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Question Number 33987 by abdo imad last updated on 28/Apr/18
find ∫_(−∞) ^(+∞) ((cos(αx))/((1+x^2 )^3 )) dx with α≥0 .
$${find}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{dx}\:{with}\:\alpha\geqslant\mathrm{0}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 02/May/18
let put f(α) = ∫_(−∞) ^(+∞) ((cos(αx))/((1+x^2 )^3 ))dx f(α) = Re( ∫_(−∞) ^(+∞) (e^(iαx) /((1+x^2 )^3 ))dx) let consider the compolex function ϕ(z)= (e^(iαz) /((1+z^2 )^3 )) the poles of ϕ are i and −i(triples) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!))((z−i)^3 ϕ(z))^(′′) =lim_(z→i) (1/2)( (e^(iαz) /((z+i)^3 )))^(′′) but ( (e^(iαz) /((z+i)^3 )))^′ = ((iα e^(iα2) (z+i)^3 −3(z+i)^2 e^(iαz) )/((z+i)^6 )) = ((iα e^(iαz) (z+i) −3 e^(iαz) )/((z+i)^4 )) ( (e^(iαz) /((z+i)^3 )))^(′′) = (((−α^2 e^(iαz) (z+i) +iα e^(iαz) −3iα e^(iαz) )(z+i)^4 −4(z+i)^3 ( iα e^(iαz) (z+i)−3 e^(iαz) ))/((z+i)^8 )) =(((−α^2 e^(iαz) (z+i) −2iα e^(iαz) )(z+i) −4(iα e^(iαz) (z+i) −3 e^(iαz) ))/((z+i)^5 )) Res(ϕ^� ,i) =((( −2i α^2 e^(−iα) −2iα e^(−α) )(2i) −4(−2α e^(−α) −3 e^(−α) ))/((2i)^5 )) ....be continued...
$${let}\:{put}\:{f}\left(\alpha\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$$${f}\left(\alpha\right)\:=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\alpha{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\right)\:{let}\:{consider}\:{the} \\ $$$${compolex}\:{function}\:\varphi\left({z}\right)=\:\frac{{e}^{{i}\alpha{z}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:{i}\:{and}\:−{i}\left({triples}\right)\:{so} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left(\left({z}−{i}\right)^{\mathrm{3}} \:\varphi\left({z}\right)\right)^{''} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left(\:\:\frac{{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right)^{''} \:\:\:{but} \\ $$$$\left(\:\frac{{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right)^{'} \:=\:\:\frac{{i}\alpha\:{e}^{{i}\alpha\mathrm{2}} \left({z}+{i}\right)^{\mathrm{3}} \:−\mathrm{3}\left({z}+{i}\right)^{\mathrm{2}} \:{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{6}} } \\ $$$$=\:\frac{{i}\alpha\:{e}^{{i}\alpha{z}} \left({z}+{i}\right)\:−\mathrm{3}\:{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$\left(\:\frac{{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right)^{''} \:=\:\:\frac{\left(−\alpha^{\mathrm{2}} \:{e}^{{i}\alpha{z}} \left({z}+{i}\right)\:+{i}\alpha\:{e}^{{i}\alpha{z}} \:−\mathrm{3}{i}\alpha\:{e}^{{i}\alpha{z}} \right)\left({z}+{i}\right)^{\mathrm{4}} \:−\mathrm{4}\left({z}+{i}\right)^{\mathrm{3}} \left(\:{i}\alpha\:{e}^{{i}\alpha{z}} \left({z}+{i}\right)−\mathrm{3}\:{e}^{{i}\alpha{z}} \right)}{\left({z}+{i}\right)^{\mathrm{8}} } \\ $$$$=\frac{\left(−\alpha^{\mathrm{2}} \:{e}^{{i}\alpha{z}} \left({z}+{i}\right)\:−\mathrm{2}{i}\alpha\:{e}^{{i}\alpha{z}} \right)\left({z}+{i}\right)\:−\mathrm{4}\left({i}\alpha\:{e}^{{i}\alpha{z}} \left({z}+{i}\right)\:−\mathrm{3}\:{e}^{{i}\alpha{z}} \right)}{\left({z}+{i}\right)^{\mathrm{5}} } \\ $$$${Res}\left(\bar {\varphi},{i}\right)\:=\frac{\left(\:−\mathrm{2}{i}\:\alpha^{\mathrm{2}} \:{e}^{−{i}\alpha} \:−\mathrm{2}{i}\alpha\:{e}^{−\alpha} \right)\left(\mathrm{2}{i}\right)\:−\mathrm{4}\left(−\mathrm{2}\alpha\:{e}^{−\alpha} \:−\mathrm{3}\:{e}^{−\alpha} \right)}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} } \\ $$$$….{be}\:{continued}… \\ $$
Commented by abdo mathsup 649 cc last updated on 03/May/18
Res(ϕ,i) =((4α^2 e^(−iα) +4α e^(−α) +8α e^(−α) +12 e^(−α) )/(32 i)) = ((4α^2 e^(−iα) + 24 e^(−α) )/(32i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((4α^2 e^(−iα) +24 e^(−α) )/(32i)) =((8π)/(32))((α^2 e^(−iα) +6 e^(−α) )/1) =(π/4)( α^2 cos(α) −iα^2 sinα +6 e^(−α) ) ⇒ f(α) = (π/4)( α^2 cosα +6 e^(−α) ) .
$${Res}\left(\varphi,{i}\right)\:=\frac{\mathrm{4}\alpha^{\mathrm{2}} \:{e}^{−{i}\alpha} \:\:+\mathrm{4}\alpha\:{e}^{−\alpha} \:+\mathrm{8}\alpha\:{e}^{−\alpha} \:+\mathrm{12}\:{e}^{−\alpha} }{\mathrm{32}\:{i}} \\ $$$$=\:\frac{\mathrm{4}\alpha^{\mathrm{2}} \:{e}^{−{i}\alpha} \:+\:\mathrm{24}\:{e}^{−\alpha} }{\mathrm{32}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{4}\alpha^{\mathrm{2}} \:{e}^{−{i}\alpha} \:+\mathrm{24}\:{e}^{−\alpha} }{\mathrm{32}{i}} \\ $$$$=\frac{\mathrm{8}\pi}{\mathrm{32}}\frac{\alpha^{\mathrm{2}} \:{e}^{−{i}\alpha} \:+\mathrm{6}\:{e}^{−\alpha} }{\mathrm{1}}\:=\frac{\pi}{\mathrm{4}}\left(\:\alpha^{\mathrm{2}} {cos}\left(\alpha\right)\:−{i}\alpha^{\mathrm{2}} {sin}\alpha\:+\mathrm{6}\:{e}^{−\alpha} \right) \\ $$$$\Rightarrow\:{f}\left(\alpha\right)\:=\:\frac{\pi}{\mathrm{4}}\left(\:\alpha^{\mathrm{2}} \:{cos}\alpha\:\:+\mathrm{6}\:{e}^{−\alpha} \right)\:. \\ $$

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