# Basic question related to Stieltjes integral

I am reading this paper. I am stuck on something, which I think is something basic but I haven't been able to figure it out yet, and I was hoping someone could explain it to me.

Let $$\sigma(u) = \frac{1}{2\pi} \int_{-T}^T F(t) \frac{u^{1/2 + it} - M^{1/2 + it}}{1/2 + it} dt + O(N (\log N)^2/T),$$ where $F$ is a analytic function and $M$, $T$ are fixed positive numbers. Then on page 52 of the paper it is stated $$\int_M^N e(u \lambda) d\sigma = \frac{1}{2\pi} \int_{-T}^T F(t) \int_M^N e(u \lambda) u^{-1/2 + it} \ du \ dt + E$$ where $$E \ll (\log N)^2 (1 + |\lambda| N)N/T.$$ I was wondering about how I can show this bound on the error term. I would greatly appreciate any explanation. Thank you very much!

PS Here $e(z)$ denote $e^{2 \pi i z}$.

It's two integrations by parts. $$\int_M^N e^{2\pi i\lambda u}\, d\sigma(u) = \sigma e^{2\pi i\lambda u}\bigr|_M^N - 2\pi i\lambda\int_M^N e^{2\pi i\lambda u}\sigma(u)\, du$$ Let me write the asymptotic formula for $\sigma$ symbolically as $\sigma(u)=I(u)+O(\ldots)$ and use it to rewrite this as $$I e^{2\pi i\lambda u}\bigr|_M^N +O(N\log^2 N/T) - 2\pi i\lambda\int_M^N e^{2\pi i\lambda u}I(u)\, du + O(|\lambda|N^2\log^2 N/T)$$ Now we just go back: $$- 2\pi i\lambda\int_M^N e^{2\pi i\lambda u}I(u)\, du = -Ie^{2\pi i\lambda u}\bigr|_M^N + \int_M^N e^{2\pi i\lambda u}I'(u)\, du ,$$ and since $I'(u)= 1/(2\pi)\int_{-T}^T F(t) u^{-1/2+it}\, dt$, this gives the desired formula, after using Fubini (note that the $I e^{2\pi i\lambda u}\bigr|_M^N$ cancel).
Note that $$d\sigma(u)=\frac{1}{2\pi}\int^{T}_{-T}F(t)u^{-1/2+it}dt+dO(N(\log N)^2/T),$$ hence the error $E$ can be estimated by an integration by parts: $$\int^N_Me(u\lambda)dO(N(\log N)^2/T)\ll O\bigg(\frac{N(\log N)^2}{T}\bigg)+\int^N_MO\bigg(\frac{N(\log N)^2}{T}\bigg)de(u\lambda)$$ and the second term can be bounded by $$O\bigg(\frac{N(\log N)^2}{T}\bigg)\int^N_M|\lambda e(u\lambda)|du\ll\frac{N(\log N)^2}{T}|\lambda|N$$ as required.