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Question Number 36056 by abdo mathsup 649 cc last updated on 28/May/18
calculate ∫_(−∞) ^(+∞) ((2x)/((x^2 +mx +1)^2 ))dx with ∣m∣<2
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} \:+{mx}\:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{with}\:\mid{m}\mid<\mathrm{2} \\ $$
Commented by abdo mathsup 649 cc last updated on 30/May/18
let consider the complex function ϕ(z) = ((2z)/((z^2 +mz +1)^2 )) poles of ϕ! roots of z^(2 ) +mz +1 Δ =m^2 −4 <0 because ∣m∣<2 ⇒ Δ =−(4−m^2 ) ={i(√(4−m^2 )) }^2 z_1 = ((−m +i(√(4−m^2 )))/2) and z_2 = ((−m−i(√(4−m^2 )))/2) the poles of ϕ are z_1 and z_2 (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ, z_1 ) ϕ(z) = ((2z)/((z−z_1 )^2 (z−z_2 )^2 )) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1):)){ (z−z_1 )^2 ϕ(z)}^′
$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{2}{z}}{\left({z}^{\mathrm{2}} \:+{mz}\:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{poles}\:{of}\:\varphi! \\ $$$${roots}\:{of}\:{z}^{\mathrm{2}\:} \:+{mz}\:+\mathrm{1} \\ $$$$\Delta\:={m}^{\mathrm{2}} \:−\mathrm{4}\:<\mathrm{0}\:\:{because}\:\mid{m}\mid<\mathrm{2}\:\Rightarrow \\ $$$$\Delta\:=−\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\:=\left\{{i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }\:\right\}^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\:\frac{−{m}\:+{i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\mathrm{2}}\:\:{and}\:{z}_{\mathrm{2}} \:=\:\frac{−{m}−{i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \:\:\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\:{z}_{\mathrm{1}} \right) \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{2}{z}}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} \:} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right):}\left\{\:\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{'} \\ $$
Commented by abdo mathsup 649 cc last updated on 30/May/18
Res(ϕ ,z_1 ) =lim_(z→z_1 ) { ((2z)/((z−z_2 )^2 ))}^′ =lim_(z→z_1 ) ((2(z−z_2 )^2 −2z 2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((2(z−z_2 ) −4z)/((z−z_2 )^3 )) = 2 ((z_1 −z_2 −2z_1 )/((z_1 −z_2 )^3 )) =2((i(√(4−m^2 )) −2((−m+i(√(4−m^2 )))/2))/((i(√(4−m^2 )))^3 )) =2 (m/(−i(4−m^2 )(√(4−m^( )))) = ((−2m)/(i(4−m^2 )(√(4−m^2 )))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((−2m)/(i(4−m^2 )(√(4−m^2 )))) = ((−4mπ)/((4−m^2 )(√(4−m^2 )))) so I = ((−4mπ)/((4−m^2 )(√(4−m^2 )))) .
$${Res}\left(\varphi\:,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\left\{\:\frac{\mathrm{2}{z}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} \:\:−\mathrm{2}{z}\:\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{2}\left({z}−{z}_{\mathrm{2}} \right)\:−\mathrm{4}{z}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\:=\:\mathrm{2}\:\frac{{z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} \:−\mathrm{2}{z}_{\mathrm{1}} }{\left({z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\frac{{i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }\:\:−\mathrm{2}\frac{−{m}+{i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\mathrm{2}}}{\left({i}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }\right)^{\mathrm{3}} } \\ $$$$=\mathrm{2}\:\:\frac{{m}}{−{i}\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{m}^{\left(\right.} }}\:\:=\:\:\frac{−\mathrm{2}{m}}{{i}\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{−\mathrm{2}{m}}{{i}\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }} \\ $$$$=\:\:\frac{−\mathrm{4}{m}\pi}{\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}\:\:{so} \\ $$$$\:{I}\:=\:\frac{−\mathrm{4}{m}\pi}{\left(\mathrm{4}−{m}^{\mathrm{2}} \right)\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}\:. \\ $$
Answered by sma3l2996 last updated on 28/May/18
I=∫_(−∞) ^(+∞) ((2x)/((x^2 +mx+1)^2 ))dx=∫_(−∞) ^(+∞) ((2x+m−m)/((x^2 +mx+1)^2 ))dx =−[(1/(x^2 +mx+1))]_(−∞) ^(+∞) −m∫_(−∞) ^(+∞) (dx/((x^2 +2×(m/2)×x+(m^2 /4)−(m^2 /4)+1)^2 )) =0−m∫_(−∞) ^(+∞) (dx/(((x+(m/2))^2 +((4−m^2 )/4))^2 )) =−m∫_(−∞) ^(+∞) (dx/((((4−m^2 )/4)((((2x+m)/( (√(4−m^2 )))))^2 +1))^2 )) / ∣m∣<2 let u=((2x+m)/( (√(4−m^2 ))))⇒dx=((√(4−m^2 ))/2)du I=−((8m(√(4−m^2 )))/((4−m^2 )^2 ))∫_(−∞) ^(+∞) (du/((u^2 +1)^2 )) tant=u⇒dt=(du/(1+u^2 )) (du/((u^2 +1)^2 ))=(dt/(1+tan^2 t))=cos^2 (t)dt=(((1+cos(2t))/2))dt ∫_(−π/2) ^(π/2) (1+cos(2t))dt=[t+(1/2)sin2t]_(−π/2) ^(π/2) =π I=−((4πm(√(4−m^2 )))/((4−m^2 )^2 ))
$${I}=\int_{−\infty} ^{+\infty} \frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +{mx}+\mathrm{1}\right)^{\mathrm{2}} }{dx}=\int_{−\infty} ^{+\infty} \frac{\mathrm{2}{x}+{m}−{m}}{\left({x}^{\mathrm{2}} +{mx}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\: \\ $$$$=−\left[\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{mx}+\mathrm{1}}\right]_{−\infty} ^{+\infty} −{m}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}×\frac{{m}}{\mathrm{2}}×{x}+\frac{{m}^{\mathrm{2}} }{\mathrm{4}}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{0}−{m}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left(\left({x}+\frac{{m}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4}−{m}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$=−{m}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left(\frac{\mathrm{4}−{m}^{\mathrm{2}} }{\mathrm{4}}\left(\left(\frac{\mathrm{2}{x}+{m}}{\:\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}\right)^{\mathrm{2}} +\mathrm{1}\right)\right)^{\mathrm{2}} }\:\:\:/\:\:\mid{m}\mid<\mathrm{2} \\ $$$${let}\:\:{u}=\frac{\mathrm{2}{x}+{m}}{\:\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}\Rightarrow{dx}=\frac{\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\mathrm{2}}{du} \\ $$$${I}=−\frac{\mathrm{8}{m}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\left(\mathrm{4}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} }\int_{−\infty} ^{+\infty} \frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${tant}={u}\Rightarrow{dt}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}={cos}^{\mathrm{2}} \left({t}\right){dt}=\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right){dt} \\ $$$$\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} \left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt}=\left[{t}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{t}\right]_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} =\pi \\ $$$${I}=−\frac{\mathrm{4}\pi{m}\sqrt{\mathrm{4}−{m}^{\mathrm{2}} }}{\left(\mathrm{4}−{m}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by abdo mathsup 649 cc last updated on 30/May/18
correct sir sma3l thanks a lots.
$${correct}\:{sir}\:{sma}\mathrm{3}{l}\:\:{thanks}\:{a}\:{lots}. \\ $$

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