GCD and the Riemann zeta funtion 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.
 A: 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.
A: 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
$$\zeta(s) \prod_p \frac{ w (1-zw) }{(w - 1) (p - z w)}
= \zeta(s) \prod_p \frac{ w (1-zw)}{(1 - 1/w) w (p - z w)}
\\= \zeta(s)\zeta(t) 
\prod_p \frac{ 1-zw } {p - z w} 
\\ = \zeta(s)\zeta(t) 
\prod_p \frac{ 1-zw}{z w \times (p/z/w - 1)}
\\= \zeta(s)\zeta(t) 
\prod_p \frac{ zw-1}{z w \times (1-p/z/w)} 
= \zeta(s)\zeta(t)\zeta(s+t-1)
\prod_p \frac{ zw-1}{z w}
\\ = \zeta(s)\zeta(t)\zeta(s+t-1)
\prod_p \left(1-\frac{1}{z w}\right)
= \zeta(s)\zeta(t)\frac{\zeta(s+t-1)}{\zeta(s+t)}.$$
