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Question Number 40155 by maxmathsup by imad last updated on 16/Jul/18
caoculate ∫_0 ^∞ ((t dt)/((1+t^4 )^2 ))
$${caoculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}\:{dt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 16/Jul/18
Residus method changement t^2 = x give I = (1/2)∫_0 ^∞ ((2t)/((1+t^4 )^2 ))dt =(1/2) ∫_0 ^∞ (dx/((1+x^2 )^2 )) =(1/4) ∫_(−∞) ^(+∞) (dx/((1+x^2 )^2 )) let ϕ(z) = (1/((z^2 +1)^2 )) we have ϕ(z) = (1/((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(z+i)^(−2) }^((1)) =lim_(z→i) −2(z+i)^(−3) =−2(2i)^(−3) = ((−2)/((2i)^3 )) =((−2)/(−8i)) =(1/(4i)) I = (1/4) (2iπ)(1/(4i)) ⇒ I =(π/8) .
$${Residus}\:{method} \\ $$$${changement}\:{t}^{\mathrm{2}} =\:{x}\:\:\:{give}\:\: \\ $$$${I}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{\mathrm{2}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left\{\:\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \left\{\left({z}+{i}\right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow{i}} \:\:\:−\mathrm{2}\left({z}+{i}\right)^{−\mathrm{3}} =−\mathrm{2}\left(\mathrm{2}{i}\right)^{−\mathrm{3}} \:=\:\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−\mathrm{8}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\mathrm{2}{i}\pi\right)\frac{\mathrm{1}}{\mathrm{4}{i}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{8}}\:. \\ $$
Answered by ajfour last updated on 16/Jul/18
I=(1/2)∫_0 ^( ∞) (dz/((1+z^2 )^2 )) where z=t^2 let z=tan θ ⇒ dz=sec^2 θdθ I=(1/4)∫_0 ^( π/2) (1+cos 2θ)dθ =(π/8) .
$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\:\:{where}\:{z}={t}^{\mathrm{2}} \\ $$$$\:\:\:{let}\:\:{z}=\mathrm{tan}\:\theta\:\:\:\Rightarrow\:\:{dz}=\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\:\:{I}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:\:\:\:=\frac{\pi}{\mathrm{8}}\:. \\ $$

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