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Question Number 69564 by mathmax by abdo last updated on 25/Sep/19
let f(a) =∫_0 ^∞   (dx/(x^4 −2x^2  +a))   with a real and a>1  1) determine a explicit form for f(a)  2)  calculate g(a) =∫_0 ^∞   (dx/((x^4 −2x^2 +a)^2 ))  3) find the values of integrals  ∫_0 ^∞   (dx/(x^4 −2x^2  +3))  and ∫_0 ^∞   (dx/((x^4 −2x^2  +3)^2 ))
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}}\:\:\:{with}\:{a}\:{real}\:{and}\:{a}>\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}} \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
1) f(a) =∫_0 ^∞   (dx/(x^4 −2x^2  +a)) ⇒2f(a) =∫_(−∞) ^(+∞)   (dx/(x^4 −2x^2  +a))  let W(z) =(1/(z^4 −2z^2  +a))     poles of W?  z^4 −2z^2 +a =0 ⇒t^2 −2t +a =0 with t=z^2   Δ^′ =1−a<0 ⇒Δ^′ =(i(√(a−1)))^2  ⇒t_1 =1+i(√(a−1))  t_2 =1−i(√(a−1)) ⇒∣t_1 ∣ =(√(1+a−1))=(√a) ⇒t_1 =(√a)e^(iarctan((√(a−1))))   t_2 =(√a)e^(−iarctan((√(a−1))))  ⇒W(z) =(1/((z^2 −t_1 )(z^2 −t_2 )))  =(1/((z^2 −(√a)e^(iarctan((√(a−1)))) )(z^2 −(√a)e^(−iarctan((√(a−1)))) )))  =(1/((z−a^(1/4)  e^((i/2)arctan((√(a−1)))) )(z+a^(1/4)  e^((i/2)arctan((√(a−1)))) )(z−a^(1/4)  e^(−(i/2)arctan((√(a−1)))) (z+a^(1/4)  e^(−(i/2)arctan((√(a−1)))) )))  the poles of W are +^−  a^(1/4)  e^((i/2) arctan((√(a−1))))  and +^− a^(1/4)  e^(−(i/2)arctan((√(a−1))))   ∫_(−∞) ^(+∞)  W−z)dz =2iπ { Res(W,a^(1/4)  e^((i/2)arcta((√(a−1)))) )+Res(W,−a^(1/4)  e^(−(i/2)arctan((√(a−1)))) }  Res(W,a^(1/4)  e^((i/2)arctan((√(a−1)))) )  =(1/(2a^(1/4)  e^((i/2)arctan((√(a−1)))) ((√a)e^(iarctan((√(a−1)))) −(√a)e^(−iarctan((√(a−1)))) )))  =(e^(−(i/2)arctan((√(a−1)))) /(2 a^(3/4) (2i sin(arctan(√(a−1))))) =(e^(−(i/2) arctan((√(a−1)))) /(4i a^(3/4)  sin{arctan((√(a−1)))}))  Res(W,−a^(1/4)  e^(−(i/2)arctan((√(a−1)))) )  =(1/(−2a^(1/4)  e^(−(i/2)arctan((√(a−1)))) ((√a)e^(−iarctan((√(a−1)))) −(√a)e^(iarctan((√(a−1)))) )))  =(e^((i/2)arctan((√(a−1)))) /(2a^(1/4)   a^(1/2) (2isin{arctan(√(a−1)))})) =(e^((i/2) arctan((√(a−1)))) /(4ia^(3/4)  sin{arctan(√(a−1))}))  but sin{arctan(√(a−1))) =((√(a−1))/( (√(1+a−1)))) =((√(a−1))/( (√a)))=(√((a−1)/a))  ∫_(−∞) ^(+∞)   W(z)dz =2iπ{(e^(−(i/2)arctan(√(a−1))) /(4i a^(3/4) ×(√((a−1)/a)))) +(e^((i/2)arctan(√(a−1))) /(4ia^(3/4) ×(√((a−1)/a))))}  =(π/(2 a^(3/4) ((√(a−1))/a^(1/2) ))){ 2cos{arctan(√(a−1))}  =(π/(a^(1/4) (√(a−1))))×cos{arctan(√(a−1))}  but  cos{arctan(√(a−1))} =(1/( (√(1+a−1)))) =(1/( (√a))) ⇒  ∫_(−∞) ^(+∞)  W(z)dz =(π/((^4 (√a))(√(a^2 −a)))) =2f(a) ⇒  f(a) =(π/(2(^4 (√a))(√(a^2 −a)))) with  a>1
$$\left.\mathrm{1}\right)\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}}\:\Rightarrow\mathrm{2}{f}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}} \\ $$$${let}\:{W}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} \:+{a}}\:\:\:\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +{a}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}\:+{a}\:=\mathrm{0}\:{with}\:{t}={z}^{\mathrm{2}} \\ $$$$\Delta^{'} =\mathrm{1}−{a}<\mathrm{0}\:\Rightarrow\Delta^{'} =\left({i}\sqrt{{a}−\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{{a}−\mathrm{1}} \\ $$$${t}_{\mathrm{2}} =\mathrm{1}−{i}\sqrt{{a}−\mathrm{1}}\:\Rightarrow\mid{t}_{\mathrm{1}} \mid\:=\sqrt{\mathrm{1}+{a}−\mathrm{1}}=\sqrt{{a}}\:\Rightarrow{t}_{\mathrm{1}} =\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \\ $$$${t}_{\mathrm{2}} =\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({z}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} −\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}^{\mathrm{2}} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}+{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\left({z}−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left({z}+{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)\right.} \\ $$$${the}\:{poles}\:{of}\:{W}\:{are}\:\overset{−} {+}\:{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \:{and}\:\overset{−} {+}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \\ $$$$\left.\int_{−\infty} ^{+\infty} \:{W}−{z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left({W},{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arcta}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)+{Res}\left({W},−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right\}\right. \\ $$$${Res}\left({W},{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left(\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{2}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{2}{i}\:{sin}\left({arctan}\sqrt{{a}−\mathrm{1}}\right)\right.}\:=\frac{{e}^{−\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{4}{i}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{sin}\left\{{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)\right\}} \\ $$$${Res}\left({W},−{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \left(\sqrt{{a}}{e}^{−{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} −\sqrt{{a}}{e}^{{iarctan}\left(\sqrt{{a}−\mathrm{1}}\right)} \right)} \\ $$$$=\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{2}{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{a}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}{isin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right)\right\}}\:=\frac{{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\sqrt{{a}−\mathrm{1}}\right)} }{\mathrm{4}{ia}^{\frac{\mathrm{3}}{\mathrm{4}}} \:{sin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}} \\ $$$${but}\:{sin}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right)\:=\frac{\sqrt{{a}−\mathrm{1}}}{\:\sqrt{\mathrm{1}+{a}−\mathrm{1}}}\:=\frac{\sqrt{{a}−\mathrm{1}}}{\:\sqrt{{a}}}=\sqrt{\frac{{a}−\mathrm{1}}{{a}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\sqrt{{a}−\mathrm{1}}} }{\mathrm{4}{i}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} ×\sqrt{\frac{{a}−\mathrm{1}}{{a}}}}\:+\frac{{e}^{\frac{{i}}{\mathrm{2}}{arctan}\sqrt{{a}−\mathrm{1}}} }{\mathrm{4}{ia}^{\frac{\mathrm{3}}{\mathrm{4}}} ×\sqrt{\frac{{a}−\mathrm{1}}{{a}}}}\right\} \\ $$$$=\frac{\pi}{\mathrm{2}\:{a}^{\frac{\mathrm{3}}{\mathrm{4}}} \frac{\sqrt{{a}−\mathrm{1}}}{{a}^{\frac{\mathrm{1}}{\mathrm{2}}} }}\left\{\:\mathrm{2}{cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\right. \\ $$$$=\frac{\pi}{{a}^{\frac{\mathrm{1}}{\mathrm{4}}} \sqrt{{a}−\mathrm{1}}}×{cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\:\:{but} \\ $$$${cos}\left\{{arctan}\sqrt{{a}−\mathrm{1}}\right\}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{a}−\mathrm{1}}}\:=\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\frac{\pi}{\left(^{\mathrm{4}} \sqrt{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{a}}}\:=\mathrm{2}{f}\left({a}\right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{{a}}\right)\sqrt{{a}^{\mathrm{2}} −{a}}}\:{with}\:\:{a}>\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
2) we have f^′ (a) =−∫_0 ^∞  (dx/((x^4 −2x^2  +a)^2 )) =−g(a) ⇒  g(a)=−f^′ (a)  rest to calculate f^′ (a)...
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+{a}\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)=−{f}^{'} \left({a}\right)\:\:{rest}\:{to}\:{calculate}\:{f}^{'} \left({a}\right)… \\ $$
Commented by mathmax by abdo last updated on 27/Sep/19
3) ∫_0 ^∞ (dx/(x^4 −2x^2  +3)) =f(3) =(π/(2(^4 (√3))(√(9−3)))) =(π/(2(^4 (√3))(√6)))
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}}\:={f}\left(\mathrm{3}\right)\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\sqrt{\mathrm{9}−\mathrm{3}}}\:=\frac{\pi}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)\sqrt{\mathrm{6}}} \\ $$

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