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Question Number 96567 by bemath last updated on 02/Jun/20

Commented by john santu last updated on 02/Jun/20

set y = (√(−1+(√((4/x)−3)))) x = (4/(3+(1+y^2 )^2 )) ∫_0 ^1 y dx = ∫_∞ ^0 xdy = ∫_∞ ^0 ((4/(3+(1+y^2 )^2 ))) dy = (1/(2(√2))) [ln (((2+y(y+(√2)))/(2+y(y−(√2))))]_∞ ^0 + (1/(√6)) [tan^(−1) (((y(√2)+1)/(√3)))+tan^(−1) (((y(√2)−1)/(√3)))]_∞ ^0 = (π/(√6))

$$\mathrm{set}\:{y}\:=\:\sqrt{−\mathrm{1}+\sqrt{\frac{\mathrm{4}}{{x}}−\mathrm{3}}}\: \\ $$$${x}\:=\:\frac{\mathrm{4}}{\mathrm{3}+\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{y}\:{dx}\:=\:\underset{\infty} {\overset{\mathrm{0}} {\int}}\:{xdy}\:=\:\underset{\infty} {\overset{\mathrm{0}} {\int}}\:\left(\frac{\mathrm{4}}{\mathrm{3}+\left(\mathrm{1}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)\:{dy} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\left[\mathrm{ln}\:\left(\frac{\mathrm{2}+{y}\left({y}+\sqrt{\mathrm{2}}\right)}{\mathrm{2}+{y}\left({y}−\sqrt{\mathrm{2}}\right)}\right]_{\infty} ^{\mathrm{0}} +\right. \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\:\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}\sqrt{\mathrm{2}}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right]_{\infty} ^{\mathrm{0}} \\ $$$$=\:\frac{\pi}{\sqrt{\mathrm{6}}}\: \\ $$

Answered by mathmax by abdo last updated on 02/Jun/20

I =∫_0 ^1 (√((√((4/x)−3))−1))dx changement (√((4/x)−3))=t give (4/x)−3 =t^(2 ) ⇒ (4/x) =t^2 +3 ⇒(x/4) =(1/(t^2 +3)) ⇒x =(4/(t^2 +3)) ⇒dx =−((4×(2t))/((t^2 +3)^2 ))dt =−((8t)/((t^2 +3)^2 ))dt ⇒ I =−∫_1 ^(+∞) (√(t−1))(((−8t)/((t^2 +3)^2 )))dt =8 ∫_1 ^∞ ((t(√(t−1)))/((t^2 +3)^2 ))dt =_((√(t−1))=u) 8 ∫_0 ^∞ (((1+u^2 )u)/(((1+u^2 )^2 +3)^2 ))(2u)du =16 ∫_0 ^∞ ((u^2 (1+u^2 ))/(((u^2 +1)^2 +3)^2 ))du =8 ∫_(−∞) ^(+∞) ((u^2 (1+u^2 ))/(((u^2 +1)^2 +3)^2 ))du let ϕ(z) =((z^2 (1+z^2 ))/({(z^2 +1)^2 +3}^2 )) ⇒ ϕ(z) =((z^2 (1+z^2 ))/((z^2 +1−i(√3))^2 (z^2 +1+i(√3))^2 )) poles of ϕ? z^2 +1−i(√3)=z^2 −(−1+i(√3)) we have −1+i(√3)=2(−(1/2) +((i(√3))/2)) =2e^((i2π)/3) ⇒z^2 +1−i(√3)=z^2 −2 e^((i2π)/3) =(z−(√2)e^((iπ)/3) )(z+(√2)e^((iπ)/3) ) z^2 +1+i(√3)=z^2 −(−1−i(√3)) =z^2 −2e^(−((i2π)/3)) =(z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) ) ⇒ϕ(z) =((z^2 (1+z^2 ))/((z−(√2)e^((iπ)/3) )^2 (z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )^2 (z+(√2)e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√2)e^((iπ)/3) ) +Res(ϕ,−(√2)e^(−((iπ)/3)) )} Res(ϕ,(√2)e^((iπ)/3) ) =lim_(z→(√2)e^((iπ)/3) ) (1/((2−1)!)) {(z−(√2)e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→(√2)e^((iπ)/3) ) {((z^2 +z^4 )/((z+(√2)e^((iπ)/3) )^2 (z^2 +1+i(√3))^2 ))}^((1)) ...be continued....

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}−\mathrm{1}}\mathrm{dx}\:\mathrm{changement}\:\sqrt{\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}}=\mathrm{t}\:\mathrm{give}\:\frac{\mathrm{4}}{\mathrm{x}}−\mathrm{3}\:=\mathrm{t}^{\mathrm{2}\:} \:\Rightarrow \\ $$$$\frac{\mathrm{4}}{\mathrm{x}}\:=\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}\:\Rightarrow\frac{\mathrm{x}}{\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{4}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}}\:\Rightarrow\mathrm{dx}\:=−\frac{\mathrm{4}×\left(\mathrm{2t}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt}\:=−\frac{\mathrm{8t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\int_{\mathrm{1}} ^{+\infty} \:\sqrt{\mathrm{t}−\mathrm{1}}\left(\frac{−\mathrm{8t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\right)\mathrm{dt}\:=\mathrm{8}\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{t}\sqrt{\mathrm{t}−\mathrm{1}}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=_{\sqrt{\mathrm{t}−\mathrm{1}}=\mathrm{u}} \:\:\:\:\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{u}}{\left(\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\left(\mathrm{2u}\right)\mathrm{du}\:=\mathrm{16}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\left(\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{du} \\ $$$$=\mathrm{8}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{u}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\left(\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{du}\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\left\{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\right\}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{z}^{\mathrm{2}} −\left(−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right)\:\mathrm{we}\:\mathrm{have}\:−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:=\mathrm{2e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{z}^{\mathrm{2}} −\mathrm{2}\:\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:=\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right) \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{z}^{\mathrm{2}} −\left(−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}\right)\:=\mathrm{z}^{\mathrm{2}} −\mathrm{2e}^{−\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:=\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right) \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)}{\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:+\mathrm{Res}\left(\varphi,−\sqrt{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left(\mathrm{z}−\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} } \:\:\:\left\{\frac{\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}^{\mathrm{4}} }{\left(\mathrm{z}+\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:...\mathrm{be}\:\mathrm{continued}.... \\ $$$$ \\ $$

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