 8, 10, 12 , 15\}$.

If $y=m+\sum a_k x_k$ where is any non-negative integer; then $\zeta (1+\frac{1}{y}) \notin \mathbb N$.

Some applications of Theorem 6 are given below.

Example 7. If $x \in \mathbb R^+$ and $2 \le n \le 5, n \in \mathbb N$ then at least one of the two numbers

$\zeta (1+\frac{1}{x})$ and $\zeta (1+\frac{1}{nx})$

is not a integer.

Example 8. If $x, y \in \mathbb R^+$ then at least one of the three numbers

$\zeta (1+\frac{1}{x}), \zeta (1+\frac{1}{y})$ and $\zeta (1+\frac{1}{x+y})$

is not a integer.