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 $e^{2 \pi i z}$. 
 A: 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.
A: 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).
