I try to understand an application of distribution of primes in arithmetic progressions

Let $$f(x) = \sum_{p \leq x p\equiv 3 \bmod 10} 1$$

So computing $f(40) = 3$ i.e. the primes: 3, 13, and 23

Now $$h(x) = \sum_{p \leq x p\equiv 3 \bmod 10} \ln p$$

My question is: what is 'the meaning of $h(x)$? and is $h(40)$ equal to: $\ln 3 + \ln 13 + \ln23$

and secondly: why is the relation between $f(x)$ and $h(x)$ equal to: $$h(x) = f(x) \ln x - \int_{2}^{x} \frac{f(t)}{t} dt$$

  • $\begingroup$ Yes. Because $\ln p =\ln x- \int_p^x \frac{dt}{t}$. As Hamidine said $h(x)$ is a weighted version of $f$, more regular because its Mellin transform is meromorphic. $\endgroup$ – reuns Apr 4 at 4:02

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