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Question Number 80925 by M±th+et£s last updated on 08/Feb/20

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

Commented by mathmax by abdo last updated on 10/Feb/20

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(x\right)}{1+x^2}dx\Rightarrow I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{ix}}{x^2+1}dx\right)let$$$$\phi \left(z\right)=\frac{e^{iz}}{z^2+1}\Rightarrow \phi \left(z\right)=\frac{e^{iz}}{(z-i)(z+i)}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi (zdz?=2i\pi Res(\phi ,i)=2i\pi \frac{e^{-1}}{2i}=\frac{\pi }{e}\Rightarrow I=\frac{\pi }{e}$$

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