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In https://academic.oup.com/blms/article-abstract/20/2/121/266256?redirectedFrom=fulltext Vaughan shows the following bounds for the $L^1$-mean of the exponential sum over primes $$\sqrt x\ll \int _0^1\left |\sum _{n\leq x}\Lambda (n)e(\alpha n)\right |d\alpha \ll \sqrt {x\log x}$$ and further says it's likely to be $\asymp \sqrt {x\log x}$. (Note that by Cauchy-Schwarz the upper bound is trivial so the question is whether the $\sqrt {\log x}$ power is really there).

Question: Is there a heuristic reason why this should be so? Or are there any numerical calculations of it?

I initially thought it may not be true based on the following reasoning: Replacing $\Lambda (n)$ by $\omega (n)$, we can get the same $\sqrt x$ lower bound whilst the trivial CS bound is $\sqrt x\log \log x$, however I believe I can show that for $\omega (n)$ the sum really is $\asymp \sqrt x$. I guess too optimistically I thought the removing the $\log \log x$ for $\omega (n)$ may amount to removing the $\sqrt {\log x}$ for $\Lambda (n)$. However, there may really be no link between $\omega (n)$ and $\Lambda (n)$ and there's no reason to expect any similarities in behaviour.

I have tried to do some numerical calculations for the sum with $\Lambda (n)$ and it does seem that the true size is indeed $\sqrt {x\log x}$, but I don't want to completely eliminate the possibility of it being smaller based on my (bad) programming skills either.

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  • $\begingroup$ A heuristic reason can probably be given by a random model of the primes or a random model of Möbius. Certainly for a random set of numbers of size $x/\log x$ between $1$ and $x$, the Cauchy-Schwarz bound for the exponential sum over that set is close to sharp. Really one should do a more sophisticated heuristic where one chooses a random set of numbers coprime to small primes so that one correctly predicts the large values of the exponential sum near ration $\alpha$ but this shouldn't affect the $L^1$ norm much. $\endgroup$ Commented Jun 24, 2024 at 18:37
  • $\begingroup$ Or one can express von Mangoldt in terms of Möbius and use a Möbius randomness heuristic. In both cases, the point will be to check that for most $\alpha$ the sum $\sum_{n\leq x }\Lambda(n) e(\alpha n)$ has an approximately Gaussian distribution whose variance does not depend much on $\alpha$ except for a few exceptional values of $\alpha$ near rational numbers. The expected $L^1$ norm of a Gaussian is a universal constant times the integral of the standard deviation which if the standard deviation is approximately constant is close to the $L^2$ norm. $\endgroup$ Commented Jun 24, 2024 at 18:40
  • $\begingroup$ $\omega(n)$ is dominated by the contribution of small prime factors (i.e. for any $\delta>0$ we can estimate $\omega(n)$ to within $O(1)$ knowing only the prime factors $<n^\delta$), so in this case almost all the $L^2$ norm of the exponential sum is concentrated near rational numbers of small denominator where it doesn't contribute much to the $L^1$ norm. $\endgroup$ Commented Jun 24, 2024 at 18:43
  • $\begingroup$ On the other hand, while small prime factors may be used to roughly approximate $\Lambda(n)$ (if a number has small prime factors it's certainly not prime, and if it has none it is more likely to be prime) the approximation is not very good, and the error can still be as large as $\log n/2$ even knowing all prime factors of size $< n^{1/2-o(1)}$ (since $n$ could be a product of two primes of size close to $\sqrt{n}$ or prime itself). $\endgroup$ Commented Jun 24, 2024 at 18:44
  • $\begingroup$ It feels like you're indeed answering my implied question "should there be a parallel between both norms" although I probably need to think longer about your comments to fully understand what you're saying, so in the meantime this comment is just a "thanks for those comments":) (f you want to write it as an answer I can happily accept it). $\endgroup$ Commented Jun 25, 2024 at 15:55

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