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Question Number 33204 by prof Abdo imad last updated on 12/Apr/18
find the value of ∫_(−∞) ^(+∞) ((cos(ax))/(1+x+x^2 )) dx.
$${find}\:{the}\:{value}\:{of}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left({ax}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\:{dx}. \\ $$
Commented by abdo imad last updated on 13/Apr/18
let put f(a) =∫_(−∞) ^(+∞) ((cos(ax))/(1+x+x^2 ))dx f(a) =Re (∫_(−∞) ^(+∞) (e^(iax) /(1+x+x^2 ))dx) let consider the complex function ϕ(z) = (e^(iaz) /(1+z+z^2 )) the poles of ϕ are j and j^− j =e^(i((2π)/3)) and ϕ(z) =(e^(iaz) /((z−j)(z−j^− ))) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ Res(ϕ,j) Res(ϕ,j) = (e^(iaj) /(j−j^− )) =(e^(ia(−(1/2)+i((√3)/2))) /(2iIm(j))) = ((e^(−((ia)/2)) .e^(−a((√3)/2)) )/(2i((√3)/2))) =(e^(−a((√3)/2)) /(i(√3))) ( cos((a/2)) −isin((a/2))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−a((√3)/2)) /(i(√3)))( cos((a/2))−isin((a/2)))⇒ f(a) =((2π)/( (√3))) cos((a/2)).e^(−a((√3)/2)) .
$${let}\:{put}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({ax}\right)}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{dx} \\ $$$${f}\left({a}\right)\:={Re}\:\left(\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{iax}} }{\mathrm{1}+{x}+{x}^{\mathrm{2}} }{dx}\right)\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{iaz}} }{\mathrm{1}+{z}+{z}^{\mathrm{2}} }\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{j}\:{and}\:\overset{−} {{j}} \\ $$$${j}\:={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\:{and}\:\varphi\left({z}\right)\:=\frac{{e}^{{iaz}} }{\left({z}−{j}\right)\left({z}−\overset{−} {{j}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{j}\right) \\ $$$${Res}\left(\varphi,{j}\right)\:=\:\frac{{e}^{{iaj}} }{{j}−\overset{−} {{j}}}\:=\frac{{e}^{{ia}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} }{\mathrm{2}{iIm}\left({j}\right)}\:=\:\frac{{e}^{−\frac{{ia}}{\mathrm{2}}} \:.{e}^{−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{\mathrm{2}{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \\ $$$$=\frac{{e}^{−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{{i}\sqrt{\mathrm{3}}}\:\left(\:{cos}\left(\frac{{a}}{\mathrm{2}}\right)\:−{isin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} }{{i}\sqrt{\mathrm{3}}}\left(\:{cos}\left(\frac{{a}}{\mathrm{2}}\right)−{isin}\left(\frac{{a}}{\mathrm{2}}\right)\right)\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\:{cos}\left(\frac{{a}}{\mathrm{2}}\right).{e}^{−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} \:. \\ $$$$ \\ $$

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