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Question Number 69110 by Fawole last updated on 20/Sep/19

$$Evaluate{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cos\left(x\right)}{x^2+1}dx$$

Commented by mathmax by abdo last updated on 20/Sep/19

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cosx}{x^2+1}dx\Rightarrow I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{x^2+1}dx\right)let$$$$W\left(z\right)=\frac{e^{iz}}{z^2+1}\Rightarrow W\left(z\right)=\frac{e^{iz}}{(z-i)(z+i)}sothepolesofWareiand-i$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i)$$$$Res(W,i)={lim}_{z\to i}(z-i)W\left(z\right)=\frac{e^{-1}}{2i}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \times \frac{e^{-1}}{2i}$$$$=\frac{\pi }{e}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(x\right)}{x^2+1}dx=\frac{\pi }{e}$$

Commented by Fawole last updated on 20/Sep/19

$$Nicesolution$$

Commented by mathmax by abdo last updated on 20/Sep/19

$$thanks$$

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