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Question Number 134301 by bramlexs22 last updated on 02/Mar/21

Ω = ∫_0 ^∞ (x^2 /((1+x^2 )^4 )) dx

$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{dx} \\ $$

Answered by EDWIN88 last updated on 02/Mar/21

replace x by (1/x) yields Ω = ∫_∞ ^( 0) (x^(−2) /((1+x^(−2) )^4 )).(−x^(−2) dx) =∫_0 ^∞ (x^4 /((x^2 +4)^4 )) dx 2Ω = ∫_0 ^∞ ((x^4 +x^2 )/((x^2 +1)^4 )) dx = ∫_0 ^∞ (x^2 /((1+x^2 )^3 )) dx Ω =(1/2)∫_0 ^∞ (x^2 /((1+x^2 )^4 )) dx . Apply integration by parts { ((u=(1/2)x⇒du=(1/2)dx)),((dv=(x/((x^2 +1)^3 )) dx⇒v=−(x/(4(1+x^2 )^2 )))) :} Ω=−(1/8)(x^2 /((1+x^2 )^2 )) ∣_0 ^∞ +∫_0 ^∞ (dx/(8(1+x^2 )^2 )) Ω=(1/8)∫_0 ^∞ (dx/((1+x^2 )^2 )) .replace x by (1/x) Ω=(1/8)∫_∞ ^0 (((−x^(−2) ))/((1+x^(−2) )^2 )) dx= (1/8)∫_0 ^∞ (x^2 /((1+x^2 )^2 )) dx adding ⇒2Ω = (1/8)∫_0 ^∞ ((1+x^2 )/((1+x^2 )^2 )) dx Ω = (1/(16))∫_0 ^∞ (dx/(1+x^2 )) = (1/(16))arctan (x) ]_0 ^∞ =(π/(32))

$$\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{yields}\: \\ $$$$\Omega\:=\:\int_{\infty} ^{\:\mathrm{0}} \frac{\mathrm{x}^{−\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{−\mathrm{2}} \right)^{\mathrm{4}} }.\left(−\mathrm{x}^{−\mathrm{2}} \:\mathrm{dx}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{4}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{4}} }\:\mathrm{dx} \\ $$$$\mathrm{2}\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\:\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\mathrm{dx} \\ $$$$\Omega\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:\mathrm{dx}\:.\:\mathrm{Apply}\:\mathrm{integration}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\begin{cases}{\mathrm{u}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\Rightarrow\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{dx}}\\{\mathrm{dv}=\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dx}\Rightarrow\mathrm{v}=−\frac{\mathrm{x}}{\mathrm{4}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }}\end{cases} \\ $$$$\Omega=−\frac{\mathrm{1}}{\mathrm{8}}\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mid_{\mathrm{0}} ^{\infty} \:+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{8}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:.\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{8}}\int_{\infty} ^{\mathrm{0}} \:\frac{\left(−\mathrm{x}^{−\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{−\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx}=\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{adding}\:\Rightarrow\mathrm{2}\Omega\:=\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\left.\Omega\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{16}}\mathrm{arctan}\:\left(\mathrm{x}\right)\:\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{32}} \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 02/Mar/21

∫_0 ^∞ (x^2 /((1+x^2 )^4 ))dx=(1/2)∫_0 ^∞ (u^(1/2) /((1+u)^4 ))du x^2 =u =(1/2)∫_0 ^∞ (u^((3/2)−1) /((1+u)^((5/2)+(3/2)) ))du=(1/2).((Γ((3/2))Γ((5/2)))/(Γ(4)))=(1/(12)).(1/2).(3/2).(1/2)π=(π/(32))

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{u}\right)^{\mathrm{4}} }{du}\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}} }{du}=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{4}\right)}=\frac{\mathrm{1}}{\mathrm{12}}.\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\pi=\frac{\pi}{\mathrm{32}} \\ $$

Answered by Ñï= last updated on 02/Mar/21

x=tanθ Ω=∫_0 ^(π/2) ((tan^2 θ)/(sec^8 θ))sec^2 θdθ =∫_0 ^(π/2) cos^4 θsin^2 θdθ =(1/2)B((5/2),(3/2)) =((Γ((5/2))Γ((3/2)))/(2Γ(4))) =(((3/2)×(1/2)×(1/2)π)/(2×3×2)) =(π/(32))

$${x}={tan}\theta \\ $$$$\Omega=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{tan}^{\mathrm{2}} \theta}{{sec}^{\mathrm{8}} \theta}{sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{4}} \theta{sin}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{4}\right)} \\ $$$$=\frac{\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\pi}{\mathrm{2}×\mathrm{3}×\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{32}} \\ $$

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