calculate: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$ using complex analysis.
My try:
$\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$
symetric therefore : $ \int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx=2\int_{0}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$
calculate instead: $2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz$
use pizza slice:$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=\int_{0}^{2\pi}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}d\theta+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR$
take limits:
$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=Lim_{R\rightarrow\infty}\int_{0}^{2\pi}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}d\theta+Lim_{\theta\searrow0}\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR+Lim_{\theta\nearrow0}\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR$
$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=0+\int_{0}^{R}\frac{1}{1-e^{\pi\theta i}R^{2}}dR+\int_{0}^{R}\frac{1}{1-e^{\pi\theta i}R^{2}}dR$
According the residue theorem at$ \int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=2\pi iRes_{z=-1}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}=0
$
therefore:$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=0$
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\underline{\underline{Complex\ Integration}}:}$ \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x} = 2\int_{0}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x = 2\,\Re\int_{0}^{\infty}{\expo{\pi x\ic/2} - \color{red}{\large\ic} \over 1 - x^{2}}\,\dd x \\[5mm] = &\ -\overbrace{\lim_{R \to \infty}\Re\int_{\large x\ \in\ R\expo{\pars{0,\pi/2}\,\ic}}{\expo{\pi x\ic/2} - \ic \over 1 - x^{2}}\,\dd x}^{\ds{=\ 0}}\ -\ 2\,\Re\int_{\infty}^{0}{\expo{\ic\pi\pars{\ic y}/2} - \ic \over 1 - \pars{\ic y}^{2}}\,\ic\,\dd y \\[5mm] = &\ 2\int_{0}^{\infty}{\dd y \over 1 + y^{2}} = 2\,{\pi \over 2} = \bbx{\large\pi} \\ & \end{align}
$\ds{\underline{\underline{Real\ Integration}}:}$ \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x} = {1 \over 2}\int_{-\infty}^{\infty}\bracks{% {\cos\pars{\pi x/2} \over 1 - x} + {\cos\pars{\pi x/2} \over 1 + x}}\,\dd x \\[5mm] = &\ -\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over x - 1}\,\dd x = \int_{-\infty}^{\infty}{\sin\pars{\pi x/2} \over x}\,\dd x = \int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[5mm] = &\ \bbx{\large\pi} \\ & \end{align}