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Question Number 27619 by abdo imad last updated on 11/Jan/18
find the value of ∫_0 ^∞ ((cos(2x))/((1+x^2 )^2 ))dx.
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}. \\ $$
Commented by abdo imad last updated on 12/Jan/18
let put I= ∫_0 ^∞ ((cos(2x))/((1+x^2 )^2 ))dx I= (1/2) ∫_R ((cos(2x))/((1+x^2 )^2 ))dx=(1/2) Re( ∫_R (e^(i2x) /((1+x^2 )^2 ))dx ) let introduce the complex function f(z)= (e^(i(2z)) /((1+z^2 )^2 )) we have f(z)= (e^(i(2z)) /((z−i)^2 (z+i)^2 )) so the poles of f are i and −i (double poles) ∫_R f(z)dz= 2iπ Res(f,i)( we dont take −i because Im(−i)<0) Re(f,i) =lim_(z−>i) (1/((2−1)!)) (d/dz)( (z−i)^2 f(z)) =lim_(z−>i ) (d/dz)((e^(i(2z)) /((z+i)^2 )))= =lim_(z−>i) ((2i e^(i(2z)) (z+i)^2 −2(z+i) e^(i(2z)) )/((z+i)^4 )) =lim_(z−>i) ((2i e^(i(2z)) (z+i) −2 e^(i(2z)) )/((z+i)^3 )) = ((2i e^(−2) (2i) −2 e^(−2) )/((2i)^3 ))= ((−4 e^(−2) −2 e^(−2) )/(−8i))=((6 e^(−2) )/(8i)) = ((3 e^(−2) )/(4i)) ∫_R f(z)dz= 2iπ.((3 e^(−2) )/(4i)) = ((3π)/2) e^(−2) I= (1/2) ∫_R ^ f(z)dz = ((3π)/4) e^(−2) = ((3π)/(4 e^2 )) .
$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\:{Re}\left(\:\int_{{R}} \:\:\frac{{e}^{{i}\mathrm{2}{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\right) \\ $$$${let}\:{introduce}\:\:{the}\:{complex}\:{function} \\ $$$${f}\left({z}\right)=\:\frac{{e}^{{i}\left(\mathrm{2}{z}\right)} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{we}\:{have}\:\:{f}\left({z}\right)=\:\:\frac{{e}^{{i}\left(\mathrm{2}{z}\right)} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles} \\ $$$${of}\:{f}\:{are}\:{i}\:{and}\:−{i}\:\left({double}\:{poles}\right) \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi\:{Res}\left({f},{i}\right)\left(\:\:\:{we}\:{dont}\:{take}\:−{i}\:{because}\:{Im}\left(−{i}\right)<\mathrm{0}\right) \\ $$$${Re}\left({f},{i}\right)\:={lim}_{{z}−>{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\frac{{d}}{{dz}}\left(\:\left({z}−{i}\right)^{\mathrm{2}} {f}\left({z}\right)\right) \\ $$$$={lim}_{{z}−>{i}\:} \frac{{d}}{{dz}}\left(\frac{{e}^{{i}\left(\mathrm{2}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{2}} }\right)= \\ $$$$={lim}_{{z}−>{i}} \:\:\:\:\:\:\frac{\mathrm{2}{i}\:{e}^{{i}\left(\mathrm{2}{z}\right)} \left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right)\:{e}^{{i}\left(\mathrm{2}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}−>{i}} \:\:\:\:\:\:\:\frac{\mathrm{2}{i}\:{e}^{{i}\left(\mathrm{2}{z}\right)} \left({z}+{i}\right)\:−\mathrm{2}\:{e}^{{i}\left(\mathrm{2}{z}\right)} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\:\:\:\:\:\frac{\mathrm{2}{i}\:{e}^{−\mathrm{2}} \left(\mathrm{2}{i}\right)\:−\mathrm{2}\:{e}^{−\mathrm{2}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }=\:\:\frac{−\mathrm{4}\:{e}^{−\mathrm{2}} −\mathrm{2}\:{e}^{−\mathrm{2}} }{−\mathrm{8}{i}}=\frac{\mathrm{6}\:{e}^{−\mathrm{2}} }{\mathrm{8}{i}}\:=\:\frac{\mathrm{3}\:{e}^{−\mathrm{2}} }{\mathrm{4}{i}} \\ $$$$\int_{{R}} {f}\left({z}\right){dz}=\:\mathrm{2}{i}\pi.\frac{\mathrm{3}\:{e}^{−\mathrm{2}} }{\mathrm{4}{i}}\:\:=\:\frac{\mathrm{3}\pi}{\mathrm{2}}\:{e}^{−\mathrm{2}} \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{{R}} ^{} {f}\left({z}\right){dz}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:{e}^{−\mathrm{2}} =\:\frac{\mathrm{3}\pi}{\mathrm{4}\:{e}^{\mathrm{2}} }\:. \\ $$

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