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This was an exercise using the first Chebyshev function, $\vartheta(x)= \sum_{p \leq x} \log p.$ The question is simply how to prove (2) below, the rest is my two thoughts on how to proceed. [Edit: trimmed after answer.]

$p_k$ is a number with k prime factors including repetitions. It is not the kth prime. $\pi_k(x)$ is the number of $p_k \leq x.$ "log" is the natural log but easier to read than "$\ln$" I think.

We know that $\sum_{p \leq x} \log p \sim x $ [Hardy and Wright p. 451]. We also know that $$\sum_{n\leq x} \log n \approx \int_2^x \log t\hspace{2mm}dt \approx x \log x - x.$$

So (omitting the indices),

$$\frac{\sum \log p}{\sum \log n} \approx \frac{\sum \log p}{x\log x - x}\sim \frac{x}{x\log x - x }\sim \frac{1}{\log x } \sim \frac{\pi(x)}{x}.$$

So we could ask whether

$$(1).....\hspace{5mm}\frac{\sum \log p_k}{\sum \log n}\hspace{2mm} ?\sim \frac{\pi_k(x)}{ x}.$$

For example: for x = 10000, $\pi_5 = 963,$ so $\frac{\pi_5(x) }{x} = 0.0963$ and $\frac{\sum \log p_5}{\sum \log 10^4} \approx 0.0974$

So from (1) via the prior line we would get $$(2)...\hspace{4mm}\frac{\pi_k(x)}{\sum \log p_k} ? \sim \frac{1}{\log x}.$$

Since $ \sum_k \pi_k(x) = x,$ we would have

$$(3)...\hspace{4mm}\sum_k \sum_{p_{k}(m) \leq x} \log p_{k}(m) \sim x \log x. $$

In case it's not clear: for each k we sum over $m = 1,2,3,... $ with $ p_k(m) \leq x$ and then we sum over k such that $p_k \leq x$ for some m.

So for $$\pi_2(x) \approx \frac{\sum \log p_2}{\log x}$$ I thought to compare it to the generalized PNT... [Edit: example omitted]

Or from (2) we could get that $\frac{\pi_k(x) = m}{\sum_{n=1}^m \log p_k(n)}\sim \frac{1}{\log x},$ so

$$ \frac{(1/m) m }{(1/m) \log \prod p_k(m)}\sim \frac{1}{\log x} $$ and

$$ \sqrt[m]{p_k(1)p_k(2)...p_k(m)} \sim x \hspace{4mm}...?$$

Suggestions/corrections welcome.

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up vote 2 down vote accepted

We have that $$\pi_k(x)\sim\sum_{p_k\leq x}1 \sim \frac{x (\log \log x)^{k-1}}{(k-1)!\log x}\ \ \ \ \ \ \ \ \ \ (A)$$ and $$\vartheta_k(x)\sim\sum_{p_k\leq x}\log (p_k) \sim \frac{x (\log \log x)^{k-1}}{(k-1)!},\ \ \ \ \ \ \ \ \ \ (B)$$ so I can reinterpret your question as how does $(B)$ follow from $(A)$, as that implies your equation $2$. Equivalently, you are asking, if I know $(A)$, can I prove that $$\theta_k(x)\sim \pi_k(x)\log x.$$ The short answer is that this follows directly from partial summation. First, lets write the sum as a Riemann Stieltjes integral $$\sum_{p_k\leq x} \log p_k=\int_1^x \log t d\left(\pi_k(t)\right),$$ and then using integration by parts, this equals $$\pi_k(x)\log x-\int_{1}^x\frac{\pi_k(t)}{t}dt.$$ Now, it is possible to use the asymptotic $(A)$ to bound the second integral from above, and we have reduced your question to an elementary real analysis problem.

You may also want to take a look at the paper: A simple proof of a theorem of Landau by E.M. Wright. This paper gives an elementary proof of $(A)$ using the hyperbola method.

Additionally, here is a note related problem which I wrote at some point. It may interest you.

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