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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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