How does this integral converge towards zero as R approaches ∞?

Question

I am struggling to understand how that, for the integral below, the result is 0 as R tends to infinity.

integral

I've tried re-arranging the integrand to make sense of it but I haven't gotten far with this. Does anyone know how to approach an integral of this form when taking limits of R?

Thank you for your time

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\verts{\ic\int_{0}^{\pi}\expo{\ic R\cos\pars{\theta} - R\sin\pars{\theta}}\,\,\,\,\,\dd\theta}} \leq \int_{0}^{\pi}\expo{- R\sin\pars{\theta}}\,\,\dd\theta \\[5mm] = & \ \int_{-\pi/2}^{\pi/2}\expo{- R\cos\pars{\theta}}\,\,\dd\theta = 2\int_{0}^{\pi/2}\expo{- R\cos\pars{\theta}}\,\,\dd\theta \\[5mm] = & \ 2\int_{0}^{\pi/2}\expo{- R\sin\pars{\theta}}\,\,\dd\theta \leq 2\int_{0}^{\pi/2} \expo{- 2R\theta/\pi}\,\,\dd\theta \\[5mm] & = \pi\,{1 - \expo{-R} \over R} \,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\, \bbx{\pi \over R} \\ & \end{align}

Answer 2

$$I(R)=i\int_0^\pi e^{iR\cos(x)-R\sin(x)}dx=i\int_0^\pi e^{iR\left(\cos(x)+i\sin(x)\right)}dx=i\int_0^\pi e^{iRe^{ix}}dx$$ $u=e^{ix}$ so $dx=\frac{du}{iu}$ so: $$I(R)=-\int_{-1}^1\frac{e^{iRu}}{u}du$$ $$I'(R)=-i\int_{-1}^1e^{iRu}du=-\left[\frac{e^{iRu}}{R}\right]_{-1}^1=-\frac{e^{iR}-e^{-iR}}{R}=-\frac{2\sinh(R)}{R}$$ and from this we could calculate: $$\lim_{R\to\infty}\left[-\frac{2\sinh(R)}{R}\right]$$ and you could use this to help answer your question