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Question Number 74349 by mathmax by abdo last updated on 22/Nov/19

calculate ∫_0 ^∞ ((cos(2πx))/((x^2 +3)^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\mathrm{2}\pi{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$

Commented by abdomathmax last updated on 23/Nov/19

let I =∫_0 ^∞ ((cos(2πx))/((x^2 +3)^2 ))dx changement x=(√3)t give I =∫_0 ^∞ ((cos(2π(√3)t))/(9(t^2 +1)^2 ))×(√3)dt =((√3)/9) ∫_0 ^∞ ((cos(2π(√3)t))/((t^2 +1)^2 ))dt ⇒2I=((√3)/9) ∫_(−∞) ^(+∞) ((cos(2π(√3)t))/((t^2 +1)^2 ))dt= ((√3)/9) Re(∫_(−∞) ^(+∞) (e^(i(2π(√3))t) /((t^2 +1)^2 ))dx) let W(z)=(e^(i(2π(√3))z) /((z^2 +1)^2 )) ⇒ W(z)=(e^(i(2π(√3))z) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i)= lim_(z→i) (1/((2−1)!)){ (z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(i(2π(√3))z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2iπ(√3)e^(i(2π(√3))z) (z+i)^2 −2(z+i)e^(2iπ(√3)z) )/((z+i)^4 )) =lim_(z→i) ((2iπ(√3)e^(i(2π(√3))z) (z+i)−2 e^(2iπ(√3)z) )/((z+i)^3 )) =((2iπ(√3)e^(−2π(√3)) (2i)−2e^(−2π(√3)) )/(−8i)) =(((−4π(√3)−2)e^(−2π(√3)) )/(+8i)) =(((2π(√3)+1)e^(−2π(√3)) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =(π/2)(2π(√3)+1)e^(−2π(√3)) ⇒ I=((√3)/(18))×(π/2)(2π(√3)+1)e^(−2π(√3))

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}\pi{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\:{changement}\:{x}=\sqrt{\mathrm{3}}{t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}{t}\right)}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }×\sqrt{\mathrm{3}}{dt}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}{t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow\mathrm{2}{I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}{t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){t}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}\right) \\ $$$${let}\:{W}\left({z}\right)=\frac{{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){z}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{W}\left({z}\right)=\frac{{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){z}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)=\:{lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\right)^{\mathrm{2}} {W}\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\left\{\:\:\frac{{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{2}{i}\pi\sqrt{\mathrm{3}}{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){z}} \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{\mathrm{2}{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\:\frac{\mathrm{2}{i}\pi\sqrt{\mathrm{3}}{e}^{{i}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}\right){z}} \left({z}+{i}\right)−\mathrm{2}\:{e}^{\mathrm{2}{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}{i}\pi\sqrt{\mathrm{3}}{e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} \left(\mathrm{2}{i}\right)−\mathrm{2}{e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} }{−\mathrm{8}{i}} \\ $$$$=\frac{\left(−\mathrm{4}\pi\sqrt{\mathrm{3}}−\mathrm{2}\right){e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} }{+\mathrm{8}{i}}\:=\frac{\left(\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{1}\right){e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} }{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{1}\right){e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} \:\:\Rightarrow \\ $$$${I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}×\frac{\pi}{\mathrm{2}}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{1}\right){e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} \\ $$

Commented by abdomathmax last updated on 23/Nov/19

I =((π(√3))/(36))(2π(√3)+1) e^(−2π(√3))

$${I}\:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{36}}\left(\mathrm{2}\pi\sqrt{\mathrm{3}}+\mathrm{1}\right)\:{e}^{−\mathrm{2}\pi\sqrt{\mathrm{3}}} \\ $$

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