I am reading [this paper](https://doi.org/10.1023/A:1022801220384). 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 $e^{2 \pi i z}$. 



Basic question related to Stieltjes integral
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).


