Criteria for irrationality of Euler's constant

Question

Define for $n\in\mathbb{N}$, $$I_n=\int_0^1\int_0^1 -\frac{(x(1-x)y(1-y))^n}{(1-xy)\log xy}dx dy$$

In this article it is proved that $$I_n=\binom{2n}{n}\gamma+L_n-A_n$$ where $L_n=d^{-1}_{2n}\log S_n$, $A_n=\sum_{i=0}^{n}\binom{n}{i}^2 H_{n+i}$ and $$S_n=\prod_{k=1}^{n}\prod_{i=0}^{\min(k-1,n-k)}\prod_{j=i+1}^{n-i}(n+k)^{\frac{2d_{2n}}{j}\binom{n}{i}^2}$$ where $d_n=\text{lcm}(1,2,...,n)$ and $H_n$ is the $n$th Harmonic number.

In Theorem $4$ it is proved using the above representation of $I_n$ that for $n$ fixed, $\{\log S_n\}=d_{2n}I_n$ if and only if $\gamma=p/q$ for some $p,q\in\mathbb{Z}$, where $q|d_{2n}\binom{2n}{n}$ and $\{x\}$ denotes the fractional part of $x$.

Question: Can we conclude that for $n$ fixed, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma$ is rational?

We have by multiplication by $d_{2n}$ $$d_{2n}I_n=d_{2n}\binom{2n}{n}\gamma+d_{2n}L_n-d_{2n}A_n$$ which can be rewritten as $$\log S_n-d_{2n}I_n=d_{2n}A_n- d_{2n}\binom{2n}{n}\gamma$$ where $d_{2n}A_n\in\mathbb{Z}$. So we have, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma=a/b$ for some $a,b\in\mathbb{Z}$, where $b|d_{2n}\binom{2n}{n}$. Can we conclude that for $n$ fixed, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma$ is rational?

Any help will be highly appreciated. Thank you.

Edit I am looking for an answer which I can accept. Thank you.

Answer 1

Unfortunately, the claim

for $n$ fixed, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma=p/q$ for some $p,q\in\mathbb{Z}$ and $q|d_{2n}{2n\choose n}$.

provides a partial answer to your question as follows.

Proposition. The following conditions are equivalent:

(1) $\gamma$ is irrational;

(2) $\log S_n-d_{2n} I_n\not\in\mathbb{Z}$ for each natural $n$;

(3) $\log S_n-d_{2n} I_n\not\in\mathbb{Z}$ for infinitely many natural $n$.

Proof. The implication $(1)\Rightarrow (2)$ directly follows from the claim.

The implication $(2)\Rightarrow (3)$ is trivial.

$(3)\Rightarrow (1)$ Suppose for a contradiction that $\gamma=p/q$ for some $p,q\in\mathbb N$. Pick any $n\ge q/2$. Then $q|d_{2n}|d_{2n}{2n\choose n}$, so $\log S_n-d_{2n} I_n\in\mathbb{Z}$, a contradiction. $\square$

Answer 2

So we have, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma=a/b$ for some $a,b\in\mathbb{Z}$, where $b|d_{2n}\binom{2n}{n}$. Can we conclude that for $n$ fixed, $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma$ is rational?

If I understand what you wrote correctly, you have already proved that $\log S_n-d_{2n} I_n\in\mathbb{Z}$ if and only if $\gamma=a/b$ for some $a,b\in\mathbb{Z}$, where $b|d_{2n}\binom{2n}{n}$.

So, I think the answer to your question is No.

We can say that $$\log S_n-d_{2n}I_n\in\mathbb Z\implies \text{$\gamma$ is rational}$$ since $$\begin{align}&\log S_n-d_{2n}I_n\in\mathbb Z \\\\&\implies d_{2n}A_n- d_{2n}\binom{2n}{n}\gamma\in\mathbb Z \\\\&\implies \underbrace{d_{2n}\binom{2n}{n}}_{\text{a positive integer}}\gamma\in\mathbb Z \\\\&\implies \text{$\gamma$ is rational}\end{align}$$

However, it is not true that $$\text{$\gamma$ is rational}\implies \log S_n-d_{2n}I_n\in\mathbb Z$$

If $\gamma$ is rational such that $d_{2n}\binom{2n}{n}\gamma$ is not an integer, then $\log S_n-d_{2n}I_n$ is not an integer since $$\log S_n-d_{2n}I_n=\underbrace{d_{2n}A_n}_{\text{integer}}- \underbrace{d_{2n}\binom{2n}{n}\gamma}_{\text{not integer}}$$

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Added :

• It is true that if $\gamma$ is irrational, then $\log S_n-d_{2n}I_n\not\in\mathbb Z$ for each natural $n$.

• It is true that if $\log S_n−d_{2n}I_n\not\in\mathbb Z$ for each natural $n$, then $\gamma$ is irrational. The reason is as follows. For any integer $b$ such that $\gamma=\frac ab$, there is $n$ such that $b\mid d_{2n}$. So, for such an $n$, $\log S_n-d_{2n}I_n\in\mathbb Z$.

Added 2 :

We can say that $\gamma$ is irrational if and only if $\log S_n-d_{2n}I_n\not\in\mathbb Z$ for each natural $n$.