2
$\begingroup$

I want to be able to show that,

\begin{equation} \sum_{p} \log(p) p^{-s} = s \int_{1}^{\infty} \frac{\theta(t)}{t^{s+1}} dt \end{equation} where $\theta(x) =\sum_{p \le x} \log p$. and $\theta(x) = O(x)$.

Using Abel's Summation formula, letting $a(n) = \log(n)$ and $\phi(n) = n^{-s}$,from the definitions here https://en.wikipedia.org/wiki/Abel%27s_summation_formula

I have,

\begin{equation} \sum_{1 \le p \le x} \log p (x^{-s}) - \int_{1}^{x} \sum_{1 \le p \le x} \log u (u^{-s})' du \end{equation}

Is this the right formula to use?

$\endgroup$
1
$\begingroup$

Yes, using the definition of theta function $\theta\left(x\right)=\sum_{p\leq x}\log\left(p\right) $ you have, using Abel summation $$\sum_{p\leq x}\frac{\log\left(p\right)}{p^{s}}=\frac{\theta\left(x\right)}{x^{s}}+s\int_{2}^{x}\frac{\theta\left(t\right)}{t^{s+1}}dt $$ and now, assuming $\textrm{Re}\left(s\right)>1 $, take the limit $x\rightarrow\infty $.

$\endgroup$
0
$\begingroup$

I think you meant to write $\theta(x) = \sum_{p \le x} \log \boldsymbol{p}$. Anyway, the derivation is actually rather straightforward.

Here is a useful trick. Let $\chi_{E}(x)$ be the function that is $1$ if $x \in E$ and is $0$ otherwise. Then you can write $$\sum_{p \le x} \log p = \sum_p \log p \chi_{[p,\infty)}(x).$$ Since everything is nonnegative you can swap summation and integral: $$ \int_1^\infty \frac{\theta(t)}{t^{s+1}} \, dt = \int_1^\infty \left( \sum_p \log p \chi_{[p,\infty)}(t) \right)t^{-s-1} \, dt = \sum_p \log p \int_1^\infty \chi_{[p,\infty)}(t) t^{-s-1}\, dt$$ where $$\int_1^\infty \chi_{[p,\infty)}(t) t^{-s-1}\, dt = \int_p^\infty t^{-s-1} \, dt = \frac{1}{s} p^{-s}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.