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Question Number 72396 by mathmax by abdo last updated on 28/Oct/19

calculate ∫_0 ^∞ ((1+x^2 )/(2+x^2 +x^4 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} }{dx} \\ $$

Commented by mathmax by abdo last updated on 29/Oct/19

let A =∫_0 ^∞ ((1+x^2 )/(x^4 +x^2 +2))dx ⇒2A =∫_(−∞) ^(+∞) ((x^2 +1)/(x^(4 ) +x^2 +2))dx let W(z)=((z^2 +1)/(z^4 +z^2 +2)) poles of W? z^4 +z^2 +2 =0 ⇒t^2 +t +2 =0 with t=z^2 Δ=1−8 =−7 ⇒t_1 =((−1+i(√7))/2) and t_2 =((−1−i(√7))/2) ∣t_1 ∣ =(1/2)(√(1+7))=(1/2)(2(√2))=(√2) ⇒t_1 =(√2)e^(−iarctan((√7))) t_2 =conj(t_1 )=(√2)e^(iarctan((√7))) ⇒W(z)=((z^2 +1)/((z^2 −(√2)e^(iarctan((√7))) )(z^2 −(√2)e^(−iarctan((√7))) ))) =((z^2 +1)/((z−(^4 (√2))e^((i/2)arctan((√7))) )(z+(^4 (√2))e^((i/2)arctan((√7))) )(z−(^4 (√2))e^(−(i/2)arctan((√7))) )(z+(^4 (√2))e^(−(i/2)arctan((√7))) ))) ∫_(−∞) ^(+∞) W(z)dz =2iπ{ Res(W,^4 (√2)e^((i/2)arctan((√7))) )+Res(W,−^4 (√2)e^(−(i/2)arctan((√7))) } ...be continued....

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{2}}{dx}\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}}{{x}^{\mathrm{4}\:} +{x}^{\mathrm{2}} \:+\mathrm{2}}{dx} \\ $$$${let}\:{W}\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{{z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{2}}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} \:+{z}^{\mathrm{2}} \:+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+{t}\:+\mathrm{2}\:=\mathrm{0}\:\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta=\mathrm{1}−\mathrm{8}\:=−\mathrm{7}\:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{7}}}{\mathrm{2}}\:{and}\:{t}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)=\sqrt{\mathrm{2}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \\ $$$${t}_{\mathrm{2}} ={conj}\left({t}_{\mathrm{1}} \right)=\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \:\Rightarrow{W}\left({z}\right)=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}^{\mathrm{2}} −\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} \right)} \\ $$$$=\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}−\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}+\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}−\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)\left({z}+\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)} \\ $$$$\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left({W},^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)+{Res}\left({W},−^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right\}\right. \\ $$$$...{be}\:{continued}.... \\ $$

Commented by mathmax by abdo last updated on 30/Oct/19

Res(W,^4 (√2)e^((i/2)arctan((√7))) )=((1+(√2)e^(iarctan((√7))) )/(2^4 (√2)e^((i/2)arctan((√7))) ((√2))(2i)sin(arctan((√7))))) =((1+(√2)e^(iarctan((√7))) )/((4i)(√2)(^4 (√2))e^((i/2)arctan((√7))) sin(arctan(√7)))) Res(W,−^4 (√2)e^(−(i/2)arctan((√7))) )=((1+(√2)e^(−iarctan((√7))) )/(−(√2)(2i)sin(arctan((√7)))(−2^4 (√2)e^(−(i/2)arctan((√7))) ))) =((1+(√2)e^(−iarctan((√7))) )/((4i)(√2)(^4 (√2)) e^(−(i/2)arctan((√7))) sin(arctan(√7)))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ{((e^(−(i/2)arctan((√7))) +(√2)e^((i/2)arctan((√7))) )/((4i)(√2)(^4 (√2))sin(arctan((√7)))) +((e^(−(i/2)arctan((√7))) +(√2) e^(−(i/2)arctan((√7))) )/((4i)(√2)(^4 (√2)) sin(arctan(√7))))} rest to complete the calculus.

$${Res}\left({W},^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)=\frac{\mathrm{1}+\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} }{\mathrm{2}^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \left(\sqrt{\mathrm{2}}\right)\left(\mathrm{2}{i}\right){sin}\left({arctan}\left(\sqrt{\mathrm{7}}\right)\right)} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{2}}{e}^{{iarctan}\left(\sqrt{\mathrm{7}}\right)} }{\left(\mathrm{4}{i}\right)\sqrt{\mathrm{2}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \:\:{sin}\left({arctan}\sqrt{\mathrm{7}}\right)} \\ $$$${Res}\left({W},−^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)=\frac{\mathrm{1}+\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} }{−\sqrt{\mathrm{2}}\left(\mathrm{2}{i}\right){sin}\left({arctan}\left(\sqrt{\mathrm{7}}\right)\right)\left(−\mathrm{2}^{\mathrm{4}} \sqrt{\mathrm{2}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} \right)} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{2}}{e}^{−{iarctan}\left(\sqrt{\mathrm{7}}\right)} }{\left(\mathrm{4}{i}\right)\sqrt{\mathrm{2}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} {sin}\left({arctan}\sqrt{\mathrm{7}}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} +\sqrt{\mathrm{2}}{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} }{\left(\mathrm{4}{i}\right)\sqrt{\mathrm{2}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right){sin}\left({arctan}\left(\sqrt{\mathrm{7}}\right)\right.}\right. \\ $$$$\left.+\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} +\sqrt{\mathrm{2}}\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{7}}\right)} }{\left(\mathrm{4}{i}\right)\sqrt{\mathrm{2}}\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\:{sin}\left({arctan}\sqrt{\mathrm{7}}\right)}\right\}\:\:\:{rest}\:{to}\:{complete}\:{the}\:{calculus}. \\ $$

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