I´m working on a book for my seminar. And I am hard stuck on the proof of one Theorem.
So the Theorem says
For $~T\ge 2$
$N(T+1)-N(T)\ll\log(T)$,
where N(T) is the amount of zeros of zeta/xi with imaginary part > T
Proof.
Wie use Jensen formula $\biggl(\int\limits_{0}^{R}\frac{n(r)}{r}dr=\frac{1}{2\pi}\int\limits_{0}^{2\pi}\log(\vert F(R~e^{i\theta})\vert)d\theta\biggl)$ to the function
$F(s):=\frac{\xi(s+2+iT)}{\xi(2+iT)}~~~$ n(r) ist the amount of zeros in the disk $\vert s\vert\leq r$
with $R=3$. For $r\in [\sqrt{5},3]$, the disk $\vert s\vert\leq r$ includes the rectangle $-2,~-2+i,~-1,~-1+i$.
So we have $n(r)\ge N(T+1)-N(T)$. <- HWO DO WE GET THIS ESTIMATE????????
With the previous Theorems we have
$\log(\vert F(s)\vert)\ll\log(T)~~~(\vert s\vert\leq 3)$ With Jensen formula follows,
$(N(T+1)-N(T))\log\biggl(\frac{3}{\sqrt{5}}\biggl)\leq\int\limits_{\sqrt{5}}^{3}\frac{n(r)}{r}dr\ll\log(T)$.
SO we got the claim.
Anyone has an idea hwo to get to the part in the middle?
Note that for, say, $T \geq 5$, $F(s)=0$ iff $s+2+iT$ is a zero of $\zeta$.
For all $z$ with imaginary part between $T$ and $T+1$, such that $\zeta(z)=0$, we can write $z=2+s+iT$ where the real part of $s$ is between $1$ and $2$ (because $\zeta$’s non-real are in the critical band) zeros are and its imaginary part is between $0$ and $1$, thus $|s|^2 \leq 5$ and $s$ is a zero of $F$ of the same order as $z$ in $\zeta$.
Therefore, $|N(T+1)-N(T)| \leq n(\sqrt{5})$.