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Question Number 36168 by abdo mathsup 649 cc last updated on 29/May/18

let A(t) = ∫_(−∞) ^(+∞) ((sin(xt))/(( x +1+i)^2 )) dx with t from R 2) calculate A(t) 2) extract Re(A(t)) and Im(A(t)) 3) find the value of ∫_(−∞) ^(+∞) ((cos(3x))/((x+1+i)^2 ))dx

$${let}\:{A}\left({t}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{sin}\left({xt}\right)}{\left(\:{x}\:+\mathrm{1}+{i}\right)^{\mathrm{2}} }\:{dx}\:\:{with}\:{t}\:{from}\:{R} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{extract}\:{Re}\left({A}\left({t}\right)\right)\:{and}\:{Im}\left({A}\left({t}\right)\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}+\mathrm{1}+{i}\right)^{\mathrm{2}} }{dx} \\ $$

Commented by maxmathsup by imad last updated on 20/Aug/18

2) we have A(t) =∫_(−∞) ^(+∞) ((sin(xt)(x+1−i)^2 )/((x+1+i)^2 (x+1−i)^2 ))dx = ∫_(−∞) ^(+∞) ((sin(xt){x^2 +2x(1−i) +(1−i)^2 })/({(x+1)^2 +1}^2 )) dx =∫_(−∞) ^(+∞) ((sin(xt){ x^2 +2x−2xi −2i })/({(x+1)^z +1}^2 ))dx =∫_(−∞) ^(+∞) (((x^2 +2x)sin(xt))/({(x+1)^2 +1}^2 ))dx + i ∫_(−∞) ^(+∞) (((−2x−2)sin(xt))/({(x+1)^2 +1}^2 ))dx ⇒ Re(A(x)) =∫_(−∞) ^(+∞) (((x^2 +2x)sin(xt))/({(x+1)^2 +1}^2 ))dx and Im (A(x)) = ∫_(−∞) ^(+∞) (((−2x−2)sin(xt))/({(x+1)^2 +1}^2 )) dx

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{A}\left({t}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({xt}\right)\left({x}+\mathrm{1}−{i}\right)^{\mathrm{2}} }{\left({x}+\mathrm{1}+{i}\right)^{\mathrm{2}} \left({x}+\mathrm{1}−{i}\right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({xt}\right)\left\{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\left(\mathrm{1}−{i}\right)\:+\left(\mathrm{1}−{i}\right)^{\mathrm{2}} \right\}}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({xt}\right)\left\{\:{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{2}{xi}\:−\mathrm{2}{i}\:\right\}}{\left\{\left({x}+\mathrm{1}\right)^{{z}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){sin}\left({xt}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\:+\:{i}\:\int_{−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{2}{x}−\mathrm{2}\right){sin}\left({xt}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${Re}\left({A}\left({x}\right)\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){sin}\left({xt}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\:\:{and} \\ $$$${Im}\:\left({A}\left({x}\right)\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{\left(−\mathrm{2}{x}−\mathrm{2}\right){sin}\left({xt}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx} \\ $$

Commented by maxmathsup by imad last updated on 21/Aug/18

3) let I = ∫_(−∞) ^(+∞) ((cos(3x))/((x+1+i)^2 ))dx ⇒I = ∫_(−∞) ^(+∞) ((cos(3x)(x+1−i)^2 )/((x+1+i)^2 (x+1−i)^2 ))dx = ∫_(−∞) ^(+∞) ((cos(3x){x^2 +2x(1−i) +(1−i)^2 ))/({(x+1)^2 +1}^2 )) dx = ∫_(−∞) ^(+∞) ((cos(3x){x^2 +2x −2xi −2i})/({(x+1)^2 +1}^2 )) dx =∫_(−∞) ^(+∞) (((x^2 +2x)cos(3x))/({(x+1)^2 +1}^2 )) dx + i∫_(−∞) ^(+∞) (((−2x−2)cos(3x))/({(x+1)^2 +1}^2 )) dx =H +iK let find H H = Re( ∫_(−∞) ^(+∞) (((x^2 +2x)e^(i3x) )/({(x+1)^2 +1}^2 ))dx) cha7gement x+1 =t give ∫_(−∞) ^(+∞) (((x^2 +2x)e^(i3x) )/({(x+1)^2 +1}^2 ))dx = ∫_(−∞) ^(+∞) (({(t−1)^2 +2(t−1)}e^(i3(t−1)) )/((t^2 +1)^2 )) = ∫_(−∞) ^(+∞) (({t^2 −2t +1 +2t−2} e^(i3(t−1)) )/((t^2 +1)^2 ))dt=e^(−3i) ∫_(−∞) ^(+∞) (((t^2 −1)e^(i3t) )/((t^2 +1)^2 ))dt let ϕ(z) =(((z^2 −1)e^(i3z) )/((z^2 +1)^2 )) we have ϕ(z) =(((z^2 −1) e^(3iz) )/((z−i)^2 (z+i)^2 )) the poles of ϕ are i and −i(doubles) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i) =lim_(z→i) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(((z^2 −1)e^(3iz) )/((z+i)^2 ))}^((1)) = lim_(z→i) (((2z e^(3iz) +3i(z^2 −1)e^(3iz) )(z+i)^2 −2(z+i)(z^2 −i)e^(3iz) )/((z+i)^4 )) =lim_(z→i) (((2z +3iz^2 −3i)e^(3iz) (z+i) −2(z^2 −i) e^(3iz) )/((z+i)^3 )) =(((2i−3i −3i)e^(−3) (2i) −2(−1−i)e^(−3) )/((2i)^3 )) = ((−4i(2i)e^(−3) +2(1+i)e^(−3) )/(−8i)) =−(((10 +2i)e^(−3) )/(8i)) =((i(5 +i)e^(−3) )/4) ⇒ e^(−3i) ∫_(−∞) ^(+∞) ϕ(z)dz =e^(−3i ) {2iπ ((i(5+i)e^(−3) )/4)} =e^(−3i) (−(π/2)(5+i)) =−(π/2)(5+i)( cos(3)−isin(3)) =−(π/2){5 cos(3)−5i sin(3) +i cos(3) +sin(3)} H =Re( ∫...) =−(π/2)( 5 cos(3) +sin(3)) let find k we have −K =Re( ∫_(−∞) ^(+∞) (((2x+2) e^(i3x) )/({(x+1)^2 +1}^2 ))dx) changement x+1 =t give ∫_(−∞) ^(+∞) (((2x+2) e^(i3x) )/({(x+1)^2 +1}^2 ))dx =∫_(−∞) ^(+∞) (({2(t−1) +2}e^(i3(t−1)) )/({t^2 +1}^2 ))dt = ∫_(−∞) ^(+∞) ((2t e^(−3i) e^(i3t) )/((t^2 +1)^2 ))dt = 2 e^(−3i) ∫_(−∞) ^(+∞) ((t e^(3it) )/((t^2 +1)^2 ))dt let ϕ(z) =((z e^(3iz) )/((z^2 +1)^2 )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Re(ϕ,i)....be continued...

$$\left.\mathrm{3}\right)\:{let}\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}+\mathrm{1}+{i}\right)^{\mathrm{2}} }{dx}\:\Rightarrow{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)\left({x}+\mathrm{1}−{i}\right)^{\mathrm{2}} }{\left({x}+\mathrm{1}+{i}\right)^{\mathrm{2}} \left({x}+\mathrm{1}−{i}\right)^{\mathrm{2}} }{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)\left\{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\left(\mathrm{1}−{i}\right)\:+\left(\mathrm{1}−{i}\right)^{\mathrm{2}} \right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)\left\{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:−\mathrm{2}{xi}\:−\mathrm{2}{i}\right\}}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx}\:+\:{i}\int_{−\infty} ^{+\infty} \:\:\:\frac{\left(−\mathrm{2}{x}−\mathrm{2}\right){cos}\left(\mathrm{3}{x}\right)}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }\:{dx}\:={H}\:+{iK} \\ $$$${let}\:{find}\:{H} \\ $$$$\:{H}\:=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){e}^{{i}\mathrm{3}{x}} }{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\right)\:\:{cha}\mathrm{7}{gement}\:{x}+\mathrm{1}\:={t}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\right){e}^{{i}\mathrm{3}{x}} }{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{\left\{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{2}\left({t}−\mathrm{1}\right)\right\}{e}^{{i}\mathrm{3}\left({t}−\mathrm{1}\right)} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\frac{\left\{{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{1}\:+\mathrm{2}{t}−\mathrm{2}\right\}\:{e}^{{i}\mathrm{3}\left({t}−\mathrm{1}\right)} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}={e}^{−\mathrm{3}{i}} \int_{−\infty} ^{+\infty} \:\:\:\frac{\left({t}^{\mathrm{2}} \:−\mathrm{1}\right){e}^{{i}\mathrm{3}{t}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$${let}\:\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} \:−\mathrm{1}\right){e}^{{i}\mathrm{3}{z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\:{e}^{\mathrm{3}{iz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i}\left({doubles}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:\:\:{but} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left\{\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \left\{\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\:{lim}_{{z}\rightarrow{i}} \:\:\:\:\:\frac{\left(\mathrm{2}{z}\:{e}^{\mathrm{3}{iz}} \:+\mathrm{3}{i}\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{\mathrm{3}{iz}} \right)\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right)\left({z}^{\mathrm{2}} −{i}\right){e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\left(\mathrm{2}{z}\:+\mathrm{3}{iz}^{\mathrm{2}} \:−\mathrm{3}{i}\right){e}^{\mathrm{3}{iz}} \left({z}+{i}\right)\:−\mathrm{2}\left({z}^{\mathrm{2}} −{i}\right)\:{e}^{\mathrm{3}{iz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{2}{i}−\mathrm{3}{i}\:−\mathrm{3}{i}\right){e}^{−\mathrm{3}} \left(\mathrm{2}{i}\right)\:−\mathrm{2}\left(−\mathrm{1}−{i}\right){e}^{−\mathrm{3}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\:\frac{−\mathrm{4}{i}\left(\mathrm{2}{i}\right){e}^{−\mathrm{3}} \:+\mathrm{2}\left(\mathrm{1}+{i}\right){e}^{−\mathrm{3}} }{−\mathrm{8}{i}} \\ $$$$\:=−\frac{\left(\mathrm{10}\:+\mathrm{2}{i}\right){e}^{−\mathrm{3}} }{\mathrm{8}{i}}\:=\frac{{i}\left(\mathrm{5}\:+{i}\right){e}^{−\mathrm{3}} }{\mathrm{4}}\:\Rightarrow \\ $$$${e}^{−\mathrm{3}{i}} \:\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:={e}^{−\mathrm{3}{i}\:} \:\left\{\mathrm{2}{i}\pi\:\frac{{i}\left(\mathrm{5}+{i}\right){e}^{−\mathrm{3}} }{\mathrm{4}}\right\}\:={e}^{−\mathrm{3}{i}} \:\left(−\frac{\pi}{\mathrm{2}}\left(\mathrm{5}+{i}\right)\right) \\ $$$$=−\frac{\pi}{\mathrm{2}}\left(\mathrm{5}+{i}\right)\left(\:{cos}\left(\mathrm{3}\right)−{isin}\left(\mathrm{3}\right)\right) \\ $$$$=−\frac{\pi}{\mathrm{2}}\left\{\mathrm{5}\:{cos}\left(\mathrm{3}\right)−\mathrm{5}{i}\:{sin}\left(\mathrm{3}\right)\:+{i}\:{cos}\left(\mathrm{3}\right)\:+{sin}\left(\mathrm{3}\right)\right\} \\ $$$${H}\:={Re}\left(\:\int...\right)\:=−\frac{\pi}{\mathrm{2}}\left(\:\mathrm{5}\:{cos}\left(\mathrm{3}\right)\:+{sin}\left(\mathrm{3}\right)\right)\:\:\:{let}\:{find}\:{k}\:\:{we}\:{have} \\ $$$$−{K}\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{\left(\mathrm{2}{x}+\mathrm{2}\right)\:{e}^{{i}\mathrm{3}{x}} }{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\right)\:\:\:{changement}\:{x}+\mathrm{1}\:={t}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{\left(\mathrm{2}{x}+\mathrm{2}\right)\:{e}^{{i}\mathrm{3}{x}} }{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dx}\:=\int_{−\infty} ^{+\infty} \:\:\frac{\left\{\mathrm{2}\left({t}−\mathrm{1}\right)\:+\mathrm{2}\right\}{e}^{{i}\mathrm{3}\left({t}−\mathrm{1}\right)} }{\left\{{t}^{\mathrm{2}} \:+\mathrm{1}\right\}^{\mathrm{2}} }{dt} \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{\mathrm{2}{t}\:{e}^{−\mathrm{3}{i}} \:{e}^{{i}\mathrm{3}{t}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\:\mathrm{2}\:{e}^{−\mathrm{3}{i}} \:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}\:{e}^{\mathrm{3}{it}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:{let} \\ $$$$\varphi\left({z}\right)\:=\frac{{z}\:{e}^{\mathrm{3}{iz}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Re}\left(\varphi,{i}\right)....{be}\:{continued}... \\ $$

Commented by math khazana by abdo last updated on 21/Aug/18

H =−(π/2) e^(−3) (5cos(3)+sin(3)) .

$${H}\:=−\frac{\pi}{\mathrm{2}}\:{e}^{−\mathrm{3}} \left(\mathrm{5}{cos}\left(\mathrm{3}\right)+{sin}\left(\mathrm{3}\right)\right)\:. \\ $$

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