Reply to thread

Message: <blockquote data-quote="AfterMath" data-source="post: 740117">I have a question regarding a conjecture due to H. L. Montgomery on the number of primes in short intervals. The conjecture apparently arises from probabilistic reasoning upon assuming the Riemann Hypothesis and some statistical randomness across the ordinates $\gamma$ of the nontrivial zeros of the Riemann zeta function. Let $\psi(x)=\sum_{n\le x}\Lambda(n)$ denote the (second) Chebyshev function. Then (considering only $\varepsilon>0$), he makes the followingConjecture. (Montgomery) $$\psi(x+h)-\psi(x)=h+O_{\varepsilon}(h^{\frac{1}{2}}x^{\varepsilon})$$ for $2\le h\le x$.My question: Which heuristic arguments support this conjecture (excluding numerical verification)?The conjecture appears in several places:(A) H. L. Montgomery, "Problems concerning prime numbers", Proceedings of symposia in pure mathematics, Vol. XXVIII, pp.307-310, Mathematical developments arising from Hilbert Problems (1976), AMS. Providence, Rhode island. [See p.309](B) D. A. Goldston, "On a result of Littlewood concerning prime numbers", Acta Arithmetica, Vol.40 (1982), pp. 263-271. [See p.269](C) H. L. Montgomery, R. C. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge University Press (2007) [Conjecture 13.4, p.422]Following the argument of Montgomery and Vaughan (p.422), is something I manage up to a certain point, but I'm not sure how to "get there". Specifically, using the explicit formula for $\psi(x)$ on the form $$\psi(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\frac{\zeta'}{\zeta}(0)-\frac{1}{2}\log(1-x^{-2})+\frac{1}{2}\Lambda(x) \hspace{5mm}(x>1),$$ we can write $$\psi(x+h)-\psi(x)=h-\sum_{|\gamma|\le T}C(\rho)+\lim_{U\to \infty}\sum_{T<|\gamma|\le U}C(\rho)+ O\big(\hspace{-0.2mm}\log\hspace{0.4mm}\max(x,h)\big)$$ for $x,h\ge 2$, say, where $$C(\rho)=\frac{(x+h)^{\rho}-x^{\rho}}{\rho} \ll \min\Big(hx^{\beta-1}, \frac{x^{\beta}}{|\gamma|}\Big).$$Now assume that the Riemann hypothesis is true, and write \begin{align*}C(\rho)=&\;\int_{x}^{x+h}t^{\rho-1}dt=\int_{x}^{x+h}x^{\rho-1}dt+\int_{x}^{x+h}t^{\rho-1}-x^{\rho-1}dt\\ =&\; hx^{\rho-1}+\int_{0}^{h}(x+t)^{\rho-1}-x^{\rho-1}dt\\ =&\; hx^{\rho-1}+\int_{0}^{h}(\rho-1)\int_{0}^{t}(x+z)^{\rho-2}dzdt\\[1mm] =&\; hx^{-\frac{1}{2}+\gamma i}+O\big(h^2x^{-\frac{3}{2}}|\gamma|\big). \hspace{30mm} (\dagger) \end{align*} Ignoring the error here for the moment and taking $T=x/h$, then $$\sum_{|\gamma|\le x/h}hx^{-\frac{1}{2}+\gamma i}=\frac{h}{\sqrt{x}}\sum_{|\gamma|\le x/h}\text{e}^{\gamma \log(x)i}.$$ Now, if we were to replace the $\gamma \log x$-s with independent and identically distributed uniform random variables $(Y_n)_{n=1}^{\infty}$ on the interval $[0,2\pi)$ , then, as this <a href="https://mathoverflow.net/questions/89478/magnitude-of-the-sum-of-complex-i-u-d-random-variables-in-the-unit-circle">this post</a> shows, we could conjecture that the above sum behaves like $$\frac{h}{\sqrt{x}}\sum_{n\ll N(x/H)}\text{e}^{Y_n i}\ll_{\varepsilon} \frac{h}{\sqrt{x}}N\big(\frac{x}{h}\big)^{1/2+\varepsilon}\ll_{\varepsilon} \frac{h}{\sqrt{x}}\Big(\frac{x}{h}\log\big(\frac{x}{h}\big)\Big)^{\frac{1}{2}+\varepsilon}=h^{\frac{1}{2}-\varepsilon}x^{\varepsilon}\big(\hspace{-0.2mm}\log \frac{x}{h}\big)^{\frac{1}{2}+\varepsilon},$$ as in Montgomery's conjecture.This is the point where I get stuck. For indeed, there are two unresolved sizes here. The first is the contribution of the large zeros: $$\sum_{|\gamma|>x/h}C(\rho), \hspace{15mm} (\ddagger)$$ and the second is the contribution of the error term in $(\dagger)$. The contribution of the error in $(\dagger)$ may not be too hard to resolve, but I am quite unable to show that the sum $(\ddagger)$ is of order $O_{\varepsilon}(h^{\frac{1}{2}}x^{\varepsilon})$. For example, using the explicit formula for $\psi(x)$ on the (commonly stated) form $$\psi(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}+O\Big(\frac{x\log^2(xT)}{T}+\log x\Big) \hspace{4mm} (x,T\ge 2),$$ I obtain $$\sum_{|\gamma|>x/h}C(\rho)\ll \frac{(x+h)\log^2((x+h)\frac{x}{h})}{x/h}\ll h\log^2 x$$ if $2\le h\le x$, which is not $O_{\varepsilon}(h^{\frac{1}{x}}x^{\varepsilon})$ unless $h\ll x^{2\varepsilon}(\log x)^{-4}$ (and this does not permit $2\le h\le x$ if, say, $0<\varepsilon<1/2$). The problem may be that the explicit formula with error term used here is unconditional, and that a better formula assuming RH should be employed.Montgomery and Vaughan say about this (on p.422), that "The contribution of zeros with $|\gamma|>x/h$ can be attenuated by employing a smoother weight, but no amount of smoothing will eliminate the smaller zeros." By smoothing, they here likely mean that the explicit formula for $\sum_{n\le x}\Lambda(n)=\sum_{n=1}^{\infty}\Lambda(n)1_{(x,x+h]}(n)$ should be replaced by an explicit formula for $$\sum_{n=1}^{\infty}\Lambda(n)w(n),$$ where $w(n)=w(n;x,h)$ is a `weight function'. This weight function should be such that it gives a useful explicit formula (meaning that the contribution of the $\rho$-s in the right hand side decays rapidly as a function of $|\gamma|$), but also approximates the indicator function $1_{(x,x+h]}(n)$ to such an extent that $\sum_{n}\Lambda(n)w(n)$ is close to $\psi(x+h)-\psi(x)$.I have been playing around with such explicit formulas lately, including the formulas \begin{align*} \frac{1}{k!}\sum_{n\le x}\Lambda(n)(x-n)^{k}=&\; \frac{x^{k+1}}{(k+1)!}-\frac{x^{k}}{k!}\frac{\zeta'}{\zeta}(0)-\sum_{\rho}\frac{x^{\rho+k}}{\rho(\rho+1)\cdots (\rho+k)}+ \sum_{0\le j\le (k-1)/2}\frac{x^{k-2j-1}}{(2j+1)!(k-2j-1)!}\frac{\zeta'}{\zeta}(-2j-1)\\[1.5mm] +&\; (-1)^{k}\sum_{j>k/2}x^{k-2j}\frac{(2j-k-1)!}{(2j)!}+\sum_{0<j\le k/2} \frac{x^{k-2j}}{(2j)!(k-2j)!}\Big(\frac{1}{2}\frac{\zeta''}{\zeta'}(-2j)-\log x+\sum_{\substack{r=-2j\\ r\ne 0}}^{k-2j}r^{-1}\Big) \hspace{5mm} (x\ge 1, k\in \mathbb{N}^{+}), \\[2mm] \frac{1}{\Gamma(\xi+1)}\sum_{n<x}\Lambda(n)(x-n)^{\xi}=&\; \frac{x^{\xi+1}}{\Gamma(\xi+2)}-\sum_{\rho}\frac{x^{\rho+\xi}\Gamma(\rho)}{\Gamma(\rho+\xi+1)}-\frac{x^{\xi}}{\Gamma(\xi+1)}\frac{\zeta'}{\zeta}(0)+\sum_{j=0}^{\infty} \frac{x^{\xi-2j-1}}{\Gamma(2j+2)\Gamma(\xi-2j+1)}\cdot \frac{\zeta'}{\zeta}(-2j-1)\\ -&\;\sum_{j=1}^{\infty} \frac{x^{\xi-2j}}{\Gamma(2j+1)\Gamma(\xi-2j+1)}\Big(-C_{\text{Eul}}+\sum_{k=1}^{2j}\frac{1}{k}+\frac{1}{2}\frac{\zeta''}{\zeta'}(-2j)+\log(x)-\psi^{(0)}(\xi-2j+1)\Big) \hspace{5mm} (x\ge 1, \text{Re }\xi>0, \xi \not \in \mathbb{Z}), \\[2mm] \sum_{n\le x}\Lambda(n)\log(x/n)=&\;x-\sum_{\rho}\frac{x^{\rho}}{\rho}-(\log 2\pi)\log x-(\frac{\zeta'}{\zeta})'(0)-\frac{1}{4}\sum_{k=1}^{\infty}\frac{x^{-2k}}{k^2} \hspace{5mm} (x>1),\\[2mm] \sum_{n=1}^{\infty}\Lambda(n)\text{e}^{-n/z}=&\; z-\sum_{\rho}\Gamma(\rho)z^{\rho}-\text{e}^{-1/z}\log 2\pi -(-1+\cosh 1/z)\log z+\sum_{k=1}^{\infty}(-1)^{k}\frac{\zeta'}{\zeta}(k+1)\frac{z^{-k}}{k!}\\ -&\;\sum_{k=0}^{\infty}\frac{\Gamma'}{\Gamma}(2k+2)\frac{z^{-2k-1}}{(2k+1)!} \hspace{5mm} (\text{Re }z>0). \end{align*} However, I am not able to get the desired result. Indeed, I am not able to combine the weight functions provided in the formulas above, to get a weight function approximating $1_{(x,x+h]}(n)$ to a reasonable extent while giving a reasonable explicit formula. This may be because:<ol> <li data-xf-list-type="ol">I have not found any literature explaining what would constitute a good smoothing of $1_{(x,x+h]}(n)$ in this case (i.e. how does the smoothness and cut-off play a role)</li> <li data-xf-list-type="ol">Taking a weighted sum of one of the explicit formulas above, one could likely approximate $1_{(x,x+h]}(n)$ by expressions of the form $\sum_{n=1}^{N}a_nw(n,x_n,h_n)$. Here I am not at all sure which function spaces I would try to do an approximation in (e.g. which norm).</li> </ol>These investigations are related to my Master's thesis on primes in short intervals, and I would highly appreciate if anyone could comment on how Montgomery's conjecture can be backed up. (And possibly also what the goal/strategy of the smoothing process should be, if you would be kind enough to explain).Sincerely, R.</blockquote>

Name

Verification