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Question Number 139036 by mnjuly1970 last updated on 21/Apr/21
prove :: 𝛗=∫_0 ^( ∞) ((√x)/(x^2 +2x+5))dx=(Ο€/(2(βˆšΟ•))) Ο•:= golden ratio ...
$$\:\:\:\: \\ $$$$\:\:\:\:\:{prove}\::: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\frac{\pi}{\mathrm{2}\sqrt{\varphi}} \\ $$$$\:\:\:\:\:\:\varphi:=\:{golden}\:{ratio}\:… \\ $$
Answered by Dwaipayan Shikari last updated on 21/Apr/21
(1/(4i))∫_0 ^∞ ((√x)/((x+1βˆ’2i)))βˆ’((√x)/((x+1+2i)))dx =(1/(2i))∫_0 ^∞ (t^2 /(t^2 +1βˆ’2i))βˆ’(t^2 /(t^2 +1+2i))dt =(1/(2i))∫_0 ^∞ ((2iβˆ’1)/(t^2 +1βˆ’2i))+((2i+1)/(t^2 +1+2i))dt=(Ο€/(4i))(((2iβˆ’1)/( (√(1βˆ’2i))))+((2i+1)/( (√(1+2i))))) =(Ο€/(4i))(βˆ’(√(1βˆ’2i))+(√(1+2i)))=(Ο€/(4i))(2i(√(((√5)βˆ’1)/2)))=(Ο€/(2(βˆšΟ•))) == (√(x+(√(βˆ’y))))=(√((x+(√(x^2 +y)))/2))+(√((xβˆ’(√(x^2 +y)))/2)) (√(xβˆ’(√(βˆ’y))))=(√((x+(√(x^2 +y)))/2))βˆ’(√((xβˆ’(√(x^2 +y)))/2)) (√(1+2i))=(√((1+(√5))/2))+i(√(((√5)βˆ’1)/2)) (√(1βˆ’2i))=(√((1+(√5))/2))βˆ’i(√(((√5)βˆ’1)/2))
$$\frac{\mathrm{1}}{\mathrm{4}{i}}\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{x}}}{\left({x}+\mathrm{1}βˆ’\mathrm{2}{i}\right)}βˆ’\frac{\sqrt{{x}}}{\left({x}+\mathrm{1}+\mathrm{2}{i}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}βˆ’\mathrm{2}{i}}βˆ’\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{i}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{i}βˆ’\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}βˆ’\mathrm{2}{i}}+\frac{\mathrm{2}{i}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{i}}{dt}=\frac{\pi}{\mathrm{4}{i}}\left(\frac{\mathrm{2}{i}βˆ’\mathrm{1}}{\:\sqrt{\mathrm{1}βˆ’\mathrm{2}{i}}}+\frac{\mathrm{2}{i}+\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}{i}}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}{i}}\left(βˆ’\sqrt{\mathrm{1}βˆ’\mathrm{2}{i}}+\sqrt{\mathrm{1}+\mathrm{2}{i}}\right)=\frac{\pi}{\mathrm{4}{i}}\left(\mathrm{2}{i}\sqrt{\frac{\sqrt{\mathrm{5}}βˆ’\mathrm{1}}{\mathrm{2}}}\right)=\frac{\pi}{\mathrm{2}\sqrt{\varphi}} \\ $$$$== \\ $$$$\sqrt{{x}+\sqrt{βˆ’{y}}}=\sqrt{\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}}+\sqrt{\frac{{x}βˆ’\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}} \\ $$$$\sqrt{{x}βˆ’\sqrt{βˆ’{y}}}=\sqrt{\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}}βˆ’\sqrt{\frac{{x}βˆ’\sqrt{{x}^{\mathrm{2}} +{y}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}{i}}=\sqrt{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}+{i}\sqrt{\frac{\sqrt{\mathrm{5}}βˆ’\mathrm{1}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{1}βˆ’\mathrm{2}{i}}=\sqrt{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}βˆ’{i}\sqrt{\frac{\sqrt{\mathrm{5}}βˆ’\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by mnjuly1970 last updated on 21/Apr/21
thanks alot...
$${thanks}\:{alot}… \\ $$
Answered by mnjuly1970 last updated on 21/Apr/21
Answered by mathmax by abdo last updated on 21/Apr/21
Ξ¦=∫_0 ^∞ ((√x)/(x^2 +2x+5))dx β‡’Ξ¦=_((√x)=t) ∫_0 ^∞ ((t(2t))/(t^4 +2t^2 +5))dt =∫_(βˆ’βˆž) ^(+∞) (t^2 /(t^4 +2t^2 +5))dt let w(z)=(z^2 /(z^4 +2z^2 +5)) poles of w! z^4 +2z^2 +5=0 β†’x^2 +2x+5=0(t^2 =x) Ξ”=4βˆ’20=βˆ’16 β‡’x_1 =((βˆ’2+i(√(20)))/2) =((βˆ’2+2i(√5))/2)=βˆ’1+i(√5) x_2 =βˆ’1βˆ’i(√5) β‡’ w(z)=(z^2 /((z^2 +1βˆ’i(√5))(z^2 +1+i(√5)))) =(z^2 /((zβˆ’(√x_1 ))(z+(√x_1 ))(zβˆ’(√x_2 ))(z+(√x_2 )))) x_1 =(√6)e^(βˆ’iarctan((√5))) and x_2 =(√6)e^(iarctan((√5))) β‡’w(z)=(z^2 /((zβˆ’(√(√6))e^(βˆ’(i/2)arctan((√5))) )(z+(√(√6))e^(βˆ’(i/2)arctan((√5))) )(zβˆ’(√(√6))e^((i/2)arctan((√5))) )(z+(√(√6))e^((i/2)arctan((√5))) ))) ∫_(βˆ’βˆž) ^(+∞) w(z)dz =2iΟ€{ Res(w,βˆ’(√(√6))e^(βˆ’(i/2)arctan((√5))) +Res(w,(√(√6))e^((i/2)arctan((√5))) } Res(w,(√(√6))e^((i/2)arctan((√5))) ) =(((√6)e^(iarctan((√5))) )/(2(√(√6))e^((i/2)arctan((√5))) (√6)(2isin(arctan((√5)))) =(1/(4i))(e^((i/2)arctan((√5))) /( (√(√6))sin(arctan((√5)))))....be continued...
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\sqrt{\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{5}}\mathrm{dx}\:\Rightarrow\Phi=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}\left(\mathrm{2t}\right)}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{2t}^{\mathrm{2}} \:+\mathrm{5}}\mathrm{dt} \\ $$$$=\int_{βˆ’\infty} ^{+\infty} \:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} \:+\mathrm{2t}^{\mathrm{2}} \:+\mathrm{5}}\mathrm{dt}\:\:\mathrm{let}\:\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{z}^{\mathrm{4}} \:+\mathrm{2z}^{\mathrm{2}} \:+\mathrm{5}}\:\:\mathrm{poles}\:\mathrm{of}\:\mathrm{w}! \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\mathrm{2z}^{\mathrm{2}} \:+\mathrm{5}=\mathrm{0}\:\rightarrow\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{5}=\mathrm{0}\left(\mathrm{t}^{\mathrm{2}} =\mathrm{x}\right) \\ $$$$\Delta=\mathrm{4}βˆ’\mathrm{20}=βˆ’\mathrm{16}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{βˆ’\mathrm{2}+\mathrm{i}\sqrt{\mathrm{20}}}{\mathrm{2}}\:=\frac{βˆ’\mathrm{2}+\mathrm{2i}\sqrt{\mathrm{5}}}{\mathrm{2}}=βˆ’\mathrm{1}+\mathrm{i}\sqrt{\mathrm{5}} \\ $$$$\mathrm{x}_{\mathrm{2}} =βˆ’\mathrm{1}βˆ’\mathrm{i}\sqrt{\mathrm{5}}\:\Rightarrow\:\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}βˆ’\mathrm{i}\sqrt{\mathrm{5}}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}βˆ’\sqrt{\mathrm{x}_{\mathrm{1}} }\right)\left(\mathrm{z}+\sqrt{\mathrm{x}_{\mathrm{1}} }\right)\left(\mathrm{z}βˆ’\sqrt{\mathrm{x}_{\mathrm{2}} }\right)\left(\mathrm{z}+\sqrt{\mathrm{x}_{\mathrm{2}} }\right)} \\ $$$$\mathrm{x}_{\mathrm{1}} =\sqrt{\mathrm{6}}\mathrm{e}^{βˆ’\mathrm{iarctan}\left(\sqrt{\mathrm{5}}\right)} \:\:\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\sqrt{\mathrm{6}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{5}}\right)} \\ $$$$\Rightarrow\mathrm{w}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}βˆ’\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{βˆ’\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right)\left(\mathrm{z}+\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{βˆ’\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right)\left(\mathrm{z}βˆ’\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right)\left(\mathrm{z}+\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right)} \\ $$$$\int_{βˆ’\infty} ^{+\infty} \:\mathrm{w}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\mathrm{w},βˆ’\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{βˆ’\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \:+\mathrm{Res}\left(\mathrm{w},\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right\}\right.\right. \\ $$$$\mathrm{Res}\left(\mathrm{w},\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \right)\:=\frac{\sqrt{\mathrm{6}}\mathrm{e}^{\mathrm{iarctan}\left(\sqrt{\mathrm{5}}\right)} }{\mathrm{2}\sqrt{\sqrt{\mathrm{6}}}\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} \sqrt{\mathrm{6}}\left(\mathrm{2isin}\left(\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)\right.\right.} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4i}}\frac{\mathrm{e}^{\frac{\mathrm{i}}{\mathrm{2}}\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)} }{\:\sqrt{\sqrt{\mathrm{6}}}\mathrm{sin}\left(\mathrm{arctan}\left(\sqrt{\mathrm{5}}\right)\right)}….\mathrm{be}\:\mathrm{continued}… \\ $$

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