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Question Number 33362 by prof Abdo imad last updated on 15/Apr/18

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

Commented by prof Abdo imad last updated on 15/Apr/18

$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\pi x}}{1+x+x^2}\right)letintroducethe$$$$complexfunction\phi \left(z\right)=\frac{e^{i\pi z}}{1+z+z^2}.thepolesof$$$$\phi arejand\stackrel{-}{j}(j=e^{i\frac{2\pi }{3}})$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\pi z}}{1+z+z^2}dz=2i\pi Res(\phi ,j)$$$$\phi \left(z\right)=\frac{e^{i\pi z}}{(z-j)(z-\stackrel{-}{j})}\Rightarrow Res(\phi ,j)=\frac{e^{i\pi j}}{j-\stackrel{-}{j}}$$$$=\frac{e^{i\pi (-\frac{1}{2}+i\frac{\sqrt{3}}{2})}}{2i\frac{\sqrt{3}}{2}}=\frac{e^{-\pi \frac{\sqrt{3}}{2}}{e}^{-i\frac{\pi }{2}}}{i\sqrt{3}}=-i\frac{e^{-\pi \frac{\sqrt{3}}{2}}}{i\sqrt{3}}$$$$=-\frac{e^{-\pi \frac{\sqrt{3}}{2}}}{\sqrt{3}}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=-2i\pi \frac{e^{-\pi \frac{\sqrt{3}}{2}}}{\sqrt{3}}$$$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)=0alsowehave$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(\pi x\right)}{1+x+x^2}dx=\frac{-2\pi }{\sqrt{3}}{e}^{-\pi \frac{\sqrt{3}}{2}}.$$

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