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On the blending of real/complex analysis with number theory. The study involves distribution of prime numbers and other problems and helps giving asymptotic estimates to these.

7 votes

Sum over characters

Instead of using van der Corput, why don't you express the sum as a complex integral? To simplify matters, consider a smooth sum, i.e., $$S(x,t) = \sum_n f(n/x) n^{-i t},$$ where $f$ is fixed $C^\inf …
H A Helfgott's user avatar
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0 votes

Square-free numbers in an interval

Also: see Cohen-Dress, "Estimations numériques du reste de la fonction sommatoire relative aux entiers sans facteurs carrés", 1988 (https://www.imo.universite-paris-saclay.fr/~biblio/numerisation/docs …
H A Helfgott's user avatar
  • 20.2k
3 votes

$L_2$ bounds for tails of $\zeta(s)$ on a vertical line

Thanks, GH! Let me have another go. I think the following is the right way to go about things, at least if one wants something self-contained and with good, explicit constants. (The latter more or les …
H A Helfgott's user avatar
  • 20.2k
3 votes

Double sum of negative powers of integers: a direct approach?

Let me carry out matters using a complex-analytical approach, as Lucia suggests, and then say where the difficulty lies. Let $0<\beta<\alpha\leq 1$. First of all, as Lucia says, $$\sum_{m\leq x} \fra …
H A Helfgott's user avatar
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5 votes

Effective, explicit bound on $L'(1,\chi)/L(1,\chi)$

Self-answer: yes, the answer is Proposition 1.12 in Bennett-Martin-O'Bryant-Rechnitzer: https://projecteuclid.org/euclid.ijm/1552442669 They show that, for $\chi$ a Dirichlet character modulo $q\geq …
H A Helfgott's user avatar
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1 vote

Truncated Perron - logarithm-free error term?

Ah, I get it - the factor of $\log x$ is really there because the truncation is sharp. If the truncation is continuous (and of bounded variation), then, not unexpectedly, the factor disappears. (See, …
H A Helfgott's user avatar
  • 20.2k
4 votes

Bound on $L^2$ norm of $1/\zeta(1+i t)$?

While @Lucia's answer is my favorite, I thought it might be worthwhile to sketch the Plancherel-based argument I alluded to in the above. First of all, let $\phi:[0,\infty)\to \mathbb{R}$ be such that …
H A Helfgott's user avatar
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3 votes

$\zeta(s) = 1 + X^{1-s}/(s-1) + ...$?

One can in fact get a $c$ better than $2/3$. Sketch: for $X\leq 11/10$, do an Euler-Maclaurin expansion up to order $3$, and use the fact that $B_3(x)$ is small for $x$ small ($0\leq x\leq 1/10$; fo …
H A Helfgott's user avatar
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2 votes

Explicit bound on $\zeta(s)$ inside a zero-free region?

Mark Coleman just answered: the result in his notes is Prop. 5.3.2 in G. J. O. Jameson's The Prime Number Theorem.
H A Helfgott's user avatar
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3 votes

Square-free numbers in an interval

Let me just show how to derive a simple bound that has been mentioned in the comments. We are trying to bound the estimate the number $Q(x,x+u)$ of squarefree integers in $(x,x+u]$. We can now apply S …
H A Helfgott's user avatar
  • 20.2k
38 votes
Accepted

A variant of the Goldbach Conjecture

Yes - the standard proof of Vinogradov's result by means of the circle method gives this result. You just need to examine an integral $$\int_{\mathbb{R}/\mathbb{Z}} (\widehat{f}(\alpha))^2 \widehat{f} …
H A Helfgott's user avatar
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2 votes

Mean value theorem for Dirichlet series - optimize?

The following is an approach inspired by Selberg's treatment of the large sieve, and more directly by a sketch in Granville-Harper-Soundararajan (on the case of prime support). There remains an optimi …
H A Helfgott's user avatar
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3 votes

$|L'(1,\chi)/L(1,\chi)|$

One way to improve the explicit bound mentioned in the question is simply to compute $L'(1,\chi)/L(1,\chi)$ for whatever characters $\chi$ are needed. The bound in the question depends on a GRH verifi …
H A Helfgott's user avatar
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1 vote

Schur's proof of Hilbert's inequality: streamlining?

One can even attempt the same argument to attempt to derive a bound of the form $$\left|\sum_{i\ne j} \frac{\overline{v_i} v_j}{r_i-r_j} \right| \leq C \sum_i \delta_i^{-1} |v_i|^2,$$ where $\delta_i …
H A Helfgott's user avatar
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1 vote

Distribution $f$ such that (a) $\widehat{f}$ has compact support, (b) $\mathbb{E}(|X|)$ is m...

Warning: the following answer is (a) maybe a bit careless in a nineteenth-century sort of way, (b) missing its final step (which may be obvious to others, and/or amount to looking things up in the rig …
H A Helfgott's user avatar
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