Demonstrate $ \int_0^1 \frac{e^{i\theta}}{1-e^{i\theta}x} dx = \sum_{n=1}^\infty \frac {e^{in\theta}}{n} $ using LDCT I am working on this problem that i can not solve:
The problem:
Demonstrate the following equality:
if $ \theta \neq 0 , 2 \pi $ then:
$ \int_0^1 \frac{e^{i\theta}}{1-e^{i\theta}x} dx = \sum_{n=1}^\infty \frac {e^{in\theta}}{n} $  
Using Lebesgue dominated convergence theorem to the partial sums:
$ \sum_{n\leq 0}  e^{i(n+1)\theta}x^n$.
What i have got:
If i could find an integrable g that dominates all this partial sums.
then interchanging sum with integral :
$$\int_0^1 \frac{e^{i\theta}}{1-e^{i\theta}x} dx  = \int_0^1 \sum_{n=0}^\infty  e^{i(n+1)\theta}x^n dx = \sum_{n=0}^\infty  \int_0^1 e^{i(n+1)\theta}x^n dx =  \sum_{n=0}^\infty e^{i\theta n+1}\int_0^1x^n = \sum_{n=1}^\infty \frac {e^{in\theta}}{n} $$
The problem is that i can not find an integrable g that dominates all the partial sums.
I am not able to dominate the serie $ \sum_{n=1}^\infty x^ne^{i\theta n} =\sum_{n=1}^\infty {(xe^{i\theta}})^n $ if i try with triangle inequality i get the geometric serie $ g(x) = \frac{x}{1-x}$ but that is not integrable on [0,1], i think the problem is i am not using the hypothesis   $ \theta \neq 0 , 2 \pi $ I guess i can obtain a better estimate of the module of the sum using this hypothesis, i have been trying to get a better estimation using polarization equalities but with no luck. 
 A: Note that
$$
\sum_{n = 0}^N e^{i(n + 1)\theta}x^n = e^{i\theta}\frac{(1 - e^{i(N + 1)\theta}x^{N + 1})}{1 - e^{i\theta}x}.
$$
But because $\theta \neq 0, 2\pi$ the map $x \mapsto 1 - e^{i\theta}x$ never vanishes. Hence, by compactness of $[0,1]$ we have that $|1 - e^{i\theta}x| \geq c$ for some $c > 0$. It follows that
$$
\bigg|\frac{1}{1 - e^{i\theta}x}\bigg| \leq \frac{1}{c}
$$
Thus
$$
\bigg|\sum_{n = 0}^N e^{i(n + 1)\theta}x^n\bigg| \leq \frac{|1 - e^{i(N + 1)\theta}x^{N + 1}|}{c} \leq \frac{2}{c}.
$$
A: Note that for $\theta\ne 2k\pi$ for integer $k$, we have
$$\begin{align}
\int_0^1 \frac{e^{i\theta}}{1-e^{i\theta}x}\,dx&=e^{i\theta}\int_0^1 \lim_{N\to \infty}\sum_{n=0}^N(e^{i\theta}x)^n\,dx\\\\
&e^{i\theta}\int_0^1 \lim_{N\to \infty}\left(\frac{1-(e^{i\theta}x)^{N+1}}{1-(e^{i\theta}x)}\right)\,dx\\\\
\end{align}$$
It is straightforward to show that 
$$\left|\frac{1-(e^{i\theta}x)^{N+1}}{1-(e^{i\theta}x)}\right|\le \frac{2}{\sqrt{x^2-2\cos(\theta)x+1}}$$
Inasmuch as $\displaystyle \int_0^1 \frac{2}{\sqrt{x^2-2\cos(\theta)x+1}}\,dx=\log\left(\frac{\sqrt{2(1-\cos(\theta))}}{1-\cos(\theta)}\right)<\infty$, then Dominated Convergence Theorem guarantees that 
$$\begin{align}
e^{i\theta}\int_0^1 \lim_{N\to \infty}\left(\frac{1-(e^{i\theta}x)^{N+1}}{1-(e^{i\theta}x)}\right)\,dx&=e^{i\theta}\lim_{N\to \infty}\int_0^1 \left(\frac{1-(e^{i\theta}x)^{N+1}}{1-(e^{i\theta}x)}\right)\,dx\\\\
&=e^{i\theta}\lim_{N\to \infty}\sum_{n=0}^N \int_0^1 (e^{i\theta}x)^n\,dx\\\\
&=\sum_{n=0}^\infty \frac{e^{i(n+1)\theta}}{n+1}\\\\
&=\sum_{n=1}^\infty \frac{e^{in\theta}}{n}
\end{align}$$
as was to be shown!
