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Question Number 146835 by mathmax by abdo last updated on 16/Jul/21
calculate ∫_0 ^∞ (dx/((x^2 +3)^2 (x^2 +4)^2 ))
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} } \\ $$
Answered by mathmax by abdo last updated on 16/Jul/21
Ψ=(1/2)∫_(−∞) ^(+∞) (dx/((x^2 +3)^2 (x^2 +4)^2 )) let ϕ(z)=(1/((z^2 +3)^2 (z^2 +4)^2 )) ⇒ϕ(z)=(1/((z−i(√3))^2 (z+i(√3))^2 (z−2i)^2 (z+2i)^2 )) the poles of ϕ are +^− i(√3) and +^− 2i(doubles) residus ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{Res(ϕ,i(√3))+Res(ϕ,2i)} Res(ϕ,i(√3))=lim_(z→i(√3)) (1/((2−1)!)){(z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) {(1/((z+i(√3))^2 (z^2 +4)^2 ))} =−lim_(z→i(√3)) ((2(z+i(√3))(z^2 +4)^2 +4z(z^2 +4)(z+i(√3))^2 )/((z+i(√3))^4 (z^2 +4)^4 )) =−lim_(z→i(√3)) ((2(z^2 +4)+4z(z+i(√3)))/((z+i(√3))^3 (z^2 +4)^3 )) =−((2(−3+4)+4i(√3)(2i(√3)))/((2i(√3))^3 (−3+4)^3 ))=−((2−24)/(−8i(3(√3))))=−((22)/(24(√3)i))=−((11)/(12(√3)i)) Res(ϕ,2i)=lim_(z→2i) (1/((2−1)!)){(z−2i)^2 ϕ(z)}^((1)) =lim_(z→2i) {(1/((z^2 +3)^2 (z+2i)^2 ))}^((1)) =−lim_(z→2i) ((4z(z^2 +3)(z+2i)^2 +2(z+2i)(z^2 +3)^2 )/((z^2 +3)^4 (z+2i)^4 )) =−lim_(z→2i) ((4z(z+2i)+2(z^2 +3))/((z^2 +3)^3 (z+2i)^3 )) =−(((8i)(4i)+2(−4+3))/((−4+3)^3 (4i)^3 ))=−((−32−2)/((−1)(−48i)))=−((−34)/(48i))=((34)/(48i)) ⇒ ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{−((11)/(12(√3)i))+((17)/(24i))} =−((11π)/(6(√3))) +((17π)/(12)) =(((17)/(12))−((11)/(6(√3))))π ⇒Ψ=(π/2)(((17)/(12))−((11)/(6(√3))))
$$\Psi=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{z}−\mathrm{2i}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\varphi\:\mathrm{are}\:\overset{−} {+}\mathrm{i}\sqrt{\mathrm{3}}\:\mathrm{and}\:\overset{−} {+}\mathrm{2i}\left(\mathrm{doubles}\right) \\ $$$$\mathrm{residus}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\mathrm{3}}\right)+\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\sqrt{\mathrm{3}}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}\sqrt{\mathrm{3}}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}\sqrt{\mathrm{3}}} \left\{\frac{\mathrm{1}}{\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }\right\} \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}\sqrt{\mathrm{3}}} \:\:\:\frac{\mathrm{2}\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} \:+\mathrm{4z}\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{4}} } \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}\sqrt{\mathrm{3}}} \:\:\:\:\frac{\mathrm{2}\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)+\mathrm{4z}\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)}{\left(\mathrm{z}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{2}\left(−\mathrm{3}+\mathrm{4}\right)+\mathrm{4i}\sqrt{\mathrm{3}}\left(\mathrm{2i}\sqrt{\mathrm{3}}\right)}{\left(\mathrm{2i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \left(−\mathrm{3}+\mathrm{4}\right)^{\mathrm{3}} }=−\frac{\mathrm{2}−\mathrm{24}}{−\mathrm{8i}\left(\mathrm{3}\sqrt{\mathrm{3}}\right)}=−\frac{\mathrm{22}}{\mathrm{24}\sqrt{\mathrm{3}}\mathrm{i}}=−\frac{\mathrm{11}}{\mathrm{12}\sqrt{\mathrm{3}}\mathrm{i}} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{2i}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{2i}\right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2i}} \:\:\:\:\left\{\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2i}} \:\:\:\:\frac{\mathrm{4z}\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)\left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{z}+\mathrm{2i}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{4}} \left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{4}} } \\ $$$$=−\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{2i}} \:\:\:\:\frac{\mathrm{4z}\left(\mathrm{z}+\mathrm{2i}\right)+\mathrm{2}\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} \left(\mathrm{z}+\mathrm{2i}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\left(\mathrm{8i}\right)\left(\mathrm{4i}\right)+\mathrm{2}\left(−\mathrm{4}+\mathrm{3}\right)}{\left(−\mathrm{4}+\mathrm{3}\right)^{\mathrm{3}} \left(\mathrm{4i}\right)^{\mathrm{3}} }=−\frac{−\mathrm{32}−\mathrm{2}}{\left(−\mathrm{1}\right)\left(−\mathrm{48i}\right)}=−\frac{−\mathrm{34}}{\mathrm{48i}}=\frac{\mathrm{34}}{\mathrm{48i}}\:\Rightarrow \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{−\frac{\mathrm{11}}{\mathrm{12}\sqrt{\mathrm{3}}\mathrm{i}}+\frac{\mathrm{17}}{\mathrm{24i}}\right\} \\ $$$$=−\frac{\mathrm{11}\pi}{\mathrm{6}\sqrt{\mathrm{3}}}\:+\frac{\mathrm{17}\pi}{\mathrm{12}}\:=\left(\frac{\mathrm{17}}{\mathrm{12}}−\frac{\mathrm{11}}{\mathrm{6}\sqrt{\mathrm{3}}}\right)\pi\:\Rightarrow\Psi=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{17}}{\mathrm{12}}−\frac{\mathrm{11}}{\mathrm{6}\sqrt{\mathrm{3}}}\right) \\ $$$$ \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jul/21
f(a,b) = ∫_0 ^∞ (dx/((x^2 +a^2 )(x^2 +b^2 ))) (1) f(a,b) = (1/(b^2 −a^2 ))∫_0 ^∞ ((1/(x^2 +a^2 ))−(1/(x^2 +b^2 ))) dx f(a,b) = (1/(b^2 −a^2 ))[(1/a)arctan(x/a)−(1/b)arctan(x/b)]_0 ^∞ f(a,b) = (1/(b^2 −a^2 )).(π/2)((1/a)−(1/b)) f(a,b) = (π/(2ab(a+b))) (2) (1) : ((∂f(a,b))/∂a) = −2a∫_0 ^∞ (dx/((x^2 +a^2 )^2 (x^2 +b^2 ))) ((∂^2 f(a,b))/(∂a∂b)) = 4ab∫_0 ^∞ (dx/((x^2 +a^2 )^2 (x^2 +b^2 )^2 )) ((∂^2 f(a,b))/(∂a∂b)) = 4abf(a,b) (3) (2) : ((∂f(a,b))/∂a) = −((π(2a+b))/(2a^2 b(a+b)^2 )) ((∂f^2 (a,b))/(∂a∂b)) = (π/(a^2 b^2 ))[((a^3 +4a^2 b+4ab^2 +b^3 )/((a+b)^4 ))] (4) (3) and (4) : f(a,b) = (π/(4a^3 b^3 ))[((a^3 +4a^2 b+4ab^2 +b^3 )/((a+b)^4 ))] Ω = ∫_0 ^∞ (dx/((x^2 +3)(x^2 +4))) = f((√3),2) ⇒ Ω = (π/( 96(√3)))[((19(√3)+32)/(((√3)+2)^4 ))] Ω = (π/( 96(√3)))(((32+19(√3))/(97+56(√3)))) Ω = (π/( 96(√3)))(51(√3)−88) Ω = (π/4)(((17)/8)−((11(√3))/9))
$${f}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\:\:\:\left(\mathrm{1}\right) \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)\:{dx} \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{{a}}\mathrm{arctan}\frac{{x}}{{a}}−\frac{\mathrm{1}}{{b}}\mathrm{arctan}\frac{{x}}{{b}}\right]_{\mathrm{0}} ^{\infty} \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }.\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right) \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\pi}{\mathrm{2}{ab}\left({a}+{b}\right)}\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:\::\:\frac{\partial{f}\left({a},{b}\right)}{\partial{a}}\:=\:−\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\: \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({a},{b}\right)}{\partial{a}\partial{b}}\:=\:\mathrm{4}{ab}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({a},{b}\right)}{\partial{a}\partial{b}}\:=\:\mathrm{4}{abf}\left({a},{b}\right)\:\:\:\left(\mathrm{3}\right) \\ $$$$\: \\ $$$$\left(\mathrm{2}\right)\::\:\frac{\partial{f}\left({a},{b}\right)}{\partial{a}}\:=\:−\frac{\pi\left(\mathrm{2}{a}+{b}\right)}{\mathrm{2}{a}^{\mathrm{2}} {b}\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\frac{\partial{f}^{\mathrm{2}} \left({a},{b}\right)}{\partial{a}\partial{b}}\:=\:\frac{\pi}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\left[\frac{{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} {b}+\mathrm{4}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} }{\left({a}+{b}\right)^{\mathrm{4}} }\right]\:\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right)\:: \\ $$$${f}\left({a},{b}\right)\:=\:\frac{\pi}{\mathrm{4}{a}^{\mathrm{3}} {b}^{\mathrm{3}} }\left[\frac{{a}^{\mathrm{3}} +\mathrm{4}{a}^{\mathrm{2}} {b}+\mathrm{4}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} }{\left({a}+{b}\right)^{\mathrm{4}} }\right] \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:\:=\:{f}\left(\sqrt{\mathrm{3}},\mathrm{2}\right) \\ $$$$\Rightarrow\:\Omega\:=\:\frac{\pi}{\:\mathrm{96}\sqrt{\mathrm{3}}}\left[\frac{\mathrm{19}\sqrt{\mathrm{3}}+\mathrm{32}}{\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)^{\mathrm{4}} }\right] \\ $$$$\Omega\:=\:\frac{\pi}{\:\mathrm{96}\sqrt{\mathrm{3}}}\left(\frac{\mathrm{32}+\mathrm{19}\sqrt{\mathrm{3}}}{\mathrm{97}+\mathrm{56}\sqrt{\mathrm{3}}}\right) \\ $$$$\Omega\:=\:\frac{\pi}{\:\mathrm{96}\sqrt{\mathrm{3}}}\left(\mathrm{51}\sqrt{\mathrm{3}}−\mathrm{88}\right) \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{17}}{\mathrm{8}}−\frac{\mathrm{11}\sqrt{\mathrm{3}}}{\mathrm{9}}\right) \\ $$

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