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Question Number 36202 by prof Abdo imad last updated on 30/May/18
calculate ∫_0 ^∞ ((x^2 dx)/((x^2 +1)^3 ))
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} {dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$
Commented by math khazana by abdo last updated on 18/Aug/18
let A = ∫_0 ^∞ (x^2 /((x^2 +1)^3 )) dx 2A = ∫_(−∞) ^(+∞) (x^2 /((x^2 +1)^3 ))dx let consider the complex function ϕ(z) =(z^2 /((z^2 +1)^3 )) we have ϕ(z) =(z^2 /((z−i)^3 (z+i)^3 )) so the poles of ϕ are i and −i (triples) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)) { (z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) { (z^2 /((z+i)^3 ))}^((2)) =(1/2) lim_(z→i) { ((2z(z+i)^3 −3(z+i)^2 z^2 )/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) { ((2z(z+i)−3z^2 )/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) { ((−z^2 +2iz)/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) { (((−2z+2i)(z+i)^4 −4(z+i)^3 (−z^2 +2iz))/((z+i)^8 ))} =(1/2)lim_(z→i) (((−2z+2i)(z+i)−4(−z^2 +2iz))/((z+i)^5 )) =(1/2) ((−4(1−2))/((2i)^5 )) = (2/(2^5 i)) =(1/(i 2^4 )) =(1/(16i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(16i)) =(π/8) =2A ⇒ A =(π/(16)) .
$${let}\:{A}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:{dx} \\ $$$$\mathrm{2}{A}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }{dx}\:{let}\:{consider}\:{the}\:{complex} \\ $$$${function}\:\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\:{we}\:{have}\: \\ $$$$\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} }{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$$\left({triples}\right)\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\:\:\left\{\:\left({z}−{i}\right)^{\mathrm{3}} \:\varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \:\:\:\:\left\{\:\frac{{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{lim}_{{z}\rightarrow{i}} \:\left\{\:\:\:\:\frac{\mathrm{2}{z}\left({z}+{i}\right)^{\mathrm{3}} \:−\mathrm{3}\left({z}+{i}\right)^{\mathrm{2}} {z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \:\:\left\{\:\:\frac{\mathrm{2}{z}\left({z}+{i}\right)−\mathrm{3}{z}^{\mathrm{2}} }{\left({z}+{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \:\left\{\:\:\frac{−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}}{\left({z}+{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{\left(−\mathrm{2}{z}+\mathrm{2}{i}\right)\left({z}+{i}\right)^{\mathrm{4}} \:−\mathrm{4}\left({z}+{i}\right)^{\mathrm{3}} \left(−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}\right)}{\left({z}+{i}\right)^{\mathrm{8}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{lim}_{{z}\rightarrow{i}} \frac{\left(−\mathrm{2}{z}+\mathrm{2}{i}\right)\left({z}+{i}\right)−\mathrm{4}\left(−{z}^{\mathrm{2}} \:+\mathrm{2}{iz}\right)}{\left({z}+{i}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\:\frac{−\mathrm{4}\left(\mathrm{1}−\mathrm{2}\right)}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\:\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{5}} {i}}\:=\frac{\mathrm{1}}{{i}\:\mathrm{2}^{\mathrm{4}} }\:=\frac{\mathrm{1}}{\mathrm{16}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{\mathrm{16}{i}}\:\:=\frac{\pi}{\mathrm{8}}\:=\mathrm{2}{A}\:\Rightarrow \\ $$$${A}\:=\frac{\pi}{\mathrm{16}}\:. \\ $$$$ \\ $$$$ \\ $$

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