Why $\sum_{\gamma>0} (\frac{\sin(\gamma/N)}{\gamma/N})^{2} = \mathcal{O} (N\log N)$, $\gamma$ is the imaginary part of the zeros of $\zeta (x)$.

Question

Why $\sum_{\gamma>0} (\frac{\sin(\gamma/N)}{\gamma/N})^{2} = \mathcal{O} (N\log N)$, $\gamma$ is the imaginary part of the zeros of $\zeta (x)$.

In Monthomery's Multiplicative Number threory I.Classical theory, chapter 15 ,p479, in the proof of littlewood's result on the error term of the prime number theorem, it gives $\sum_{\gamma>0} (\frac{\sin(\gamma/N)}{\gamma/N})^{2} = \mathcal{O} (N\log N)$.

I want to know how does he do it? I feel I can only get as far as $\mathcal{O} (N\log^2N)$, by $\sum_{\gamma>0} (\frac{\sin(\gamma/N)}{\gamma/N})^{2} \le\sum_{T>\gamma>0} (\frac{1}{\gamma/N})+\sum_{\gamma>T} (\frac{1}{\gamma/N})^{2} (T=N\log N).$

Any pointers are appreciated!

Answer 1

The density of zeros theorem $$\sum_{\gamma<T}\gamma=O(\sum_{n< T} \log n)$$ Gives $$\sum_{\gamma<N} (\frac{sin(\gamma/N)}{\gamma/N})^2 =O(\sum_{n< N} \log n)= O(N\log N)$$

$$\sum_{\gamma\ge N} (\frac{sin(\gamma/N)}{\gamma/N})^2 = O(\sum_{n\ge N}\log n (\frac{1}{n/N})^2 ) = O(N^2\frac{\log N}{N})$$