TAOCP 4.5.2 Exercise 14
Let G=\gcd(u,v).
Section 4.5.2: The Greatest Common Divisor
Exercise 14. ▶ [**] [HM25] What is the expected value of $\lg \gcd(u, v)$ when $u$ and $v$ are (a) random positive integers; (b) random positive odd integers?
Verified: yes
Solve time: 1m19s
Let
$$ G=\gcd(u,v). $$
By Theorem 4.5.2D, for random positive integers,
$$ \Pr(G=d)=\frac{1}{\zeta(2)d^2}=\frac{6}{\pi^2d^2}, $$
since
$$ \Pr(\gcd(u',v')=1)=\frac1{\zeta(2)}. $$
Therefore
$$ E(\lg G) =\sum_{d\ge1}\frac{\lg d}{\zeta(2)d^2} =\frac1{\zeta(2)}\sum_{d\ge1}\frac{\lg d}{d^2}. $$
Using
$$ \zeta'(s)=-\sum_{n\ge1}\frac{\ln n}{n^s}, $$
we have
$$ \sum_{d\ge1}\frac{\lg d}{d^2} =\frac{-\zeta'(2)}{\ln 2}, $$
hence
$$ E(\lg \gcd(u,v)) = -\frac{\zeta'(2)}{\zeta(2)\ln 2} \approx 0.822467. $$
Thus
$$ \boxed{E(\lg \gcd(u,v)) = -\frac{\zeta'(2)}{\zeta(2)\ln 2} \approx 0.822467.} $$
For random positive odd integers, exercise 13 shows that
$$ \Pr(\gcd(u,v)=1)=\frac8{\pi^2}. $$
If $d$ is odd,
$$ \Pr(\gcd(u,v)=d) =\frac8{\pi^2d^2}, $$
and the probability is $0$ for even $d$. Hence
$$ E(\lg G) =\frac8{\pi^2}\sum_{\substack{d\ge1\ d\ \text{odd}}}\frac{\lg d}{d^2}. $$
Since
$$ \sum_{\substack{d\ge1\ d\ \text{odd}}}\frac{\ln d}{d^2} = -\zeta'(2)+2^{-2}!\left(\ln 2,\zeta(2)+\zeta'(2)\right) = -\frac34\zeta'(2)+\frac14\ln 2,\zeta(2), $$
it follows that
$$ E(\lg \gcd(u,v)) = -\frac{\zeta'(2)}{\zeta(2)\ln 2} +\frac13 \approx 1.155800. $$
Therefore
$$ \boxed{E(\lg \gcd(u,v)) = -\frac{\zeta'(2)}{\zeta(2)\ln 2} +\frac13 \approx 1.155800} $$
when $u$ and $v$ are random positive odd integers. ∎