# Approximating $\vartheta(x)=\sum_{p\le x} \log(p)?$

Consider the first Chebyshev function $$\vartheta(x)=\sum_{p\le x} \log(p)$$ where the sum runs over the primes less than or equal to $$x$$.

I wanted to approximate $$\vartheta(x).$$

My attempt was $$f(x)=\sum_{n \ge 2}^x n^\frac{-1}{n}.$$ It overcounts by about $$2$$ at $$x=101$$ giving a value of $$90.177$$ whereas $$\vartheta(x)$$ gives $$88.344.$$ I'm not sure how $$f(x)$$ performs as $$x$$ increases.

Is $$f(x)\sim \vartheta(x)?$$

• – lhf
Aug 21, 2020 at 11:03

Per $$n^{-1/n} = \exp \biggl(-\frac{\log n}{n}\biggr) = 1 - \frac{\log n}{n} + O\biggl(\frac{(\log n)^2}{n^2}\biggr)$$ we have $$f(x) = x - \frac{1}{2}(\log x)^2 + O(1)\,.$$ Thus we have $$f(x) \sim \vartheta(x) \sim x$$.

But $$f(x)$$ stays much closer to $$x$$ than $$\vartheta(x)$$. By a result of Littlewood we have $$\vartheta(x) - x \in \Omega_{\pm}(\sqrt{x}\, \log \log \log x)$$ while $$x - f(x)$$ is of much smaller magnitude, and always positive.