Bounding summation

Question

Let $T$ be a sufficiently large real number and $\gamma$ be imaginary part of non-trivial zeros of Riemann zeta function. I would like to bound the sum like $$\sum_{0<\gamma\leq T}\gamma^{-1/4}.$$ I see the paper proceeds as follows. By partial summation, we have $$\sum_{0<\gamma\leq T}\gamma^{-1/4}=T^{-1/4}N(T)+\frac{1}{4}\int_{0}^{T}t^{-\frac{5}{4}}N(t)dt.$$ Note that, as $T\rightarrow\infty$, $$N(T)=\sum_{0<\gamma\leq T}1=\frac{T}{2\pi}\log\frac{T}{2\pi e}+O(\log T).$$ Then we have \begin{align} \sum_{0<\gamma\leq T}\gamma^{-1/4}\ll T^{3/4}\log T+\int_{1}^{T}t^{-5/4}t\log tdt. \end{align} I don't understand why the interval of integration change from [0,T] to [1,T] and why we can use $N(t)=O(t\log t)$ for $1\leq t\leq T$. Could someone please explain this?