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Question Number 33704 by math khazana by abdo last updated on 22/Apr/18

$$letf\left(t\right)=\frac{1}{a^2+t^2}witha>0givethefourier$$$$transformforf.$$$$$$

Commented by prof Abdo imad last updated on 27/Apr/18

$$weknowthatF\left(f\left(x\right)\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(t\right)e^{-ixt}dtso$$$$forf\left(x\right)=\frac{1}{a^2+x^2}weget$$$$F\left(f\left(x\right)\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{-ixt}}{a^2+t^2}dtletconsiderthe$$$$complexfunction\phi \left(z\right)=\frac{e^{-ixz}}{z^2+a^2}thepolesof$$$$\phi areiaand-ia\left(simples\right)so$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,ia)$$$$\phi \left(z\right)=\frac{e^{-ixz}}{(z-ia)(z+ia)}\Rightarrow Res(\phi ,ia)=\frac{e^{-ix\left(ia\right)}}{2ia}$$$$=\frac{e^{ax}}{2ia}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{ax}}{2ia}=\frac{\pi }{a}{e}^{ax}\Rightarrow$$$$F\left(f\left(x\right)\right)=\frac{1}{\sqrt{2\pi }}\frac{\pi }{a}{e}^{ax}=\sqrt{\frac{{\pi }^2}{2\pi }}\frac{e^{ax}}{a}$$$$★F(f\left(x\right)\left(t\right)=\sqrt{\frac{\pi }{2}}\frac{e^{ax}}{a}★$$

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