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

calculate by residus theorem I = ∫_(−∞) ^(+∞) ((cos(πx))/((1+x +x^2 )))dx .

$${calculate}\:{by}\:{residus}\:{theorem} \\ $$$${I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\pi{x}\right)}{\left(\mathrm{1}+{x}\:+{x}^{\mathrm{2}} \right)}{dx}\:. \\ $$

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

I = Re( ∫_(−∞) ^(+∞) (e^(iπx) /(1+x+x^2 ))) let introduce the complexfunction ϕ(z)= (e^(iπz) /(1+z +z^2 )) .the poles of ϕ are j and j^− ( j =e^(i((2π)/3)) ) ∫_(−∞) ^(+∞) (e^(iπz) /(1+z+z^2 ))dz =2iπ Res(ϕ,j) ϕ(z) = (e^(iπz) /((z−j)(z−j^− ))) ⇒Res(ϕ,j) = (e^(iπj) /(j−j^− )) = (e^(iπ(−(1/2)+i((√3)/2))) /(2i((√3)/2))) = ((e^(−π((√3)/2)) e^(−i(π/2)) )/(i(√3))) = −i (e^(−π((√3)/2)) /(i(√3))) = −(e^(−π((√3)/2)) /(√3)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz = −2iπ (e^(−π((√3)/2)) /(√3)) I =Re( ∫_(−∞) ^(+∞) ϕ(z)dz) =0 also we have ∫_(−∞) ^(+∞) ((sin(πx))/(1+x+x^2 ))dx = ((−2π)/(√3)) e^(−π((√3)/2)) .

$${I}\:=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{e}^{{i}\pi{x}} }{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)\:{let}\:{introduce}\:{the} \\ $$$${complexfunction}\:\varphi\left({z}\right)=\:\frac{{e}^{{i}\pi{z}} }{\mathrm{1}+{z}\:+{z}^{\mathrm{2}} }\:.{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:{j}\:{and}\:\overset{−} {{j}}\:\:\:\left(\:{j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{i}\pi{z}} }{\mathrm{1}+{z}+{z}^{\mathrm{2}} }{dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{j}\right) \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\pi{z}} }{\left({z}−{j}\right)\left({z}−\overset{−} {{j}}\right)}\:\Rightarrow{Res}\left(\varphi,{j}\right)\:=\:\frac{{e}^{{i}\pi{j}} }{{j}−\overset{−} {{j}}} \\ $$$$=\:\frac{{e}^{{i}\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{{e}^{−\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:\:{e}^{−{i}\frac{\pi}{\mathrm{2}}} }{{i}\sqrt{\mathrm{3}}}\:=\:−{i}\:\frac{{e}^{−\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{{i}\sqrt{\mathrm{3}}} \\ $$$$=\:−\frac{{e}^{−\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{\sqrt{\mathrm{3}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:−\mathrm{2}{i}\pi\:\frac{{e}^{−\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{\sqrt{\mathrm{3}}} \\ $$$${I}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:=\mathrm{0}\:{also}\:{we}\:{have} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{{sin}\left(\pi{x}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{dx}\:=\:\frac{−\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:{e}^{−\pi\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:. \\ $$

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