Question Number 29038 by abdo imad last updated on 03/Feb/18
$${find}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{cos}\left({at}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}. \\ $$
Commented by abdo imad last updated on 11/Feb/18
![let put I(a)= ∫^(+∞) _(−∞) ((cos(at))/(1+t^4 ))dt =Re(∫_(−∞) ^(+∞) (e^(iat) /(1+t^4 ))dt)let introduce the complex functionϕ(z)= (e^(iaz) /(1+z^4 )) poles of ϕ? z^4 =−1 ⇔z^4 =e^(iπ) if z= r e^(iθ) ⇒ r=1 and 4θ =(2k+1)π ⇒θ=(2k+1)(π/4) so the poles of ϕ are z_k =e^(i(2k+1)(π/4)) and k∈[[o,3]] z_0 = e^(i(π/4)) , z_1 =e^(i((3π)/4)) , z_2 =e^(i((5π)/4)) , z_3 =e^(k((7π)/4)) ϕ(z)= (e^(iaz) /((z−z_0 )(z−z_1 )(z−z_2 )(z−z_3 ))) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ(Re(ϕ,z_0 ) +Re(ϕ,z_1 )) Res(ϕ,z_0 )= (e^(iaz_0 ) /(4z_0^ ^3 )) =−(1/4)z_0 e^(iaz_0 ) Res(ϕ,z_1 )= (e^(iaz_1 ) /(4z_1 ^3 ))=−(1/4)z_1 e^(iaz_1 ) but z_1 =z_0 ^− ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ(−(1/4))( z_0 e^(iaz_0 ) +z_0 ^− e^(iaz_0 ^− ) ) =−((iπ)/2)( z_0 e^(iaz_0 ) +z_0 ^− e^(iaz_9 ^− ) )....be continued....](Q29640.png)
$${let}\:{put}\:{I}\left({a}\right)=\:\underset{−\infty} {\int}^{+\infty} \frac{{cos}\left({at}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:={Re}\left(\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{iat}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\right){let} \\ $$$${introduce}\:{the}\:{complex}\:{function}\varphi\left({z}\right)=\:\frac{{e}^{{iaz}} }{\mathrm{1}+{z}^{\mathrm{4}} } \\ $$$${poles}\:{of}\:\varphi?\:{z}^{\mathrm{4}} =−\mathrm{1}\:\Leftrightarrow{z}^{\mathrm{4}} ={e}^{{i}\pi} \:\:{if}\:{z}=\:{r}\:{e}^{{i}\theta} \Rightarrow \\ $$$${r}=\mathrm{1}\:\:{and}\:\mathrm{4}\theta\:=\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\Rightarrow\theta=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}\:{so}\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:{z}_{{k}} ={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}} \:\:{and}\:{k}\in\left[\left[{o},\mathrm{3}\right]\right] \\ $$$${z}_{\mathrm{0}} =\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\:\:,\:\:{z}_{\mathrm{1}} ={e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\:\:\:\:,\:{z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:\:\:\:,\:{z}_{\mathrm{3}} ={e}^{{k}\frac{\mathrm{7}\pi}{\mathrm{4}}} \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{iaz}} }{\left({z}−{z}_{\mathrm{0}} \right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)\left({z}−{z}_{\mathrm{3}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left({Re}\left(\varphi,{z}_{\mathrm{0}} \right)\:+{Re}\left(\varphi,{z}_{\mathrm{1}} \right)\right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{0}} \right)=\:\frac{{e}^{{iaz}_{\mathrm{0}} } }{\mathrm{4}{z}_{\mathrm{0}^{} } ^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}{z}_{\mathrm{0}} \:{e}^{{iaz}_{\mathrm{0}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\:\frac{{e}^{{iaz}_{\mathrm{1}} } }{\mathrm{4}{z}_{\mathrm{1}} ^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{4}}{z}_{\mathrm{1}} \:{e}^{{iaz}_{\mathrm{1}} } \:\:\:{but}\:{z}_{\mathrm{1}} ={z}_{\mathrm{0}} ^{−} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\:{z}_{\mathrm{0}} {e}^{{iaz}_{\mathrm{0}} } \:+{z}_{\mathrm{0}} ^{−} \:{e}^{{iaz}_{\mathrm{0}} ^{−} } \right) \\ $$$$=−\frac{{i}\pi}{\mathrm{2}}\left(\:{z}_{\mathrm{0}} {e}^{{iaz}_{\mathrm{0}} } \:+{z}_{\mathrm{0}} ^{−} \:{e}^{{iaz}_{\mathrm{9}} ^{−} } \right)....{be}\:{continued}.... \\ $$