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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
up vote 4 down vote accepted

What if you don't integrate over all of $(s,\infty)$? Say, fix $S$ large enough, and...

$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\int_s^S\frac{1}{\sigma-1}+O(1)d\sigma=\log\frac{1}{s-1}+O(1) $$


$$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\sum_p\int_s^S\frac{\log p}{p^\sigma}d\sigma=\sum_{p}\left(\frac{1}{p^s}-\frac{1}{p^S}\right)=\sum_p\frac{1}{p^s}~+O(1) $$

for $s<S$ as $s\to1^+$.

(Some uniform convergence stuff needs to be checked so that the interchange is justified.)

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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.

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Mobius inversion way is pretty nice! (+1) – user 1618033 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 (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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