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Question Number 124201 by mnjuly1970 last updated on 01/Dec/20
::: nice calculus ::: please prove ::: Ω = ∫_0 ^( ∞) (x^(1/2) /(x^2 +2x+5))dx=(π/( (√ϕ))) where ϕ is Golden ratio...
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\::::\:\:{nice}\:\:{calculus}\:::: \\ $$$$\:\:\:\:\:{please}\:\:{prove}\:::: \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\frac{\pi}{\:\sqrt{\varphi}} \\ $$$$\:\:\:\:\:{where}\:\:\varphi\:\:{is}\:{Golden}\:{ratio}… \\ $$
Answered by ajfour last updated on 02/Dec/20
I=∫(((√x)dx)/((x+1)^2 +(2ϕ−1)^2 )) let (√x)=t ⇒ (√x)dx=2t^2 dt I=∫((2t^2 dt)/((t^2 +1)^2 +(2ϕ−1)^2 )) let t=(1/z) ⇒ dt=−(dz/z^2 ) I=∫((2dz)/((1+z^2 )^2 +(2ϕ−1)^2 z^4 )) = ∫((2dz)/(6z^4 +2z^2 +1)) = (1/3)∫(dz/((z^2 +(1/6))^2 +(((√5)/6))^2 )) ....
$${I}=\int\frac{\sqrt{{x}}{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\varphi−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:\:\sqrt{{x}}={t}\:\Rightarrow\:\:\sqrt{{x}}{dx}=\mathrm{2}{t}^{\mathrm{2}} {dt} \\ $$$${I}=\int\frac{\mathrm{2}{t}^{\mathrm{2}} {dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\varphi−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${let}\:\:{t}=\frac{\mathrm{1}}{{z}}\:\:\Rightarrow\:\:{dt}=−\frac{{dz}}{{z}^{\mathrm{2}} } \\ $$$${I}=\int\frac{\mathrm{2}{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{2}\varphi−\mathrm{1}\right)^{\mathrm{2}} {z}^{\mathrm{4}} } \\ $$$$\:\:=\:\int\frac{\mathrm{2}{dz}}{\mathrm{6}{z}^{\mathrm{4}} +\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dz}}{\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}\right)^{\mathrm{2}} } \\ $$$$…. \\ $$
Commented by mindispower last updated on 02/Dec/20
sir ajfour your methode worcks since 6 z^4 +2z^2 +1 can bee factorised in polynoms of degree 2 and∫(dx/(x^2 +ax+b)) can bee solved
$${sir}\:{ajfour}\:{your}\:{methode} \\ $$$${worcks}\:{since}\:\mathrm{6}\:{z}^{\mathrm{4}} \:+\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1}\:{can}\:{bee}\:{factorised} \\ $$$${in}\:{polynoms}\:{of}\:{degree}\:\mathrm{2} \\ $$$${and}\int\frac{{dx}}{{x}^{\mathrm{2}} +{ax}+{b}}\:{can}\:{bee}\:{solved} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 03/Dec/20
thanking for your effort sir ajfor..
$${thanking}\:\:{for}\:{your}\:{effort} \\ $$$${sir}\:{ajfor}.. \\ $$
Commented by mnjuly1970 last updated on 03/Dec/20
Answered by mindispower last updated on 02/Dec/20
Δ=∫_(−∞) ^∞ (z^(1/2) /(z^2 +2z+5))dz=∫_(−∞) ^0 (z^(1/2) /(z^2 +2z+5))dz+Ω =i∫_0 ^∞ (z^(1/2) /(z^2 −2z+5))dz+Ω=Ω+it,Ω,t∈R Δ=2iπRes((z^(1/2) /(z^2 +2z+5)),z=−1+2i) .=2iπ(((−1+2i)^(1/2) )/(2(−1+2i)+2))=(π/2)(−1+2i)^(1/2) (−1+2i)^(1/2) =x+iy ⇒xy=1,x^2 +y^2 =(√5),x^2 −y^2 =−1 x=((((√5)−1)/2))^(1/2) ,y=(2/( (√5)−1))=((((√5)+1)/2))^(1/2) x=(√(ϕ−1)),y=(√ϕ) Δ=(π/2)(√(ϕ−1))+(π/2)i(√ϕ) Ω=Re(Δ)=(π/2)(√(ϕ−1)) ϕ^2 =ϕ+1⇒ϕ(ϕ−1)=1 ⇒(√(ϕ−1))=(1/( (√ϕ))) Ω=(π/2)(√(ϕ−1))=(π/(2(√ϕ)))
$$\Delta=\int_{−\infty} ^{\infty} \frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{5}}{dz}=\int_{−\infty} ^{\mathrm{0}} \frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{5}}{dz}+\Omega \\ $$$$={i}\int_{\mathrm{0}} ^{\infty} \frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}}{dz}+\Omega=\Omega+{it},\Omega,{t}\in\mathbb{R} \\ $$$$\Delta=\mathrm{2}{i}\pi{Res}\left(\frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{5}},{z}=−\mathrm{1}+\mathrm{2}{i}\right) \\ $$$$.=\mathrm{2}{i}\pi\frac{\left(−\mathrm{1}+\mathrm{2}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}\left(−\mathrm{1}+\mathrm{2}{i}\right)+\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left(−\mathrm{1}+\mathrm{2}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(−\mathrm{1}+\mathrm{2}{i}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={x}+{iy} \\ $$$$\Rightarrow{xy}=\mathrm{1},{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\sqrt{\mathrm{5}},{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\mathrm{1} \\ $$$${x}=\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ,{y}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}=\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}=\sqrt{\varphi−\mathrm{1}},{y}=\sqrt{\varphi} \\ $$$$\Delta=\frac{\pi}{\mathrm{2}}\sqrt{\varphi−\mathrm{1}}+\frac{\pi}{\mathrm{2}}{i}\sqrt{\varphi} \\ $$$$\Omega={Re}\left(\Delta\right)=\frac{\pi}{\mathrm{2}}\sqrt{\varphi−\mathrm{1}} \\ $$$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1}\Rightarrow\varphi\left(\varphi−\mathrm{1}\right)=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\varphi−\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{\varphi}} \\ $$$$\Omega=\frac{\pi}{\mathrm{2}}\sqrt{\varphi−\mathrm{1}}=\frac{\pi}{\mathrm{2}\sqrt{\varphi}} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 02/Dec/20
excellent
$${excellent} \\ $$

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