# From $\sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1)$ conclude that $\sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1)$

I'm reading a book on analytic number theory. It asks me to prove:

$$\sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1) \tag{A}$$ and conclude, via integration, that: $$\sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1) \tag{B}$$

Now, I know how to prove $(A)$ via Abel Summation. However, when it comes to $(B),$ I have the problem that although:

$$\frac{d}{dx} \log(x-1) = \frac{1}{x-1}$$ and

$$\frac{d}{ds} p^{-s} = (-\log p)p^{-s}$$

I have the problem that when I integrate over $O(1)$, I get $\infty$, not $O(1)$.

What am I doing wrong? How do I get from $(A)$ to $(B)$?

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@PeterTamaroff: Nice. I didn't know > and $$could stack. – user36739 Aug 16 '12 at 1:07 I just removed the word "prime". – Pedro Tamaroff Aug 16 '12 at 1:09 Why do you "integrate over O(1)"? The way I see it, O(1) does not denote dependence on s but on the number of terms of the sum. – leonbloy Aug 16 '12 at 1:54 @leonbloy But the number of terms is fixed: it's infinity. The big O is for s\to1^+. – anon Aug 16 '12 at 1:58 @anon: ah, ok, my bad. – leonbloy Aug 16 '12 at 2:02 ## 3 Answers What if you don't integrate over all of (s,\infty)? Say, fix S large enough, and...$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\int_s^S\frac{1}{\sigma-1}+O(1)d\sigma=\log\frac{1}{s-1}+O(1) $$whereas$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\sum_p\int_s^S\frac{\log p}{p^\sigma}d\sigma=\sum_{p}\left(\frac{1}{p^s}-\frac{1}{p^S}\right)=\sum_p\frac{1}{p^s}~+O(1) $$for s<S as s\to1^+. (Some uniform convergence stuff needs to be checked so that the interchange is justified.) - The first idea I had when you posted in chat was to use infinite Möbius inversion. We have$$P(s)=\sum_p\frac{1}{p^s}=\sum_{n\ge1}\frac{\mu(n)}{n}\log\zeta(ns), $$where P(\cdot) is the prime zeta function and \zeta(\cdot) is the Riemann zeta function. (For those who want to check this: write \zeta in the Euler product form and then expand each Euler factor with a series expansion for \log individually; group terms appropriately.) As s\to1^+ all of the n\ge2 terms are already O(1) (put together), so we need only put \zeta(s)=\frac{1}{s-1}+O(1) inside the first logarithm. This route requires more work and preliminary information though. - Mobius inversion way is pretty nice! (+1) – Chris's sis the artist Aug 16 '12 at 5:56 Well we know that$$\zeta(\sigma)= \frac{1}{\sigma-1}$$when$$ \sigma \to1^+$$since zeta has a simple pole in \sigma=1 with residual one. Now your function$$\sum_p\frac{\log p}{p^s}$$can be expressed in terms of the zeta function. see http://people.math.jussieu.fr/~demarche/enseignements/2011-2012/M1-TDN/MM020-TD6-corrige.pdf (it's the exercise number 7, question g, in french :)). The tricky part is why could you integrate two functions which are equivalent in the sense that$$\sum_p \frac{\log p}{p^s} ∼ \frac{1}{s-1} ?

Here it is possible since $1/s-1$ is not integrable in a neighborhood of 1. Why?

here is an sketch of how you could procede. Let f, g be continuous on R such that f~g and $\int_0^\infty$fdx diverges. I will write this integral I in what follows and I$x$ when the upper bound is $x$ instead of + $\infty$. we would like to show that I (f) ~ I (g)

In what follows we have f ~ g at infinity. (I guess you could repeat that proof for an equivalence that holds in another neighborhood)

f ~ g then f (x) = g (x) (1 + h (x)) such that $gh \to0$ when x tends to + infinity (and h tends to $0$ as well).

So you've got I$x$ (f) / I$x$ (g) = 1 + I$x$ (gh) / I$x$ (g) then show that | I$x$ (gh) / I$x$ (g) | tends to $0$ in + infinity

since |I$x$ (gh) / I$x$ (g) | = I$a$ (gh) / I$x$ (g) + I$a-x$ (gh) / Ix (g) then | I$x$ (gh) / I$x$ (g) | $\leq$ I$a$(gh) / I$x$ (g) + sup (h) on [$a$, x]

and since h tends to $0$, you can choose $a$ large enough such that $sup (h) \leq \epsilon / 2$

Once $a$ is chosen, since I$a$ (gh) is a number and I$x$ (g) diverges then there exists $x$ large enough such that I$a$ (gh) / I$x$ (g) $\leq \epsilon / 2$. Finally, for $x$ large enough everything is inferior to a certain epsilon, and you can conclude.

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