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Question Number 66694 by mathmax by abdo last updated on 18/Aug/19
let f(a) =∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +a))) with a∈](1/4),+∞[ 1) calculate f(a) 2)find also g(a) =∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +a)^2 )) 3) find the value of integrals ∫_0 ^∞ (dx/((x^4 +x^2 +3))) and ∫_0 ^∞ (dx/((x^4 +x^2 +1)^2 )) 4) developp f at integrserie.
$$\left.{let}\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} \:+{a}\right)}\:{with}\:{a}\in\right]\frac{\mathrm{1}}{\mathrm{4}},+\infty\left[\right. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{also}\:{g}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{3}\right)}\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{4}\right)\:{developp}\:{f}\:{at}\:{integrserie}. \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
1) f(a) =∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +a)) x^4 +x^2 +a =0 ⇒t^2 +t +a=0 (t=x^2 ) Δ=1−4a<0 ⇒Δ =(i(√(4a−1)))^2 ⇒t_1 =((−1+i(√(4a−1)))/2) and t_2 =((−1−i(√(4a−1)))/2) ⇒t^2 +t+a =(t−t_1 )(t−t_2 )=(x^2 −t_1 )(x^2 −t_2 ) ⇒ f(a) =∫_(−∞) ^(+∞) (dx/((x^2 −t_1 )(x^2 −t_2 ))) = ∫_(−∞) ^(+∞) {(1/(x^2 −t_1 ))−(1/(x^2 −t_2 ))}dx =(1/(i(√(4a−1)))){ ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )) −∫_(−∞) ^(+∞) (dx/(x^2 −t_1 ^− ))} =(1/(i(√(4a−1)))){2i Im( ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )))} =(2/( (√(4a−1)))) Im(∫_(−∞) ^(+∞) (dx/(x^2 −t_1 ))) let calculate ∫_(−∞) ^(+∞) (dx/(x^2 −t_1 )) let w(z) =(1/(z^2 −t_1 )) poles of w? ∣t_1 ∣ =(1/2)(√(1+4a−1))=(√a) ⇒t_1 =(√a)e^(−iarctan((√(4a−1)))) ⇒(√t_1 )=^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ⇒ w(z) =(1/((z−^4 (√a)e^(−(i/2)arctan((√(4a−1)))) )(z+^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ))) residus theorem give ∫_(−∞) ^(+∞) w(z)dz =2iπ Res(w,−^4 (√a)e^(−(i/2)arctan((√(4a−1)))) ) =2iπ×(1/(−2(^4 (√a))e^(−(i/2)arctan((√(4a−1)))) )) =−((iπ)/((^4 (√a)))) e^((i/2)rctan((√(4a−1)))) =((−iπ)/((^4 (√a)))){ cos(((arctan((√(4a−1))))/2))+i sin(((arctan((√(4a−1)))/2))} ⇒ f(a) =(2/( (√(4a−1))))(((−π)/((^4 (√a)))) cos(((arctan((√(4a−1))))/2))) =((−2π)/( (√(4a−1))(^4 (√a)))) cos(((arctan((√(4a−1))))/2)) cos^2 ((α/2)) =((1+cos(α))/2) ⇒cos((α/2))=+^− (√((1+cosα)/2)) here α=arctan(√(4a−1)) ⇒ cosα =(1/( (√(1+((√(4a−1)))^2 )))) =(1/(2(√a))) ⇒ cos((α/2)) =+^− (√((1+(1/(2(√a))))/2))=+^− (√((1+2(√a))/(4(√a))))=+^− (1/2)(√((1+2(√a))/( (√a)))) but f(a)>0 ⇒cos((α/2))=−(1/2)((√(1+2(√a)))/((^4 (√a)))) ⇒ f(a) =((−2π)/( (√(4a−1))(^4 (√a))^2 ))×((−1)/2)(√(1+2(√a)))=((π(√(1+2(√a))))/( (√a)(√(4a−1)))) ⇒ f(a) =(π/( (√a)))(√((1+2(√a))/(4a−1))) with a>(1/4)
$$\left.\mathrm{1}\right)\:{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+{a}} \\ $$$${x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+{a}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+{t}\:+{a}=\mathrm{0}\:\left({t}={x}^{\mathrm{2}} \right) \\ $$$$\Delta=\mathrm{1}−\mathrm{4}{a}<\mathrm{0}\:\Rightarrow\Delta\:=\left({i}\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:{and} \\ $$$${t}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}{\mathrm{2}}\:\Rightarrow{t}^{\mathrm{2}} \:+{t}+{a}\:=\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)=\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)}\:=\:\int_{−\infty} ^{+\infty} \left\{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{t}_{\mathrm{2}} }\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}\left\{\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }\:−\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −\overset{−} {{t}}_{\mathrm{1}} }\right\} \\ $$$$=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{4}{a}−\mathrm{1}}}\left\{\mathrm{2}{i}\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }\right)\right\}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:{Im}\left(\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }\right) \\ $$$${let}\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} −{t}_{\mathrm{1}} }\:{let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −{t}_{\mathrm{1}} }\:\:\:{poles}\:{of}\:{w}? \\ $$$$\mid{t}_{\mathrm{1}} \mid\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{4}{a}−\mathrm{1}}=\sqrt{{a}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \\ $$$$\Rightarrow\sqrt{{t}_{\mathrm{1}} }=^{\mathrm{4}} \sqrt{{a}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \:\Rightarrow\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−^{\mathrm{4}} \sqrt{{a}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \right)\left({z}+^{\mathrm{4}} \sqrt{{a}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give}\: \\ $$$$\int_{−\infty} ^{+\infty} {w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},−^{\mathrm{4}} \sqrt{{a}}{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mathrm{1}}{−\mathrm{2}\left(^{\mathrm{4}} \sqrt{{a}}\right){e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} }\:=−\frac{{i}\pi}{\left(^{\mathrm{4}} \sqrt{{a}}\right)}\:{e}^{\frac{{i}}{\mathrm{2}}{rctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)} \\ $$$$=\frac{−{i}\pi}{\left(^{\mathrm{4}} \sqrt{{a}}\right)}\left\{\:{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)}{\mathrm{2}}\right)+{i}\:{sin}\left(\frac{{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right.}{\mathrm{2}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}}\left(\frac{−\pi}{\left(^{\mathrm{4}} \sqrt{{a}}\right)}\:{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)}{\mathrm{2}}\right)\right) \\ $$$$=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}\left(^{\mathrm{4}} \sqrt{{a}}\right)}\:{cos}\left(\frac{{arctan}\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)}{\mathrm{2}}\right) \\ $$$${cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)\:=\frac{\mathrm{1}+{cos}\left(\alpha\right)}{\mathrm{2}}\:\Rightarrow{cos}\left(\frac{\alpha}{\mathrm{2}}\right)=\overset{−} {+}\sqrt{\frac{\mathrm{1}+{cos}\alpha}{\mathrm{2}}} \\ $$$${here}\:\alpha={arctan}\sqrt{\mathrm{4}{a}−\mathrm{1}}\:\Rightarrow\:{cos}\alpha\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\sqrt{\mathrm{4}{a}−\mathrm{1}}\right)^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}\:\Rightarrow \\ $$$${cos}\left(\frac{\alpha}{\mathrm{2}}\right)\:=\overset{−} {+}\sqrt{\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}}{\mathrm{2}}}=\overset{−} {+}\sqrt{\frac{\mathrm{1}+\mathrm{2}\sqrt{{a}}}{\mathrm{4}\sqrt{{a}}}}=\overset{−} {+}\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}+\mathrm{2}\sqrt{{a}}}{\:\sqrt{{a}}}} \\ $$$${but}\:{f}\left({a}\right)>\mathrm{0}\:\Rightarrow{cos}\left(\frac{\alpha}{\mathrm{2}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\frac{\sqrt{\mathrm{1}+\mathrm{2}\sqrt{{a}}}}{\left(^{\mathrm{4}} \sqrt{{a}}\right)}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{−\mathrm{2}\pi}{\:\sqrt{\mathrm{4}{a}−\mathrm{1}}\left(^{\mathrm{4}} \sqrt{{a}}\right)^{\mathrm{2}} }×\frac{−\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{2}\sqrt{{a}}}=\frac{\pi\sqrt{\mathrm{1}+\mathrm{2}\sqrt{{a}}}}{\:\sqrt{{a}}\sqrt{\mathrm{4}{a}−\mathrm{1}}}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\pi}{\:\sqrt{{a}}}\sqrt{\frac{\mathrm{1}+\mathrm{2}\sqrt{{a}}}{\mathrm{4}{a}−\mathrm{1}}}\:\:{with}\:{a}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
2) we have f^′ (a) =−∫_(−∞) ^(+∞) (dx/((x^4 +x^2 +a)^2 )) =−g(a) ⇒ g(a)=−f^′ (a) f(a) is known rest to calculate f^′ (a)
$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)=−{f}^{'} \left({a}\right)\:\:\:\:\:{f}\left({a}\right)\:{is}\:{known}\:{rest}\:{to}\:{calculate}\:{f}^{'} \left({a}\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 19/Aug/19
3) ∫_0 ^∞ (dx/(x^4 +x^2 +3)) =(1/2)∫_(−∞) ^(+∞) (dx/(x^4 +x^2 +3)) =(1/2)f(3) =(π/(2(√3)))(√((1+2(√3))/(11)))
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{2}}{f}\left(\mathrm{3}\right) \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{11}}} \\ $$

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