How should I evaluate the following sums
1, $\sum_{p\leq t}\frac{log^2(p)}{p}$
where the sum is taken over all prime numbers.
2, $\sum_{n\leq X}\frac{\Lambda^2(n)}{n}$
where $\Lambda(\cdot)$ is the Mangoldt function
RK: By evaluating, I mean to find an approximating formula as the following one
$\sum_{p\leq t}\frac{log(p)}{p}=log(t)+O(1)$
You can write using partial summation formula,
$\displaystyle \begin{align} \sum\limits_{p \le x} \dfrac{\log^2 p}{p} = \sum\limits_{p \le x} \log p\dfrac{\log p}{p} &= \log x\sum\limits_{p \le x} \dfrac{\log p}{p} - \int_2^{x} \frac{\sum\limits_{p \le t} \frac{\log p}{p}}{t}\,dt \\&= \log x(\log x + \mathcal{O}(1)) - \int_2^{x} \frac{\log t + \mathcal{O}(1)}{t}\,dt \\&= \frac{1}{2}\log^2 x + \mathcal{O}(\log x) \end{align}$
As for the second one you can use partial summation like the first one or see that,
$\displaystyle \sum\limits_{n \le x} \frac{\Lambda^2(n)}{n} \le \sum\limits_{p \le x} \log^2 p\left(\frac{1}{p}+\frac{1}{p^2}+\cdots\right) = \sum\limits_{p \le x} \dfrac{\log^2 p}{p} +\mathcal{O}\left(\sum\limits_{p} \dfrac{\log^2 p}{p^2}\right)$
Since, $\displaystyle \sum\limits_{p} \dfrac{\log^2 p}{p^2}$ is a constant.
Thus, $\displaystyle \sum\limits_{n \le x} \frac{\Lambda^2(n)}{n} = \sum\limits_{p \le x} \dfrac{\log^2 p}{p} +\mathcal{O}(1)$.