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Question Number 168446 by amin96 last updated on 10/Apr/22

Answered by Mathspace last updated on 11/Apr/22

$$\mathrm{\Psi }=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(nx\right)}{x^2+1}dx$$$$=Re\left(\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{inx}}{x^2+1}dx\right)$$$$\phi \left(z\right)=\frac{e^{inz}}{z^2+1}\Rightarrow \phi \left(z\right)=\frac{e^{inz}}{(z-i)(z+i)}$$$$sothepolesof\phi areiand-i$$$${\int }_R\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$=2i\pi .\frac{e^{in\left(i\right)}}{2i}=\pi e^{-n}\Rightarrow$$$$\mathrm{\Psi }=\frac{\pi }{2}{e}^{-n}$$$$$$

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