# Abel's Summation formula help.

I want to be able to show that,

$$\sum_{p} \log(p) p^{-s} = s \int_{1}^{\infty} \frac{\theta(t)}{t^{s+1}} dt$$ 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,

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

Is this the right formula to use?

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 $x\rightarrow\infty$.
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 $$\int_1^\infty \chi_{[p,\infty)}(t) t^{-s-1}\, dt = \int_p^\infty t^{-s-1} \, dt = \frac{1}{s} p^{-s}.$$