Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 80451 by abdomathmax last updated on 03/Feb/20

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

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

Commented by abdomathmax last updated on 04/Feb/20

let I=∫_0 ^∞ ((cos(πx))/((x^2 +3)^2 ))dx changement x=(√3)t give I=∫_0 ^∞ ((cos(π(√3)t))/(9(t^2 +1)))×(√3)dt=((√3)/9)∫_0 ^∞ ((cos(π(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18))∫_(−∞) ^(+∞) ((cos(π(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18))Re( ∫_(−∞) ^(+∞) (e^(iπ(√3)t) /((t^2 +1)^2 ))dt) letϕ(z)=(e^(iπ(√3)z) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(iπ(√3)z) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(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)(z+i)e^(iπ(√3)z) −2e^(iπ(√3)z) )/((z+i)^3 )) =lim_(z→i) (({iπ(√3)(z+i)−2}e^(iπ(√3)z) )/((z+i)^3 )) =(({(2i)iπ(√3)−2}e^(−π(√3)) )/((2i)^3 )) =(({−2π(√3)−2}e^(−π(√3)) )/(−8i)) =(((π(√3)+1)e^(−π(√3)) )/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((π(√3)+1)e^(−π(√3)) )/(4i)) =(π/2)(1+π(√3))e^(−π(√3)) ⇒I=((√3)/(18))×(π/2)(1+π(√3))e^(−π(√3)) =((π(√3))/(36))(1+π(√3))e^(−π(√3))

$${let}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\:{changement}\:{x}=\sqrt{\mathrm{3}}{t}\:{give} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left(\pi\sqrt{\mathrm{3}}{t}\right)}{\mathrm{9}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}×\sqrt{\mathrm{3}}{dt}=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\pi\sqrt{\mathrm{3}}{t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}\int_{−\infty} ^{+\infty} \:\frac{{cos}\left(\pi\sqrt{\mathrm{3}}{t}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}{Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\pi\sqrt{\mathrm{3}}{t}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}\right) \\ $$$${let}\varphi\left({z}\right)=\frac{{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\varphi\left({z}\right)=\frac{{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\left\{\:\frac{{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{{i}\pi\sqrt{\mathrm{3}}{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} \left({z}+{i}\right)^{\mathrm{2}} −\mathrm{2}\left({z}+{i}\right){e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{{i}\pi\sqrt{\mathrm{3}}\left({z}+{i}\right){e}^{{i}\pi\sqrt{\mathrm{3}}{z}} −\mathrm{2}{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left\{{i}\pi\sqrt{\mathrm{3}}\left({z}+{i}\right)−\mathrm{2}\right\}{e}^{{i}\pi\sqrt{\mathrm{3}}{z}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{\left(\mathrm{2}{i}\right){i}\pi\sqrt{\mathrm{3}}−\mathrm{2}\right\}{e}^{−\pi\sqrt{\mathrm{3}}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left\{−\mathrm{2}\pi\sqrt{\mathrm{3}}−\mathrm{2}\right\}{e}^{−\pi\sqrt{\mathrm{3}}} }{−\mathrm{8}{i}}\:=\frac{\left(\pi\sqrt{\mathrm{3}}+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}} }{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\left(\pi\sqrt{\mathrm{3}}+\mathrm{1}\right){e}^{−\pi\sqrt{\mathrm{3}}} }{\mathrm{4}{i}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\pi\sqrt{\mathrm{3}}\right){e}^{−\pi\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{I}=\frac{\sqrt{\mathrm{3}}}{\mathrm{18}}×\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\pi\sqrt{\mathrm{3}}\right){e}^{−\pi\sqrt{\mathrm{3}}} \\ $$$$=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{36}}\left(\mathrm{1}+\pi\sqrt{\mathrm{3}}\right){e}^{−\pi\sqrt{\mathrm{3}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com