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Question Number 35440 by prof Abdo imad last updated on 19/May/18

find the value of ∫_0 ^∞ (dx/((2x^2 +1)^2 ))

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} \:\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by prof Abdo imad last updated on 19/May/18

let put I = ∫_0 ^∞ (dx/((2x^2 +1)^2 )) 2I = ∫_(−∞) ^(+∞) (dx/((2x^2 +1)^2 )) =(1/4) ∫_(−∞) ^(+∞) (dx/((x^2 +(1/2))^2 )) let condider tbe complex function ϕ(z) = (1/((z^2 +(1/2))^2 )) we have ϕ(z) = (1/((z−(i/(√2)))^2 (z +(i/(√2)))^2 )) so the?polrs of ϕ are (i/(√2)) and ((−i)/(√2)) (doubles) and Residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/(√2))) Res(ϕ,(i/(√2))) =lim_(z→(i/(√2))) { (z−(i/(√2)))^2 ϕ(z)}^′ =lim_(z→(i/(√2))) { (z+(i/(√2)))^(−2) }^′ =lim_(z→(i/(√2))) −2(z +(i/(√2)))^(−3) =−2 (((2i)/(√2)))^(−3) =−2((√2) i)^(−3) =2 ((√2))^(−3) .(1/i) = (1/(i(√2))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(i(√2))) = π(√2) 2I =(1/4) π(√2) ⇒ I = ((π(√2))/8) = ((2π)/(8(√2))) ⇒ I = (π/(4(√2))) .

$${let}\:{put}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:\:=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${let}\:{condider}\:{tbe}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{\mathrm{1}}{\left({z}−\frac{{i}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \left({z}\:+\frac{{i}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }\:{so}\:{the}?{polrs}\:{of}\:\varphi\:{are} \\ $$$$\frac{{i}}{\sqrt{\mathrm{2}}}\:{and}\:\frac{−{i}}{\sqrt{\mathrm{2}}}\:\left({doubles}\right)\:{and}\:{Residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\sqrt{\mathrm{2}}}\right) \\ $$$${Res}\left(\varphi,\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{2}}}} \:\:\left\{\:\left({z}−\frac{{i}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{2}}}} \:\:\:\left\{\:\left({z}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)^{−\mathrm{2}} \right\}^{'} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{2}}}} \:\:\:−\mathrm{2}\left({z}\:+\frac{{i}}{\sqrt{\mathrm{2}}}\right)^{−\mathrm{3}} \\ $$$$=−\mathrm{2}\:\left(\frac{\mathrm{2}{i}}{\sqrt{\mathrm{2}}}\right)^{−\mathrm{3}} \:=−\mathrm{2}\left(\sqrt{\mathrm{2}}\:{i}\right)^{−\mathrm{3}} =\mathrm{2}\:\left(\sqrt{\mathrm{2}}\right)^{−\mathrm{3}} .\frac{\mathrm{1}}{{i}} \\ $$$$=\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}}\:\Rightarrow\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}}\:=\:\pi\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}{I}\:\:=\frac{\mathrm{1}}{\mathrm{4}}\:\pi\sqrt{\mathrm{2}}\:\:\Rightarrow\:{I}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{8}}\:=\:\frac{\mathrm{2}\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:\Rightarrow\: \\ $$$${I}\:=\:\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\:. \\ $$

Commented by prof Abdo imad last updated on 19/May/18

I = ∫_0 ^∞ (dx/((2x^2 +1)^2 )) let use the changement x=(1/(√2)) tant ⇒ I = (1/(√2))∫_0 ^(π/2) ((1+tan^2 t)/((1+tan^2 )^2 ))dt = (1/(√2)) ∫_0 ^(π/2) (dt/(1+tan^2 t)) = (1/(√2)) ∫_0 ^(π/2) cos^2 t dt =(1/(√2)) ∫_0 ^(π/2) (((1+cos(2t))/2))dt = (1/(2(√2))) (π/2) + (1/(4(√2)))[ sin(2t)]_0 ^(π/2) ★ I = (π/(4(√2))) ★

$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{let}\:{use}\:{the}\:{changement} \\ $$$${x}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{tant}\:\Rightarrow\:{I}\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\frac{\pi}{\mathrm{2}}\:\:+\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}\left[\:{sin}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\bigstar\:\:\:{I}\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\bigstar \\ $$

Answered by ajfour last updated on 19/May/18

I=∫_0 ^( ∞) (dx/((2x^2 +1)^2 )) let x=(1/(√2))tan θ ⇒ dx=(1/(√2))sec^2 θdθ ⇒ I=(1/(√2))∫cos^2 θdθ I=(1/(2(√2))) ∫_0 ^( π/2) (1+cos 2θ)dθ =(1/(2(√2))) (θ+((sin 2θ)/2))∣_0 ^(π/2) = (π/(4(√2))) .

$${I}=\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:\:{x}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{tan}\:\theta\:\:\Rightarrow\:\:{dx}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow\:{I}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\int\mathrm{cos}\:^{\mathrm{2}} \theta{d}\theta \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left(\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\:=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\:\:. \\ $$

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