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Question Number 161839 by mathmax by abdo last updated on 23/Dec/21
calculate ∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))
$$\mathrm{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$
Answered by ArielVyny last updated on 23/Dec/21
I=∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))=∫_(−∞) ^(+∞) (dx/([(x+1)^2 +1]^2 )) let us consider u=x+1 ∫_(−∞) ^(+∞) (du/((u^2 +1)^2 ))=2∫_0 ^∞ (1/((u^2 +1)^2 ))du u^2 =t→2udu=dt→du=(dt/(2(√t))) I=∫_0 ^∞ (t^(−(1/2)) /((1+t)^2 ))dt knowing that ∫_0 ^∞ (x^(m−1) /((1+x)^(m+n) ))=β(m,n) m−1=−(1/2)→m=(1/2) m+n=2→n=2−(1/2)=(3/2) I=∫_0 ^∞ (t^(−(1/2)) /((1+t)^2 ))dt=β((1/2),(3/2))=Γ((3/2))Γ((1/2)) I=(1/2)π
$${I}=\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }=\int_{−\infty} ^{+\infty} \frac{{dx}}{\left[\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right]^{\mathrm{2}} } \\ $$$${let}\:{us}\:{consider}\:\:{u}={x}+\mathrm{1} \\ $$$$\int_{−\infty} ^{+\infty} \frac{{du}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$${u}^{\mathrm{2}} ={t}\rightarrow\mathrm{2}{udu}={dt}\rightarrow{du}=\frac{{dt}}{\mathrm{2}\sqrt{{t}}} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$${knowing}\:{that}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{m}−\mathrm{1}} }{\left(\mathrm{1}+{x}\right)^{{m}+{n}} }=\beta\left({m},{n}\right) \\ $$$${m}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}\rightarrow{m}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${m}+{n}=\mathrm{2}\rightarrow{n}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:{I}=\frac{\mathrm{1}}{\mathrm{2}}\pi \\ $$
Answered by Ar Brandon last updated on 23/Dec/21
I=∫(dx/((x^2 +2x+2)^2 )), Ostrogradsky gives; =((ax+b)/(x^2 +2x+2))+∫(c/(x^2 +2x+2))dx =((x+1)/(2(x^2 +2x+2)))+(1/2)∫(dx/(x^2 +2x+2)) =((x+1)/(2(x^2 +2x+2)))+(1/2)∫(dx/((x+1)^2 +1)) =((x+1)/(2(x^2 +2x+2)))+((arctan(x+1))/2)+C ⇒∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))=(π/2)
$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} },\:\mathrm{Ostrogradsky}\:\mathrm{gives}; \\ $$$$\:\:\:=\frac{{ax}+{b}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}+\int\frac{{c}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$\:\:\:=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}} \\ $$$$\:\:\:=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{arctan}\left({x}+\mathrm{1}\right)}{\mathrm{2}}+{C} \\ $$$$\:\:\:\Rightarrow\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}} \\ $$
Answered by Ar Brandon last updated on 24/Mar/22
I=∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 )) , ϕ(z)=(1/((z^2 +2z+2)^2 )) Poles: z_1 =−1+i=(√2)e^(((3π)/4)i) , z_2 =(√2)e^(−(π/4)i) I=∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ, z_1 ) Res (ϕ, z_1 )=lim_(z→z_1 ) (d/dz){(z−z_1 )^2 ϕ(z)}=lim_(z→z_1 ) (d/dz){(1/((z−z_2 )^2 ))} =lim_(z→z_1 ) {−(2/((z−z_2 )^3 ))}=−(2/((z_1 −z_2 )^3 ))=−(2/((2i)^3 ))=(1/(4i)) ⇒∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))=2πi((1/(4i)))=(π/2)
$${I}=\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\:\:,\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\:\mathrm{Poles}:\:{z}_{\mathrm{1}} =−\mathrm{1}+{i}=\sqrt{\mathrm{2}}{e}^{\frac{\mathrm{3}\pi}{\mathrm{4}}{i}} ,\:{z}_{\mathrm{2}} =\sqrt{\mathrm{2}}{e}^{−\frac{\pi}{\mathrm{4}}{i}} \\ $$$${I}=\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left(\varphi,\:{z}_{\mathrm{1}} \right) \\ $$$${Res}\:\left(\varphi,\:{z}_{\mathrm{1}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{1}} } {\mathrm{lim}}\frac{\mathrm{d}}{\mathrm{d}{z}}\left\{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}=\underset{{z}\rightarrow{z}_{\mathrm{1}} } {\mathrm{lim}}\frac{\mathrm{d}}{\mathrm{d}{z}}\left\{\frac{\mathrm{1}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{2}} }\right\} \\ $$$$=\underset{{z}\rightarrow{z}_{\mathrm{1}} } {\mathrm{lim}}\left\{−\frac{\mathrm{2}}{\left({z}−{z}_{\mathrm{2}} \right)^{\mathrm{3}} }\right\}=−\frac{\mathrm{2}}{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)^{\mathrm{3}} }=−\frac{\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}{i}} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{2}\pi{i}\left(\frac{\mathrm{1}}{\mathrm{4}{i}}\right)=\frac{\pi}{\mathrm{2}} \\ $$

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