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I am trying to show that the prime number theorem is equivalent to the statement that $\sum_{p\le x} \log(p) \sim x \implies \pi(x) \sim x/\log(x)$. It was suggested that I use Abel summation with $a_n = \begin{cases}\log(n), \ \text{$n$ prime}, \\ 0, \ \text{otherwise.} \end{cases}$ and $b_n = 1/\log(x)$. It was also recommended I begin by showing $$\int_2^x \frac{1}{\log^2(t)} \sim O\left(\frac{x}{\log^2(x)}\right)$$ and splitting up this latter integral to $\int_2^x = \int_2^\sqrt{x} + \int_\sqrt{x}^x$. I've done the latter, and I end up with $$\int_2^x \frac{dt}{\log^2(t)}dt = \frac{2}{\log(2)} + \int_2^x \frac{1}{\log(t)}dt - \frac{x}{\log(x)}.$$ Using Abel summation, I can say that $$\sum_{1 \le k \le x} \frac{a_k}{\log(k)} = \left(\sum_{1 \le k \le x} a_k\right) \frac{1}{\log(x)} + \int_1^x \left(\sum_{1 \le k \le x}a_k \right)\frac{1}{t\log^2(t)} dt.$$ Beyond this, I'm a tad stuck...

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  • $\begingroup$ I side the last integral the bounds for $k$ should be $1\leq k\leq x$ $\endgroup$ – Julian Mejia May 3 at 4:09
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You are almost there. Note that $$\sum_{1\leq k\leq x} \frac{a_k}{\log k}= \pi(x)$$

Now, if $$\sum_{1\leq k\leq x} a_k=\sum_{p\leq x}\log p\sim x$$ Then $$(\sum_{1\leq k\leq x} a_k)\frac{1}{\log x}\sim x/\log x$$ And $$\int_1^x(\sum_{1\leq k\leq t} a_k)\frac{1}{t\log^2t}=O(\int_1^x t\frac{1}{t\log^2t})=O(x/\log^2x)=o(x/\log x)$$

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