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Question Number 76146 by john santuy last updated on 24/Dec/19

Commented by john santuy last updated on 24/Dec/19

$h o w t o p r o o f ?$
Commented by mr W last updated on 25/Dec/19

$v a l u e g i v e n i n q u e s t i o n (\frac{π}{\sqrt{6}}) i s$ $c o r r e c t, s i r .$ $i t h i n k y o u h a d a m i s t a k e i n y o u r$ $c a l c u l a t i o n . i t s h o u l d b e$ $A = - \int_{1}^{+ \infty} \sqrt{- 1 + \sqrt{t}} \frac{- 4}{{(t + 3)}^{2}} d t$ $p l e a s e r e c h e c k .$
Commented by mr W last updated on 25/Dec/19

$i h a v e p o i n t e d t h e e r r o r i n y o u r$ $a n s w e r . {i t}^{'} s i n t h e t h i r d l i n e :$ $A = - \int_{1}^{+ \infty} \sqrt{- 1 + t} \frac{- 4}{{(t + 3)}^{2}} d t = . . . . . .$ $i t s h o u l d b e$ $A = - \int_{1}^{+ \infty} \sqrt{- 1 + \sqrt{t}} \frac{- 4}{{(t + 3)}^{2}} d t = . . . . . .$
Commented by abdomathmax last updated on 25/Dec/19

$y e s y o u a r e r i g h t s i r . . . i w i l l c o r e e c t t h e a n s w e r . . .$
Commented by mathmax by abdo last updated on 26/Dec/19
$let I =∫_0 ^1 (√(−1+(√((4/x)−3))))dx cha7gement (√((4/x)−3))=t give (4/x) −3 =t^2 ⇒4−3x=t^2 x ⇒4=(t^2 +3)x ⇒x=(4/(t^2 +3)) ⇒ dx =((−8t)/((t^2 +3)^2 )) ⇒ I =−∫_1 ^(+∞) (√(t−1))×((−8t)/((t^2 +3)^2 ))dt =8 ∫_1 ^(+∞) ((t(√(t−1)))/((t^(2 ) +3)^2 ))dt =_((√(t−1))=u) 8 ∫_o ^(+∞) (((1+u^2 )u)/(((1+u^2 )^2 +3)^2 ))(2u)du =16 ∫_0 ^∞ ((u^2 (1+u^2 ))/((u^4 +2u^2 +4)^2 ))du =8 ∫_(−∞) ^(+∞) ((u^4 +u^2 )/((u^4 +2u^2 +4)^2 ))du letϕ(z)=((z^4 +z^2 )/((z^4 +2z^2 +4)^2 )) poles of ϕ? z^4 +2z^2 +4 =0 ⇒t^2 +2t +4=0 with t=z^2 Δ^′ =1−4 =−3 ⇒t_1 =−1+(√3)=2(−(1/2)+((√3)/2)) =2e^((i2π)/3) t_1 =−1−(√3)=2 e^(−((i2π)/3)) ⇒z^4 +2z^2 +4 =(z^2 −2e^((i2π)/3) )(z^2 −2e^(−((i2π)/3)) ) =(z−(√2)e^((iπ)/3) )(z+(√2)e^((iπ)/3) )(z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) )⇒ ϕ(z) =((z^4 +z^2 )/((z−(√2)e^((iπ)/3) )^2 (z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,(√2)e^((iπ)/3) )+Res(ϕ,−(√2)e^(−((iπ)/3)) )} Res(ϕ,(√2)e^((iπ)/3) ) =lim_(z→(√2)e^((iπ)/3) ) (1/((2−1)!)){(z−(√2)e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→(√2)e^((iπ)/3) ) {((z^4 +z^2 )/((z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )^2 (z+(√2)e^(−((iπ)/3)) )^2 ))}^((1)) ...be continued...$
$l e t I = \int_{0}^{1} \sqrt{- 1 + \sqrt{\frac{4}{x} - 3}} d x c h a 7 g e m e n t \sqrt{\frac{4}{x} - 3} = t g i v e$ $\frac{4}{x} - 3 = t^{2} \Rightarrow 4 - 3 x = t^{2} x \Rightarrow 4 = (t^{2} + 3) x \Rightarrow x = \frac{4}{t^{2} + 3} \Rightarrow$ $d x = \frac{- 8 t}{{(t^{2} + 3)}^{2}} \Rightarrow I = - \int_{1}^{+ \infty} \sqrt{t - 1} \times \frac{- 8 t}{{(t^{2} + 3)}^{2}} d t$ $= 8 \int_{1}^{+ \infty} \frac{t \sqrt{t - 1}}{{(t^{2} + 3)}^{2}} d t =_{\sqrt{t - 1} = u} 8 \int_{o}^{+ \infty} \frac{(1 + u^{2}) u}{{({(1 + u^{2})}^{2} + 3)}^{2}} (2 u) d u$ $= 16 \int_{0}^{\infty} \frac{u^{2} (1 + u^{2})}{{(u^{4} + 2 u^{2} + 4)}^{2}} d u = 8 \int_{- \infty}^{+ \infty} \frac{u^{4} + u^{2}}{{(u^{4} + 2 u^{2} + 4)}^{2}} d u$ $l e t φ (z) = \frac{z^{4} + z^{2}}{{(z^{4} + 2 z^{2} + 4)}^{2}} p o l e s o f φ ?$ $z^{4} + 2 z^{2} + 4 = 0 \Rightarrow t^{2} + 2 t + 4 = 0 w i t h t = z^{2}$ $Δ^{^{'}} = 1 - 4 = - 3 \Rightarrow t_{1} = - 1 + \sqrt{3} = 2 (- \frac{1}{2} + \frac{\sqrt{3}}{2}) = 2 e^{\frac{i 2 π}{3}}$ $t_{1} = - 1 - \sqrt{3} = 2 e^{- \frac{i 2 π}{3}} \Rightarrow z^{4} + 2 z^{2} + 4 = (z^{2} - 2 e^{\frac{i 2 π}{3}}) (z^{2} - 2 e^{- \frac{i 2 π}{3}})$ $= (z - \sqrt{2} e^{\frac{i π}{3}}) (z + \sqrt{2} e^{\frac{i π}{3}}) (z - \sqrt{2} e^{- \frac{i π}{3}}) (z + \sqrt{2} e^{- \frac{i π}{3}}) \Rightarrow$ $φ (z) = \frac{z^{4} + z^{2}}{{(z - \sqrt{2} e^{\frac{i π}{3}})}^{2} {(z + \sqrt{2} e^{\frac{i π}{3}})}^{2} (z - \sqrt{2} e^{- \frac{i π}{3}}) (z + \sqrt{2} e^{- \frac{i π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \sqrt{2} e^{\frac{i π}{3}}) + R e s (φ, - \sqrt{2} e^{- \frac{i π}{3}})}$ $R e s (φ, \sqrt{2} e^{\frac{i π}{3}}) = {l i m}_{z \to \sqrt{2} e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - \sqrt{2} e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to \sqrt{2} e^{\frac{i π}{3}}} {\frac{z^{4} + z^{2}}{{(z + \sqrt{2} e^{\frac{i π}{3}})}^{2} {(z - \sqrt{2} e^{- \frac{i π}{3}})}^{2} {(z + \sqrt{2} e^{- \frac{i π}{3}})}^{2}}}^{(1)}$ $. . . b e c o n t i n u e d . . .$
	Answered by mr W last updated on 24/Dec/19

	$y = \sqrt{- 1 + \sqrt{\frac{4}{x} - 3}}$ $\Rightarrow x = \frac{4}{3 + {(1 + y^{2})}^{2}}$ $\int_{0}^{1} y d x = \int_{0}^{\infty} x d y = \int_{0}^{\infty} \frac{4}{3 + {(1 + y^{2})}^{2}} d y$ $= \frac{1}{2 \sqrt{2}} {[\ln \frac{2 + y (y + \sqrt{2})}{2 + y (y - \sqrt{2})}]}_{0}^{\infty} + \frac{1}{\sqrt{6}} {[\tan^{- 1} \frac{\sqrt{2} y + 1}{\sqrt{3}} + \tan^{- 1} \frac{\sqrt{2} y - 1}{\sqrt{3}}]}_{0}^{\infty}$ $= \frac{1}{2 \sqrt{2}} (0) + \frac{1}{\sqrt{6}} (\frac{π}{2} + \frac{π}{2} - (\frac{π}{6} - \frac{π}{6}))$ $= \frac{π}{\sqrt{6}}$
	Commented by john santuy last updated on 24/Dec/19

	$t h a n k s s i r$
	Commented by mr W last updated on 06/Jun/20

	Commented by mr W last updated on 06/Jun/20

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