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Question Number 67008 by mathmax by abdo last updated on 21/Aug/19
let f(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^2 )) with x>0 1) find a explicit form of (x) 2)find also g(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^3 )) 3)find the values of integrals ∫_0 ^∞ (dt/((t^2 +3)^2 )) and ∫_0 ^∞ (dt/((t^2 +3)^3 )) 4) calculate U_θ =∫_0 ^∞ (dt/((t^2 +cos^2 θ)^2 )) with 0<θ<(π/2) 5) find f^((n)) (x) and f^((n)) (0) 6) developp f at integr serie
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:\left({x}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{also}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{U}_{\theta} =\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\:\:{with}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{5}\right)\:{find}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{6}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by mathmax by abdo last updated on 25/Aug/19
1) f(x) =∫_0 ^∞ (dt/((x^2 +t^2 )^2 )) ⇒2f(x) =∫_(−∞) ^(+∞) (dt/((t^2 +x^2 )^2 )) let =_(t=xz) ∫_(−∞) ^(+∞) ((xdz)/(x^4 (z^2 +1)^2 )) =(1/x^3 ) ∫_(−∞) ^(+∞) (dz/((z^2 +1)^2 )) let ϕ(z)=(1/((z^2 +1)^2 )) we have ϕ(z) =(1/((z−i)^2 (z+i)^2 )) resi_ dus tbeorem hive ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(z+i)^(−2) }^((1)) =lim_(z→i) −2(z+i)^(−3) =−2(2i)^(−3) =((−2)/((2i)^3 )) =((−2)/(−8i)) =(1/(4i)) ⇒ ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ×(1/(4i)) =(π/2) ⇒2f(x)=(π/(2x^3 )) ⇒f(x) =(π/(4x^3 )) (x>0) 2) we have f^′ (x) =−∫_0 ^∞ ((2(2x)(x^(2 ) +t^2 ))/((x^2 +t^2 )^4 ))dx =−4x ∫_0 ^∞ (dx/((x^2 +t^2 )^3 )) =−4x g(x) ⇒g(x) =−(1/(4x))f^′ (x) f^′ (x) =(π/4)(x^(−3) )^′ =(π/4)(−3)x^(−4) =((−3π)/(4x^4 )) ⇒ g(x) =−(1/(4x))×((−3π)/(4x^4 )) =((3π)/(16x^5 ))
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{f}\left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:{let} \\ $$$$=_{{t}={xz}} \:\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{xdz}}{{x}^{\mathrm{4}} \left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:\int_{−\infty} ^{+\infty} \:\:\frac{{dz}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:\:{resi}_{} {dus}\:{tbeorem}\:{hive} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \:={lim}_{{z}\rightarrow{i}} \:\:\left\{\left({z}+{i}\right)^{−\mathrm{2}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:−\mathrm{2}\left({z}+{i}\right)^{−\mathrm{3}} \:=−\mathrm{2}\left(\mathrm{2}{i}\right)^{−\mathrm{3}} \:=\frac{−\mathrm{2}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{−\mathrm{2}}{−\mathrm{8}{i}}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty\:} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\frac{\pi}{\mathrm{2}{x}^{\mathrm{3}} }\:\Rightarrow{f}\left({x}\right)\:=\frac{\pi}{\mathrm{4}{x}^{\mathrm{3}} }\:\:\:\:\:\left({x}>\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}\left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}\:} +{t}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{4}} }{dx}\:=−\mathrm{4}{x}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=−\mathrm{4}{x}\:{g}\left({x}\right)\:\Rightarrow{g}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}{x}}{f}^{'} \left({x}\right) \\ $$$${f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{4}}\left({x}^{−\mathrm{3}} \right)^{'} \:=\frac{\pi}{\mathrm{4}}\left(−\mathrm{3}\right){x}^{−\mathrm{4}} \:=\frac{−\mathrm{3}\pi}{\mathrm{4}{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$${g}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}{x}}×\frac{−\mathrm{3}\pi}{\mathrm{4}{x}^{\mathrm{4}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}{x}^{\mathrm{5}} } \\ $$
Commented by mathmax by abdo last updated on 25/Aug/19
3) ∫_0 ^∞ (dt/((t^2 +3)^2 )) =f((√3)) =(π/(4((√3))^3 )) =(π/(12(√3))) =((π(√3))/(36)) ∫_0 ^∞ (dt/((t^2 +3)^3 )) =g((√3)) =((3π)/(16((√3))^5 )) =((3π)/(16×9×(√3))) =(π/(48(√3))) =((π(√3))/(144))
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\frac{\pi}{\mathrm{4}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:=\frac{\pi}{\mathrm{12}\sqrt{\mathrm{3}}}\:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{36}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} }\:={g}\left(\sqrt{\mathrm{3}}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\left(\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}×\mathrm{9}×\sqrt{\mathrm{3}}}\:=\frac{\pi}{\mathrm{48}\sqrt{\mathrm{3}}}\:=\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{144}} \\ $$

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