Closed form expression for this zeta-like series involving GCD and LCM

Question

I am looking for a closed form for this function $\Lambda:\mathbb{Q}^+\to\mathbb{R}^+$:

$$\Lambda(q) = \sum_{m,n\geq 1}\left(\frac{q\wedge\frac{m}{n}}{q\vee\frac{m}{n}}\right)^\alpha\left(\frac{m \wedge n}{m\vee n}\right)^\beta$$ where $m \wedge n = \textrm{gcd}(m,n)$ and $m \vee n = \textrm{lcm}(m,n)$.

For $\beta>\frac{1}{2}$ and $q=\frac{c}{d}$ with $c\wedge d=1$ we can factor out a zeta function and write: $$\Lambda(\tfrac{c}{d}) = \zeta(2\beta)\sum_{a,b\geq 1 \mid a\wedge b=1}\left(\frac{a\wedge c}{a \vee c}\cdot\frac{b\wedge d}{b \vee d}\right)^\alpha(ab)^{-\beta}$$

The case $q=1$ reduces, for $\alpha,\beta>\frac{1}{2}$, to: $$\Lambda(1)=\zeta(2\beta)\frac{\zeta(\alpha+\beta)^2}{\zeta(2\alpha+2\beta)}$$

Can we say more?

Answer 1

I shall use the standard notaton $(a,b) = \gcd(a,b)$ and $[a,b] = \operatorname{lcm}(a,b)$. Since $[a,b] = ab/(a,b)$ and $\sum_{m \mid (a,b)} \mu(m) = 1_{(a,b) = 1}$, we have that $$\zeta(2\beta) \sum_{\substack{a,b = 1 \\ (a,b) = 1}}^{\infty} \frac{(a,c)^{\alpha} (b,d)^{\alpha}}{[a,c]^{\alpha} [b,d]^{\alpha} a^{\beta} b^{\beta}} = \frac{\zeta(2\beta)}{c^{\alpha} d^{\alpha}} \sum_{a,b = 1}^{\infty} \frac{(a,c)^{2\alpha} (b,d)^{2\alpha}}{a^{\alpha + \beta} b^{\alpha + \beta}} \sum_{m \mid (a,b)} \mu(m).$$ Making the change of variables $a \mapsto am$ and $b \mapsto bm$, this becomes $$\frac{\zeta(2\beta)}{c^{\alpha} d^{\alpha}} \sum_{a,b,m = 1}^{\infty} \frac{(am,c)^{2\alpha} (bm,d)^{2\alpha} \mu(m)}{a^{\alpha + \beta} b^{\alpha + \beta} m^{2\alpha + 2\beta}}.$$ We break up the sums over $a,b$ based on their greatest common divisors with $c,d$, so that this is $$\frac{\zeta(2\beta)}{c^{\alpha} d^{\alpha}} \sum_{\substack{c_1 c_2 = c \\ d_1 d_2 = d}} \sum_{\substack{a,b,m = 1 \\ (a,c) = c_1 \\ (b,d) = d_1}}^{\infty} \frac{(am,c)^{2\alpha} (bm,d)^{2\alpha} \mu(m)}{a^{\alpha + \beta} b^{\alpha + \beta} m^{2\alpha + 2\beta}}.$$ We make the change of variables $a \mapsto ac_1$ and $b \mapsto bd_1$, yielding $$\frac{\zeta(2\beta)}{c^{\beta} d^{\beta}} \sum_{\substack{c_1 c_2 = c \\ d_1 d_2 = d}} c_2^{\beta - \alpha} d_2^{\beta - \alpha} \sum_{\substack{a,b,m = 1 \\ (a,c_2) = 1 \\ (b,d_2) = 1}}^{\infty} \frac{(m,c_2)^{2\alpha} (m,d_2)^{2\alpha} \mu(m)}{a^{\alpha + \beta} b^{\alpha + \beta} m^{2\alpha + 2\beta}}.$$ This is $$\frac{\zeta(\alpha + \beta)^2 \zeta(2\beta)}{c^{\beta} d^{\beta}} \sum_{\substack{c_1 c_2 = c \\ d_1 d_2 = d}} c_2^{\beta - \alpha} d_2^{\beta - \alpha} \prod_{p \mid c_2} (1 - p^{-\alpha - \beta}) \prod_{p \mid d_2} (1 - p^{-\alpha - \beta}) \sum_{m = 1}^{\infty} \frac{(m,c_2)^{2\alpha} (m,d_2)^{2\alpha} \mu(m)}{m^{2\alpha + 2\beta}}.$$

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We break up the sum over $m$ based in its greatest divisor with $c_2,d_2$, noting that $(c_2,d_2) = 1$ as $(c,d) = 1$. We find that $$\sum_{m = 1}^{\infty} \frac{(m,c_2)^{2\alpha} (m,d_2)^{2\alpha} \mu(m)}{m^{2\alpha + 2\beta}} = \sum_{\substack{c_3 c_4 = c_2 \\ d_3 d_4 = d_2}} c_3^{2\alpha} d_3^{2\alpha} \sum_{\substack{m = 1 \\ (m,c_2) = c_3 \\ (m,d_2) = d_3}}^{\infty} \frac{\mu(m)}{m^{2\alpha + 2\beta}}.$$ We make the change of variables $m \mapsto c_3 d_3 m$, so that this becomes $$\sum_{\substack{c_3 c_4 = c_2 \\ d_3 d_4 = d_2}} \frac{1}{c_3^{2\beta} d_3^{2\beta}} \sum_{\substack{m = 1 \\ (m,c_4 d_4) = 1}}^{\infty} \frac{\mu(c_3 d_3 m)}{m^{2\alpha + 2\beta}}.$$ Since $\mu(c_3 d_3 m) = \mu(c_3) \mu(d_3) \mu(m) 1_{(m,c_3 d_3) = 1}$, this in turn is $$\sum_{\substack{c_3 c_4 = c_2 \\ d_3 d_4 = d_2}} \frac{\mu(c_3) \mu(d_3)}{c_3^{2\beta} d_3^{2\beta}} \sum_{\substack{m = 1 \\ (m,c_2 d_2) = 1}}^{\infty} \frac{\mu(m)}{m^{2\alpha + 2\beta}},$$ which is $$\frac{1}{\zeta(2\alpha + 2\beta)} \prod_{p \mid c_2} \frac{1 - p^{-2\beta}}{1 - p^{-2\alpha - 2\beta}} \prod_{p \mid d_2} \frac{1 - p^{-2\beta}}{1 - p^{-2\alpha - 2\beta}}.$$

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Inserting this into the earlier identity, we arrive at $$\frac{\zeta(\alpha + \beta)^2 \zeta(2\beta)}{\zeta(2\alpha + 2\beta)} \frac{1}{c^{\beta} d^{\beta}}\sum_{c_1 c_2 = c} c_2^{\beta - \alpha} \prod_{p \mid c_2} \frac{1 - p^{-2\beta}}{1 + p^{-\alpha - \beta}} \sum_{d_1 d_2 = d} d_2^{\beta - \alpha} \prod_{p \mid d_2} \frac{1 - p^{-2\beta}}{1 + p^{-\alpha - \beta}}.$$