If I want to evaluate $$\sum _{[r,r']\leq x}\log r\log r'$$ I could write it as an integral using Perron's formula, pick up a pole, and get a main term which involves looking at (the derivatives at $\mathbf w=\mathbf 0$ and evaluation at $s=1$ of) $$\sum _{r,r'=1}^\infty \frac {1}{[r,r']^sr^wr'^{w'}}.$$ This is ok because it is not hard to show that this series has an Euler product $$\prod _p\frac {1-p^{-w-w'-2s}}{(1-p^{-w-w'-s})(1-p^{-w-s})(1-p^{-w'-s})}=\frac {\zeta (s+w)\zeta (s+w')\zeta (s+w+w')}{\zeta (2s+w+w')}$$ so the main term is a sum of terms involving $\zeta (2),\zeta '(2),\zeta ''(2)$ snd the first three coefficients in the Laurent expansion for $\zeta (s)$ about $s=1$.
If I change the problem to looking at $$\sum _{[r,r']\leq x}\log r\log r'\phi (r)\phi (r')/rr'$$ (which is the actual sum I have) then shouldn't a similar argument work? This is what I thought but it seems I can't get any sensible factorisation for the Euler product (and WolframAlpha seems to tell me that the polynomial in the Euler product won't factorise).
My question is how do I evaluate the sum with the $\phi (r)/r$ terms? Does it make sense that I can't write the zeta function in a nice way? Of course without the $\log $'s this argument would work so am I just missing something basic?
Anyway, any thoughts would be greatly appreciated!