(This question has been edited)
So I am stuck in a proof and all I need for everything to fall into place is to prove this:
$$ \lim_{n\to \infty} \int_{\pi}^{2\pi}f(ne^{it})ine^{it}dt = \int_{\pi}^{2\pi}\lim_{n\to \infty}\Big(f(ne^{it})ine^{it}\Big)dt $$
Where $f$ is a meromorphic function in $\Omega$, an open set which contains the inferior complex semiplane (i.e. $\{{\bf Im}(z) \leq 0$}).
I also know that there exists real number $a > 1$ for which $\displaystyle\lim_{z \to \infty} f(z)z^a=0$.
So if I define $\{f_n\}_{n \in \mathbb{N}}$, where $f_n(z)= f(ne^{it})ine^{it}$,
It is clear that $f_n$ converges point-wise to the null function. (i.e. $f(z)=0 \, \, \forall z\in \Omega$).
So, I thought of using dominated convergence, but since $f$ can have poles in $\Omega$ I cannot bound $f_n$.
I appreciate any tips, thanks in advanced and sorry for my sloppy english.
(From here I edited it)
I found a solution to my problem, after I took a break, I realized it was pretty trivial. Maybe my mind was too worked over before and could not see it.
Since $\displaystyle\lim_{z \to \infty} f(z)z^a=0$ there exists a positive interger $M$ such that for every $z \in B(0,M)^c$, $\mid f(z) \mid \leq 1$.
Defining $g_n(t)=f_{M+n}(t)$, I can use dominated convergence on $g_n$ since the function $h(t)=1 \, \, \forall t$, bounds $g_n$ for every n.
Then, all I got to do is twerk things a little in my proof.
Actually you can bound $f_n$. $f$ may have poles, but the assumptions imply that $f(z)z^a$, has only finitely many poles. Consider $$f(ne^{it})ine^{it} = [f(ne^{it})(ne^{it})^a] (ne^{it})^{1-a}i.$$ For $n$ large the bracketed quantity is $<\epsilon$ in magnitude since $\lim_{z \to \infty} f(z)z^a = 0$, and since $a>1$, the quantity $(ne^{it})^{1-a}$ also $<\epsilon$ in magnitude for $n$ large enough. Since the interval $[\pi,2\pi]$ has finite measure, boundedness of the integrand suffices to apply dominated convergence.