Is it true that $$\int_{-\pi}^{\pi}\left |\frac{iR^{x}e^{i \theta x}}{1-Re^{i\theta}}\right |\,\mathrm{d}\theta \leq 2\pi R\frac{R^{x}}{1-R}\xrightarrow{R\to\infty}0$$ $$\int_{-\pi}^{\pi}\left |\frac{ir^{x}e^{i \theta x}}{re^{i\theta}-1}\right |\,\mathrm{d}\theta\leq 2\pi r \frac{r^{x}}{r-1}\xrightarrow{r\to 0^+}0$$ if $|x|<1$ and $x>0$, respectively?
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What did you try? – Marco Cantarini May 17 '16 at 13:29
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@MarcoCantarini I tried to find an estimation but I am not sure if it's correct. I generally used this, $||z|-|w||\leq |z-w|\leq |z|+|w|$, where $z$ and $w$ are complex numbers. Here is, for example, $|1-Re^{i\theta}|\geq |1 - R|e^{i\theta}||\stackrel{?}\geq 1-R$. It should be noted that $0<r\leq R$. – Hopeless May 17 '16 at 13:34
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We note that $$\left|1-Re^{i\theta}\right|=\left|Re^{i\theta}-1\right|\geq\left|R\left|e^{i\theta}\right|-1\right|=R-1 $$ assuming $R>1 $, hence $$\int_{-\pi}^{\pi}\left|\frac{iR^{x}e^{i\theta x}}{1-Re^{i\theta}}\right|d\theta\leq\frac{R^{x}}{R-1}\int_{-\pi}^{\pi}d\theta=2\pi\frac{R^{x}}{R-1}\longrightarrow0 $$ as $R\longrightarrow\infty $ since $\left|x\right|<1 $. In a simil way $$\left|1-re^{i\theta}\right|\geq1-r $$ assuming $0<r<1 $ hence $$\int_{-\pi}^{\pi}\left|\frac{ir^{x}e^{i\theta x}}{1-re^{i\theta}}\right|d\theta\leq2\pi\frac{r^{x}}{1-r}\longrightarrow0 $$ as $r\longrightarrow0^{+} $ since $x>0.$
Marco Cantarini
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