Approximation of the Lerch zeta function

Question

I come across the statement in this paper, title "Garunkstis, R. and Laurincikas, A. (1996). On the Lerch zeta-function", pp. 342. It states that \begin{align*} \sum_{y<n\leq N} \frac{e^{2\pi i\lambda n}}{(n+u)^s} &= \frac{1}{(N+u)^{\sigma}}\color{red}{\sum_{y<n\leq N}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}} - \int_y^N \color{red}{\sum_{y<n\leq x}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}}\,d\frac{1}{(x+u)^{\sigma}}\\ &= \int_y^N \frac{e^{2\pi i\lambda v}}{(v+u)^s}\,dv+\frac{B_\lambda}{N^{\sigma}}+\color{blue}{\frac{B_{\lambda}}{\sigma}\frac{1}{y^{\sigma}}} \end{align*} using partial summation. However when I apply Abel's summation formula, I find that \begin{align*} \sum_{y<n\leq N} \frac{e^{2\pi i\lambda n}}{(n+u)^s} &= \sum_{y<n\leq N} \overbrace{\frac{e^{2\pi i\lambda n}}{(n+u)^{it}}}^{a_n} \overbrace{\frac{1}{(n+u)^{\sigma}}}^{\phi(n)}\\ &= \frac{1}{(N+u)^{\sigma}}\color{red}{\sum_{0\leq n\leq N}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}}-\frac{1}{(y+u)^{\sigma}}\color{red}{\sum_{0\leq n\leq y}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}} - \int_y^N \color{red}{\sum_{0\leq n\leq x}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}}\frac{-\sigma}{(x+u)^{\sigma+1}}\,dx\\ &=?\\ &= \frac{1}{(N+u)^{\sigma}}\color{red}{\sum_{y<n\leq N}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}} + \int_y^N \color{red}{\sum_{y<n\leq x}} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}}\frac{\sigma}{(x+u)^{\sigma+1}}\,dx\\ &= \frac{1}{(N+u)^{\sigma}}\bigg(\int_y^N \frac{e^{2\pi i\lambda x}}{(x+u)^{it}}\,dx+B_{\lambda}\bigg)+\int_y^N \bigg(\int_y^x \frac{e^{2\pi i\lambda v}}{(v+u)^{it}}\,dv+B_{\lambda}\bigg)\frac{\sigma}{(x+u)^{\sigma+1}}\,dx\\ &= \frac{1}{(N+u)^{\sigma}}\int_y^N \frac{e^{2\pi i\lambda x}}{(x+u)^{it}}\,dx+\frac{B_{\lambda}}{(N+u)^{\sigma}}\\ &\quad+ \int_{y}^N \frac{e^{2\pi i\lambda v}}{(v+u)^{it}} \int_v^N \frac{\sigma}{(x+u)^{\sigma+1}}\,dx\,dv+B_{\lambda}\int_y^N \frac{\sigma}{(x+u)^{\sigma+1}}\,dx\\ &= \frac{1}{(N+u)^{\sigma}}\int_y^N \frac{e^{2\pi i\lambda x}}{(x+u)^{it}}\,dx+\frac{B_{\lambda}}{(N+u)^{\sigma}}\\ &\quad+ \int_{y}^N \frac{e^{2\pi i\lambda v}}{(v+u)^{it}} \bigg(-\frac{1}{(N+u)^{\sigma}}+\frac{1}{(v+u)^{\sigma}}\bigg)\,dv+B_{\lambda}\bigg(-\frac{1}{(N+u)^{\sigma}}+\frac{1}{(y+u)^{\sigma}}\bigg)\\ &= \int_y^N \frac{e^{2\pi i\lambda v}}{(v+u)^s}\,dv+\frac{B_\lambda}{N^{\sigma}}+\color{blue}{\frac{B_{\lambda}}{y^{\sigma}}} \end{align*} here we have used the property in the paper, namely \begin{align*} \sum_{y<n\leq N} \frac{e^{2\pi i\lambda n}}{(n+u)^{it}}= \int_y^N \frac{e^{2\pi i\lambda x}}{(x+u)^{it}}\,dx+B_{\lambda} \end{align*} The question mark is on why the sum starts from $y+1$ instead of $0$? The second is on why the blue color part is different from the statement they give? A side note: The constant $B_{\lambda}$ is like a bound and could be written as $B_{\lambda}/(N+u)^{\sigma}=O(N^{-\sigma})$, hence I could absorb the negative sign previously, and $x$ and $v$ are just dummy variables. In that sense $\frac{B_{\lambda}}{\sigma}\frac{1}{y^{\sigma}}=O(y^{-\sigma})$ and will be the same as what I have. I would be appreciated if someone could help me with the missing $\color{red}{\text{red}}$ part and the missing $\color{blue}{\frac{1}{\sigma}}$. Thanks in advance!