(Cross-posted from [StackExchange](https://math.stackexchange.com/questions/1600159/euler-product-for-sum-of-multiplicative-function-times-log)). Let $g$ be a multiplicative function which satisfies $0 \le g(p) \ll 1/p$ and


$$ \sum\_{p\le x} g(p) = \log \log x + C + O((\log x)^{-10}). $$


[Iwaniec and Friedlander](http://arxiv.org/pdf/math/9811186.pdf) claim the following product identity (equation 1.14). Why is it true?


$$-\sum\_n \mu(n)g(n)\log n = \prod\_{p} (1-g(p))(1-p^{-1})^{-1}$$


Even the case $g(n)=1/n$ is not clear to me. 



Euler product for sum of multiplicative function times log
Generally speaking, Euler product type formulae involving logarithms can be derived (formally, at least) from Euler product formulae without logarithms via differentiation in the $s$ parameter.


In the specific situation, one can argue as follows.
For $s>0$ one has


$$ \sum\_n \frac{\mu(n) g(n)}{n^s} = \prod\_p (1 - \frac{g(p)}{p^s}) $$
$$ = \zeta(1+s)^{-1} \prod\_p (1-\frac{g(p)}{p^s}) (1-\frac{1}{p^{1+s}})^{-1}.$$
The latter product equals $\prod\_p (1-g(p)) (1-p^{-1})^{-1} + O(|s|)$ for $s>0$ small by the hypothesis on $\sum\_{p \leq x} g(p)$. Thus the RHS is $s \prod\_p (1-g(p)) (1-p^{-1})^{-1} + O(|s|^2)$ for $s>0$ small. This implies that $\sum\_n \mu(n) g(n) = 0$ and $\sum\_n \mu(n) g(n) \log n = - \prod\_p (1-g(p)) (1-p^{-1})^{-1}$. Formally, this follows by differentiating both sides with respect to $s$ at $s=0$. To justify the calculation rigorously, first use (2.4) of Friedlander-Iwaniec to truncate $n$ to some large finite range $n \leq x$ and then perform a Taylor expansion of $\sum\_{n \leq x} \frac{\mu(n) g(n)}{n^s}$ for some suitably small $s>0$ (e.g. $s = 1/\log^2 x$).


