# Question on de la Vallee Poussin's simplified proof of Dirichlet's theorem on primes in arithmetic progressions

I've been trying to understand de la Vallee Poussin's "Demonstration Simplifiee du Theorem de Dirichlet sur la Progression Arithmetique" and I've got stuck at the following step on pg 18 where Poussin takes the logarithmic derivative of: $$\Sigma_{n}\frac{\chi(n)}{n^{s}} = \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$$ in order to obtain: $$-D\log\sum_{n}\frac{\chi(n)}{n^{s}} = \Sigma_{q}\frac{\chi(q)\log(q)}{q^{s}-\chi(q)}$$

Specifically, when I try to work through the calculation, I don't see where the -1 on the left hand side of the equation comes from. When I tried to take logs and then differentiate the first equation, I ended up with the following:

$$D\log\sum_{n}\frac{\chi(n)}{n^{s}} = \Sigma_{q}\frac{\chi(q)\log(q)}{q^{s}-\chi(q)}$$

Could someone help me understand where I'm going wrong please?

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Thanks for your comment and for working through the derivation. I've just worked through it again and seen my mistake-each time I attempted it there was one particular negative sign that I lost track of. – Lea M Mar 8 '11 at 5:12
Could you add the solution as an answer and accept it (instead of changing the title)? – Yuval Filmus Mar 8 '11 at 6:16

In what follows, D denotes differentiation with respect to $s$.

We have that

$\Sigma_{n}\frac{\chi(n)}{n^{s}} = \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$

We take the log of each side and then proceed to take the derivative. Thus we have

$Dlog\Sigma_{n}\frac{\chi(n)}{n^{s}} = Dlog \prod_{q}(1-\frac{\chi(q)}{q^{s}})^{-1}$

Now the right hand side of the equation is equal to:

$D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})^{-1} = -D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})$ using properties of log.

Now we differentiate term by term. For an arbitrary term in the sum, we have:

$Dlog(1-\frac{\chi(q)}{q^{s}})=-(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)Dq^{-s}$ by the chain rule.

But we know that

$Dq^{-s}=-log(q)q^{-s}$

Thus we have $Dlog(1-\frac{\chi(q)}{q^{s}})=(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)log(q)q^{-s}$ because the negative signs cancel.

But then $-D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}}) = \Sigma(1-\frac{\chi(q)}{q^{s}})^{-1}\chi(q)log(q)q^{-s}$

And after re-arranging, this gives us $-D\Sigma_{q}log(1-\frac{\chi(q)}{q^{s}})=-\Sigma_{q}\frac{\chi(q)log(q)}{q^{s}-\chi(q)}$. Thus we have

$Dlog\Sigma_{n}\frac{\chi(n)}{n^{s}}= -\Sigma_{q}\frac{\chi(q)log(q)}{q^{s}-\chi(q)}$ and so swapping the negative sign from the right hand side to the left hand side gives us the result.

(P.S. Please forgive me if there are typos in the above-I'm not very good with latex code!)

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Don't forget to accept your answer! – Yuval Filmus Mar 8 '11 at 13:01
@Yuval-I tried to accept it but it told me I have to wait 2 days. – Lea M Mar 8 '11 at 15:40