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I have been studying the proof given by Dirichlet for infinitely many primes in an arithmetic progression and I cannot help but wonder this :

Why did Dirichlet need to first show $$\sum_{p\leq x, \hspace{2mm} p \hspace{1mm} \equiv \hspace{1mm} h (\text{mod} \hspace{1mm} k)} \frac{\log (p)}{p} = \frac{1}{\phi(k)}\log(x) +O(1)$$ in order to prove his point? I understand how this implies the infinitude primes in an arithmetic progression but why use $\frac{\log (p)}{p}$ in the first place?

Isn't there a better way? What motivated Dirichlet to do this?

Gary
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    Mertens' first theorem (OP's asymptotic formula at the case $k=1$) is the first asymptotic formula related to prime numbers that emerges from decomposing the factorial. Perhaps this motivated Dirichlet to choose the weight $\log p/p$ – TravorLZH Jun 16 '21 at 12:00
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    You ask why Dirichlet was motivated to use that sum involving terms $\log(p)/p$. Dirichlet did no such thing. The proof he gave involved infinite series over all $p \equiv h \bmod k$, not finite sums of such $p \leq x$, and his weighting in the $p$th term was $1/p^s$ with $s > 1$, and he then let $s \to 1^+$. We call the result "Dirichlet's theorem" no matter what, but that doesn't mean a particular proof you read was actually the one due to him. – KCd Mar 22 '24 at 05:42

2 Answers2

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This is an excellent question, and it is worth understanding. The specific choice of weight function $\frac{\log p}{p}$ takes a bit of effort to explain, so I'll try to be as clear and organized as I can.

Let $\Lambda(n)$ denote the von Mangoldt function: $$ \Lambda(n) = \begin{cases} \log p & \text{if $n = p^k$}, \\ 0 & \text{otherwise}. \end{cases} $$ In general, when working with sums over prime numbers, say something of the form $$ \sum_{p\leq x} f(p), $$ it is generally easier to work with the weighted sums $$ \sum_{n\leq x} \Lambda(n) f(n). $$ The contribution of prime powers is generally small, so really the only terms contributing to the sum are $n=p$, and we get the sum over primes with a weight of $\log p$ at each prime. Since one can usually remove the weight $\log p$ via partial summation, there is really no technical reason to avoid using it.

The reasons one works with sums involving $\Lambda$ is that the Dirichlet series associated to $f(n)\Lambda(n)$ can often be written in terms of familiar $L$-functions, say $\zeta(s)$, the Riemann zeta function, or $L(s,\chi)$, the Dicihlet $L$-function associated to a Dirichlet character $\chi$.

Okay, so from what I've said so far, it seems clear that one should consider the sums $$ \sum_{\substack{n\leq x\\ n \equiv h(mod \hspace{1mm} k)}} \Lambda(n). $$ This begs the question, why include the extra weight $\frac{1}{n}$ (which is the same as $\frac{1}{p}$ because of the presence of $\Lambda$)? This is a more delicate question. In general, the logarithmic sums $$ \sum_{n\leq x} \frac{f(n)}{n} $$ are easier to handle than the summatory functions $$ \sum_{n\leq x} f(n). $$ The opening paragraphs of this article by Terry Tao go into detail about this. To quote him directly,

From a complex-analytic or Fourier-analytic viewpoint, the problem is that the logarithmic sums can usually be controlled just from knowledge of the Dirichlet series $\sum_{n} \frac{f(n)}{n^s}$ near $s=1$, but the summatory functions require control of the Dirichlet series for $s$ on or near a large portion of the line $\left\{1+it:t\in\boldsymbol{R}\right\}$.

A further discussion of this phenomenon can be found in these notes, linked by Tao in his article.

To illustrate this, it should be noted that the estimate

$$ \sum_{\substack{n\leq x\\ n \equiv h(mod \hspace{1mm} k)}} \Lambda(n) \sim \frac{x}{\phi(k)} $$ is the classic Siegel–Walfisz theorem, and is more difficult to prove than Dirichlet's theorem.

TL;DR - The specific weight $\frac{\log p}{p}$ is used because one simply needs to know less about the complex-analytic behavior of the relevant Dirichlet series (in this case the Dirichlet $L$-functions $L(s,\chi)$). Dirichlet only set out to prove an infinitude of primes in coprime residue classes; his theorem does not give an asymptotic count of such primes. The asymptotic count of such primes is precisely the content of the Siegel–Walfisz theorem, and one needs a deeper understanding of the Dirichlet $L$-functions $L(s,\chi)$ in order to prove it.

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    Just a small nitpick. The asymptotic for $\sum \Lambda(n)$ looks more like PNT for arithmetic progressions rather than Siegel-Walfisz (which gives some level of uniformity in $q$ including the error term, presumably by dealing with potential Siegel zeroes). – Erick Wong Jun 14 '21 at 07:12
  • @erick-wong Good observation. Yeah I thought about mentioning the range of $q$ for which SW is valid, but since I linked the wiki article, I didn't want add more to my answer than necessary. – Joshua Stucky Jun 14 '21 at 14:46
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To prove that there are infinitely many primes of the form $mod(a,b)$, consider the sieve:

$$\sum_\chi \sum_p^\infty \frac{\bar{\chi(a)}\chi(p)}{p} $$

$$=\sum_p \frac{1}{p} \sum_\chi \bar{\chi(a)}\chi(p) $$

$$=\sum_{p=mod(a,b)} \frac{1}{p} \phi(b) $$

since $\sum_\chi\bar{\chi(a)}\chi(p) = \phi(b)$ if $p=mod(a,b)$ and $0$ otherwise.

Hence

$$\boxed{\sum_{p=mod(a,b)} \frac{1}{p} = \frac{1}{\phi(b)} \sum_\chi \sum_p \frac{\bar{\chi(a)}\chi(p)}{p} }$$


$$= \frac{1}{\phi(b)} \sum_\chi \left(\log \sum_n \frac{ \bar{\chi(a)}\chi(n)}{n}\right) + O(1)$$

since $ \sum_p \frac{\bar{\chi(a)}\chi(p)}{p} = \log \sum_n \frac{\bar{\chi(a)}\chi(n)}{n} + O(1)$

$$= \frac{1}{\phi(b)} \left[\left( \log \sum_n \frac{\bar{\chi_0(a)}\chi_0(n)}{n}\right) + \sum_{\chi \ne \chi_0} \left(\log \sum_n \frac{\bar{\chi(a)}\chi(n)}{n}\right) \right]+ O(1)$$

$$= \frac{1}{\phi(b)} \left[\left( \log \sum_n \frac{1}{n}\right) + O(1) \right]+ O(1)$$

since $\chi_0 (k) = \bar{\chi_0 (k)} = 1$ for all $k$

$$= \infty$$

since $\sum_n \frac{1}{n}$ diverges.

Hence

$$\boxed{\sum_{p=mod(a,b)} \frac{1}{p} = \infty }$$

and so there are infinitely many primes of the form $am + b$.


The sum $\sum_p \frac{\log p}{p}$ was used for proving one of the lemmas to the theorem.

James
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  • One should really mention which part of the answer one is unhappy with before downvote stalking. Else it might be interpreted as petty bitterness. But then there is no valid reason to be bitter as long as one believes the sky is limitless and one's capability to explore it is limitless. So then which is it... – James May 22 '25 at 01:15