I am studying the residue theorem and its applications, and I don't understand the following inequalities: (this is from a solved problem)
$$\bigg|\int_0^{\pi}\frac{e^{iRe^{i\theta}}iRe^{i\theta}}{1+R^2e^{2i\theta}}d\theta\bigg|\leq \bigg|\int_0^{\pi}\frac{e^{-Rsin\theta}iRe^{i\theta}}{1+R^2e^{2i\theta}}d\theta\bigg|\leq\frac{1}{R}\int_0^\pi e^{-Rsin\theta}d\theta\leq\frac{1}{R}\pi\rightarrow0$$ as $R\rightarrow \infty$
Could anyone elaborate on why the inequalities are true?
On
One response answered part of your question, so I'll answer the rest. Note that $$ \left|e^{iRe^{i\theta}}\right|=\left|e^{iR\left(\cos \theta+i\sin \theta\right)}\right|=\left|e^{-R\sin \theta}e^{iR\cos \theta}\right|=\left|e^{-R\sin \theta}\right|\leq 1.$$ To see the last piece, observe that if $\theta\in [0,\pi],$ then $\sin\theta\in [0,1],$ and so $-R\sin\theta\in [-R,0].$ Hence, $|e^{-R\sin\theta}|\leq 1.$
For the first one its just that the imaginary part has magnitude $1$, so when the $e^{i\theta}$ in the exponent is expanded you can keep just the real part. Technically, you could also have gotten rid of the $ie^{i\theta}$ we see in the numerator after the first inequality. For the second inequality its that $\frac{R}{1+R^2} \leq \frac{R}{R^2} = \frac{1}{R}$. The last one comes from taylor expanding $e^{\sin(x)}$ I believe.