GCD and the Riemann zeta funtion

Question

I'm completely stuck on this one, as I'm just starting with analytic number theory: How to write $$\sum_{a\in\mathbb{N}}\sum_{b\in\mathbb{N}}\frac{(a,b)}{a^sb^t}$$ in terms of the Riemann zeta function? $(a,b)$ denotes the greatest common divisor.

Answer 1

Here's a sketch. First sort by the greatest common divisor, setting $g=(a,b)$: $$ \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{(a,b)}{a^s b^t} = \sum_{g=1}^\infty g \mathop{\sum_{a=1}^\infty \sum_{b=1}^\infty}_{(a,b)=g} \frac1{a^s b^t}. $$ Now use the fact that $(a,b)=g$ if and only if $a=cg$ and $b=dg$ for some integers $c,d$ with $(c,d)=1$. Rewriting in terms of the new variables $c,d$, you should get a $g^{1-s-t}$ which you can pull out of the inner double sum.

In the inner double sum, detect the condition $(c,d)=1$ by inserting the sum $\sum_{e\mid(c,d)} \mu(e)$, which equals $1$ if $(c,d)=1$ and $0$ otherwise. (By the way, this is the most valuable trick in all of analytic number theory, I believe.) Then rearrange the sums so that you have a sum over $g$, then $e$, then over $c,d$ such that $e\mid(c,d)$; note this latter condition is equivalent to $c=ef$, $d=eh$ where $f$ and $h$ now have no constraints.

With this last change of variable, you should end up with $$ \sum_{g=1}^\infty g^{1-s-t} \sum_{e=1}^\infty \frac{\mu(e)}{e^{s+t}} \sum_{f=1}^\infty \frac1{f^s} \sum_{h=1}^\infty \frac1{h^t}, $$ which you should be able to evaluate in terms of the Riemann zeta function.

Answer 2

It would appear from the wording of the question that perhaps a more elementary approach should be used.

Rewrite the sum as $$\sum_{a\ge 1} \frac{1}{a^t} \sum_{b\ge 1} \frac{\gcd(a,b)}{b^s}.$$ Now let the factorization of $a$ be $$ a = p_1^{v_1} p_2^{v_2} \cdots p_r^{v_r}$$ and let the set of primes that appear be denoted as $P(a).$

Then the Euler product for the inner sum is given by $$\prod_{k=1}^r \left(1 + \frac{p_k}{p_k^s} + \frac{p^2_k}{p_k^{2s}} + \frac{p^3_k}{p_k^{3s}} + \cdots + \frac{p^{v_k}_k}{p_k^{v_k s}} + \frac{p^{v_k}_k}{p_k^{(v_k+1) s}} + \frac{p^{v_k}_k}{p_k^{(v_k+2) s}} + \cdots\right) \prod_{p\notin P(a)} \frac{1}{1-p^{-s}}.$$ This simplifies to $$\prod_{k=1}^r \left(\frac{1-(p_k/p_k^s)^{v_k}}{1-p_k/p_k^s} + \frac{p_k^{v_k}}{p_k^{v_k s}}\frac{1}{1-(1/p_k^s)} \right)\prod_{p\notin P(a)} \frac{1}{1-p^{-s}}$$ which in terms of the Riemann zeta function is $$\zeta(s) \prod_{k=1}^r \left(\frac{1-(p_k/p_k^s)^{v_k}}{1-p_k/p_k^s}(1-(1/p_k^s)) + \frac{p_k^{v_k}}{p_k^{v_k s}}\right).$$ Note that the term in the product has the value $1$ when $v=0.$ Therefore the sum is given by $$\zeta(s) \prod_p \sum_{v\ge 0} \frac{1}{p^{vt}} \left(\frac{1-(p/p^s)^v}{1-p/p^s}(1-(1/p^s)) + \frac{p^v}{p^{v s}}\right).$$ This is $$\zeta(s) \prod_p \left(\frac{1}{1-1/p^{s+t-1}} + \frac{1-(1/p^s)}{1-p/p^s} \sum_{v\ge 0} \frac{1-(p/p^s)^v}{p^{tv}} \right)$$ which is in turn $$\zeta(s) \prod_p \left(\frac{1}{1-1/p^{s+t-1}} + \frac{1-(1/p^s)}{1-p/p^s} \frac{1}{1-1/p^t} - \frac{1-(1/p^s)}{1-p/p^s} \frac{1}{1-1/p^{s+t-1}} \right)$$ Now put $z=p^s$ and $w=p^t$ to get the intermediate result $$\zeta(s) \prod_p \left(\frac{1}{1-p/z/w} + \frac{1-1/z}{1-p/z} \frac{1}{1-1/w} - \frac{1-z}{1-p/z} \frac{1}{1-p/z/w} \right).$$ This factors as $$\zeta(s) \prod_p \frac{ w (1-zw) }{(w - 1) (p - z w)} = \zeta(s) \prod_p \frac{ w (1-zw)}{(1 - 1/w) w (p - z w)} \\= \zeta(s)\zeta(t) \prod_p \frac{ 1-zw } {p - z w} \\ = \zeta(s)\zeta(t) \prod_p \frac{ 1-zw}{z w \times (p/z/w - 1)} \\= \zeta(s)\zeta(t) \prod_p \frac{ zw-1}{z w \times (1-p/z/w)} = \zeta(s)\zeta(t)\zeta(s+t-1) \prod_p \frac{ zw-1}{z w} \\ = \zeta(s)\zeta(t)\zeta(s+t-1) \prod_p \left(1-\frac{1}{z w}\right) = \zeta(s)\zeta(t)\frac{\zeta(s+t-1)}{\zeta(s+t)}.$$