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Question Number 68876 by mathmax by abdo last updated on 16/Sep/19

calculate ∫_0 ^∞ (((−1)^x )/((3+x^2 )^2 ))dx

$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{x}} }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 17/Sep/19

the Q.is calculate ∫_(−∞) ^(+∞) (((−1)^x )/((3+x^2 )^2 ))dx

$${the}\:{Q}.{is}\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\left(−\mathrm{1}\right)^{{x}} }{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 17/Sep/19

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

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

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