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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

$set y = \sqrt{- 1 + \sqrt{\frac{4}{x} - 3}}$ $x = \frac{4}{3 + {(1 + y^{2})}^{2}}$ $\underset{0}{\int^{1}} y d x = \underset{\infty}{\int^{0}} x d y = \underset{\infty}{\int^{0}} (\frac{4}{3 + {(1 + y^{2})}^{2}}) d y$ $= \frac{1}{2 \sqrt{2}} [\ln {(\frac{2 + y (y + \sqrt{2})}{2 + y (y - \sqrt{2})}]}_{\infty}^{0} +$ $\frac{1}{\sqrt{6}} {[\tan^{- 1} (\frac{y \sqrt{2} + 1}{\sqrt{3}}) + \tan^{- 1} (\frac{y \sqrt{2} - 1}{\sqrt{3}})]}_{\infty}^{0}$ $= \frac{π}{\sqrt{6}}$
	Answered by mathmax by abdo last updated on 02/Jun/20
	$I =∫_0 ^1 (√((√((4/x)−3))−1))dx changement (√((4/x)−3))=t give (4/x)−3 =t^(2 ) ⇒ (4/x) =t^2 +3 ⇒(x/4) =(1/(t^2 +3)) ⇒x =(4/(t^2 +3)) ⇒dx =−((4×(2t))/((t^2 +3)^2 ))dt =−((8t)/((t^2 +3)^2 ))dt ⇒ I =−∫_1 ^(+∞) (√(t−1))(((−8t)/((t^2 +3)^2 )))dt =8 ∫_1 ^∞ ((t(√(t−1)))/((t^2 +3)^2 ))dt =_((√(t−1))=u) 8 ∫_0 ^∞ (((1+u^2 )u)/(((1+u^2 )^2 +3)^2 ))(2u)du =16 ∫_0 ^∞ ((u^2 (1+u^2 ))/(((u^2 +1)^2 +3)^2 ))du =8 ∫_(−∞) ^(+∞) ((u^2 (1+u^2 ))/(((u^2 +1)^2 +3)^2 ))du let ϕ(z) =((z^2 (1+z^2 ))/({(z^2 +1)^2 +3}^2 )) ⇒ ϕ(z) =((z^2 (1+z^2 ))/((z^2 +1−i(√3))^2 (z^2 +1+i(√3))^2 )) poles of ϕ? z^2 +1−i(√3)=z^2 −(−1+i(√3)) we have −1+i(√3)=2(−(1/2) +((i(√3))/2)) =2e^((i2π)/3) ⇒z^2 +1−i(√3)=z^2 −2 e^((i2π)/3) =(z−(√2)e^((iπ)/3) )(z+(√2)e^((iπ)/3) ) z^2 +1+i(√3)=z^2 −(−1−i(√3)) =z^2 −2e^(−((i2π)/3)) =(z−(√2)e^(−((iπ)/3)) )(z+(√2)e^(−((iπ)/3)) ) ⇒ϕ(z) =((z^2 (1+z^2 ))/((z−(√2)e^((iπ)/3) )^2 (z+(√2)e^((iπ)/3) )^2 (z−(√2)e^(−((iπ)/3)) )^2 (z+(√2)e^(−((iπ)/3)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√2)e^((iπ)/3) ) +Res(ϕ,−(√2)e^(−((iπ)/3)) )} Res(ϕ,(√2)e^((iπ)/3) ) =lim_(z→(√2)e^((iπ)/3) ) (1/((2−1)!)) {(z−(√2)e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→(√2)e^((iπ)/3) ) {((z^2 +z^4 )/((z+(√2)e^((iπ)/3) )^2 (z^2 +1+i(√3))^2 ))}^((1)) ...be continued....$
	$I = \int_{0}^{1} \sqrt{\sqrt{\frac{4}{x} - 3} - 1} dx changement \sqrt{\frac{4}{x} - 3} = t give \frac{4}{x} - 3 = t^{2} \Rightarrow$ $\frac{4}{x} = t^{2} + 3 \Rightarrow \frac{x}{4} = \frac{1}{t^{2} + 3} \Rightarrow x = \frac{4}{t^{2} + 3} \Rightarrow dx = - \frac{4 \times (2 t)}{{(t^{2} + 3)}^{2}} dt = - \frac{8 t}{{(t^{2} + 3)}^{2}} dt \Rightarrow$ $I = - \int_{1}^{+ \infty} \sqrt{t - 1} (\frac{- 8 t}{{(t^{2} + 3)}^{2}}) dt = 8 \int_{1}^{\infty} \frac{t \sqrt{t - 1}}{{(t^{2} + 3)}^{2}} dt$ $=_{\sqrt{t - 1} = u} 8 \int_{0}^{\infty} \frac{(1 + u^{2}) u}{{({(1 + u^{2})}^{2} + 3)}^{2}} (2 u) du = 16 \int_{0}^{\infty} \frac{u^{2} (1 + u^{2})}{{({(u^{2} + 1)}^{2} + 3)}^{2}} du$ $= 8 \int_{- \infty}^{+ \infty} \frac{u^{2} (1 + u^{2})}{{({(u^{2} + 1)}^{2} + 3)}^{2}} du let φ (z) = \frac{z^{2} (1 + z^{2})}{{{(z^{2} + 1)}^{2} + 3}^{2}} \Rightarrow$ $φ (z) = \frac{z^{2} (1 + z^{2})}{{(z^{2} + 1 - i \sqrt{3})}^{2} {(z^{2} + 1 + i \sqrt{3})}^{2}} poles of φ ?$ $z^{2} + 1 - i \sqrt{3} = z^{2} - (- 1 + i \sqrt{3}) we have - 1 + i \sqrt{3} = 2 (- \frac{1}{2} + \frac{i \sqrt{3}}{2}) = {2 e}^{\frac{i 2 π}{3}}$ $\Rightarrow z^{2} + 1 - i \sqrt{3} = z^{2} - 2 e^{\frac{i 2 π}{3}} = (z - \sqrt{2} e^{\frac{i π}{3}}) (z + \sqrt{2} e^{\frac{i π}{3}})$ $z^{2} + 1 + i \sqrt{3} = z^{2} - (- 1 - i \sqrt{3}) = z^{2} - {2 e}^{- \frac{i 2 π}{3}} = (z - \sqrt{2} e^{- \frac{i π}{3}}) (z + \sqrt{2} e^{- \frac{i π}{3}})$ $\Rightarrow φ (z) = \frac{z^{2} (1 + z^{2})}{{(z - \sqrt{2} e^{\frac{i π}{3}})}^{2} {(z + \sqrt{2} e^{\frac{i π}{3}})}^{2} {(z - \sqrt{2} e^{- \frac{i π}{3}})}^{2} (z + \sqrt{2} e^{- \frac{i π}{3}})}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, \sqrt{2} e^{\frac{i π}{3}}) + Res (φ, - \sqrt{2} e^{- \frac{i π}{3}})}$ $Res (φ, \sqrt{2} e^{\frac{i π}{3}}) = \lim_{z \to \sqrt{2} e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - \sqrt{2} e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to \sqrt{2} e^{\frac{i π}{3}}} {\frac{z^{2} + z^{4}}{{(z + \sqrt{2} e^{\frac{i π}{3}})}^{2} {(z^{2} + 1 + i \sqrt{3})}^{2}}}^{(1)} . . . be continued . . . .$
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