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Question Number 29000 by abdo imad last updated on 03/Feb/18

$$provethst{\int }_{\mathbb{R}}\frac{e^{i\xi x}}{1+x^2}dx=\pi e^{-\mid \xi \mid }.$$

Commented by abdo imad last updated on 04/Feb/18

$$letputI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\xi x}}{1+x^2}dx$$$$case1for\xi ⩾0letputf\left(z\right)=\frac{e^{i\xi z}}{1+z^2}thepolesoffare$$$$iand-i\left(simples\right)sowehave$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Re(f,i)=2i\pi \frac{e^{i\left(\xi i\right)}}{2i}=\pi .e^{-\xi }$$$$case2if\xi <oI={\int }_R\frac{e^{-i\xi (-x)}}{1+x^2}dx={\int }_R\frac{e^{-i\xi t}}{1+t^2}dt(ch.-x=t)then$$$$wetakeg\left(x\right)=\frac{e^{-i\xi t}}{1+t^2}\Rightarrow I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}g\left(z\right)dz=2i\pi Re(g,i)$$$$=2i\pi \frac{e^{\xi }}{2i}=\pi .e^{\xi }inbothcase{\int }_R\frac{e^{i\xi x}}{1+x^2}dx=\pi e^{-\mid \xi \mid }.$$

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