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Question Number 26360 by abdo imad last updated on 24/Dec/17

$$findthevalueof{\int }_0^{\propto }\frac{cos\left(\alpha x\right)}{1+x^2}dx.$$

Commented by abdo imad last updated on 25/Dec/17

$$letputI={\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{1+x^2}dx\Rightarrow 2I={\int }_{\mathbb{R}}\frac{cos\left(\alpha x\right)}{1+x^2}dx$$$$=Re\left({\int }_{\mathbb{R}}\frac{e^{i\alpha x}}{1+x^2}dx\right)andbyresidustherem$$$${\int }_{\mathbb{R}}f\left(z\right)dz=2i\pi Res(f,i)/f\left(z\right)=\frac{e^{i\alpha z}}{1+z^2}$$$$Re(f,i)={lim}_{z->i}(z-i)f\left(z\right)=\frac{e^{-\alpha }}{2i}$$$${\int }_{\mathbb{R}}f\left(z\right)dz=2i\pi \frac{e^{-\alpha }}{2i}=\pi \stackrel{-\alpha }{e}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{1+x^2}dx=\frac{\pi }{2}{e}^{-\alpha }.(lookthatIm\left({\int }_{\mathbb{R}}\frac{e^{i\alpha x}}{1+x^2}dx\right)=0)$$

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