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Question Number 57899 by maxmathsup by imad last updated on 13/Apr/19
let f(x) =∫_0 ^(+∞) (dt/((t^2 +x^2 )^3 )) with x>0 1) find a explicit form off (x) 1) calculate ∫_0 ^∞ (dx/((t^2 +3)^3 )) and ∫_0 ^∞ (dt/((t^2 +4)^3 )) 2) find the value of A(θ) =∫_0 ^∞ (dt/((t^2 +sin^2 θ)^3 )) with 0<θ<π.
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{off}\:\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} }\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }\:\:{with}\:\mathrm{0}<\theta<\pi. \\ $$
Commented by maxmathsup by imad last updated on 17/Apr/19
1) we have f(x)=(1/2) ∫_(−∞) ^(+∞) (dt/((t^2 +x^2 )^3 )) let consider the complex function ϕ(z) =(1/((z^2 +x^2 )^3 )) poles of ϕ? we have ϕ(z) =(1/((z−ix)^3 (z+ix)^3 )) so the poles of are +^− ix (triples) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,ix) Res(ϕ,ix) =lim_(z→ix) (1/((3−1)!)){ (z−ix)^3 ϕ(z)}^((2)) =lim_(z→ix) (1/2){ (1/((z+ix)^3 ))}^((2)) we have {(1/((z+ix)^3 ))}^((1)) =−((3(z+ix)^2 )/((z+ix)^6 )) =−(3/((z+ix)^4 )) ⇒{(1/((z+ix)^3 ))}^((2)) =−3 ((−4(z+ix)^3 )/((z+ix)^8 )) =((12)/((z+ix)^5 )) ⇒Res(ϕ,ix) =lim_(z→ix) (6/((z+ix)^5 )) =(6/((2ix)^5 )) =(6/(2^5 x^5 i)) =(6/(32ix^5 )) =(3/(16ix^5 )) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (3/(16 x^5 i)) =((3π)/(8 x^5 )) ⇒ f(x) =((3π)/(16 x^5 )) ( with x>0) 2) ∫_0 ^∞ (dt/((t^2 +3)^3 )) =f((√3)) = ((3π)/(16 ((√3))^5 )) =((3π)/(16 ((√3))^4 (√3))) =((3π)/(9×16 (√3))) ∫_0 ^∞ (dt/((t^2 +4)^3 )) =f(2) =((3π)/(16 (2)^5 )) =((3π)/(16×32))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{poles}\:{of}\:\varphi? \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{ix}\right)^{\mathrm{3}} \left({z}+{ix}\right)^{\mathrm{3}} }\:\:{so}\:\:{the}\:{poles}\:{of}\:{are}\:\overset{−} {+}{ix}\:\left({triples}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ix}\right) \\ $$$${Res}\left(\varphi,{ix}\right)\:={lim}_{{z}\rightarrow{ix}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\:\left({z}−{ix}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{ix}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:\:{we}\:{have}\: \\ $$$$\left\{\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \:=−\frac{\mathrm{3}\left({z}+{ix}\right)^{\mathrm{2}} }{\left({z}+{ix}\right)^{\mathrm{6}} }\:=−\frac{\mathrm{3}}{\left({z}+{ix}\right)^{\mathrm{4}} }\:\Rightarrow\left\{\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=−\mathrm{3}\:\frac{−\mathrm{4}\left({z}+{ix}\right)^{\mathrm{3}} }{\left({z}+{ix}\right)^{\mathrm{8}} }\:=\frac{\mathrm{12}}{\left({z}+{ix}\right)^{\mathrm{5}} }\:\Rightarrow{Res}\left(\varphi,{ix}\right)\:={lim}_{{z}\rightarrow{ix}} \:\:\:\:\frac{\mathrm{6}}{\left({z}+{ix}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{6}}{\left(\mathrm{2}{ix}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {x}^{\mathrm{5}} {i}}\:=\frac{\mathrm{6}}{\mathrm{32}{ix}^{\mathrm{5}} }\:=\frac{\mathrm{3}}{\mathrm{16}{ix}^{\mathrm{5}} }\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{3}}{\mathrm{16}\:{x}^{\mathrm{5}} {i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}\:{x}^{\mathrm{5}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:{x}^{\mathrm{5}} }\:\:\:\left(\:\:{with}\:{x}>\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} }\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\:\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \sqrt{\mathrm{3}}}\:=\frac{\mathrm{3}\pi}{\mathrm{9}×\mathrm{16}\:\sqrt{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} }\:={f}\left(\mathrm{2}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\mathrm{2}\right)^{\mathrm{5}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}×\mathrm{32}} \\ $$
Commented by maxmathsup by imad last updated on 17/Apr/19
3) ∫_0 ^∞ (dt/((t^2 +sin^2 θ)^3 )) =f(sinθ) =((3π)/(16 sin^5 θ))
$$\left.\mathrm{3}\right)\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }\:={f}\left({sin}\theta\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:{sin}^{\mathrm{5}} \theta} \\ $$

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