φ=∫_1 ^( ∞) ((x^4 −2x^2 +2)/(x(√(x^2 −1))))dx=?


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Question Number 127929 by mnjuly1970 last updated on 03/Jan/21

$ϕ = \int_{1}^{\infty} \frac{x^{4} - 2 x^{2} + 2}{x \sqrt{x^{2} - 1}} d x = ?$
	Answered by Ar Brandon last updated on 03/Jan/21
	$∅=∫_1 ^∞ ((x^4 −2x^2 +2)/(x(√(x^2 −1))))dx=∫_1 ^∞ {(((x^2 −1)^2 )/(x(√(x^2 −1))))+(1/(x(√(x^2 −1))))}dx =∫_1 ^∞ {(((x^2 −1)^(3/2) )/x)+(((x^2 −1)^(−(1/2)) )/x)}dx x^2 =u ⇒2xdx=du =(1/2)∫_1 ^∞ {(((u−1)^(3/2) )/u)+(((u−1)^(−(1/2)) )/u)}du (1/u)=v ⇒−(du/u^2 )=dv ⇒du=−(dv/v^2 ) ∅=(1/2)∫_0 ^1 {v((1/v)−1)^(3/2) +v((1/v)−1)^(−(1/2)) }(dv/v^2 ) =(1/2)∫_0 ^1 {v^(−(5/2)) (1−v)^(3/2) +v^(−(1/2)) (1−v)^(−(1/2)) }dv =(1/2){β(−(3/2),(5/2))+β((1/2),(1/2))} But for β(m,n), n,m>0... Maybe some error occured. Please check.$
	$\emptyset = \int_{1}^{\infty} \frac{x^{4} - {2 x}^{2} + 2}{x \sqrt{x^{2} - 1}} dx = \int_{1}^{\infty} {\frac{{(x^{2} - 1)}^{2}}{x \sqrt{x^{2} - 1}} + \frac{1}{x \sqrt{x^{2} - 1}}} dx$ $= \int_{1}^{\infty} {\frac{{(x^{2} - 1)}^{\frac{3}{2}}}{x} + \frac{{(x^{2} - 1)}^{- \frac{1}{2}}}{x}} dx$ $x^{2} = u \Rightarrow 2 xdx = du$ $= \frac{1}{2} \int_{1}^{\infty} {\frac{{(u - 1)}^{\frac{3}{2}}}{u} + \frac{{(u - 1)}^{- \frac{1}{2}}}{u}} du$ $\frac{1}{u} = v \Rightarrow - \frac{du}{u^{2}} = dv \Rightarrow du = - \frac{dv}{v^{2}}$ $\emptyset = \frac{1}{2} \int_{0}^{1} {v {(\frac{1}{v} - 1)}^{\frac{3}{2}} + v {(\frac{1}{v} - 1)}^{- \frac{1}{2}}} \frac{dv}{v^{2}}$ $= \frac{1}{2} \int_{0}^{1} {v^{- \frac{5}{2}} {(1 - v)}^{\frac{3}{2}} + v^{- \frac{1}{2}} {(1 - v)}^{- \frac{1}{2}}} dv$ $= \frac{1}{2} {β (- \frac{3}{2}, \frac{5}{2}) + β (\frac{1}{2}, \frac{1}{2})}$ $But for β (m, n), n, m > 0 . . .$ $Maybe some error occured . Please check .$
	Commented by Ar Brandon last updated on 03/Jan/21

	$\emptyset = \frac{1}{2} {β (- \frac{3}{2}, \frac{5}{2}) + \frac{Γ (\frac{1}{2}) Γ (\frac{1}{2})}{Γ (1)}}$ $= \frac{1}{2} β (- \frac{3}{2}, \frac{5}{2}) + \frac{π}{2}$
	Commented by mnjuly1970 last updated on 03/Jan/21

	$t h a n k y o u s o m u v h$ $y o u a r e r i g h t$ $i n t e g r a l i s d i v e r g e n t . .$ $g r a t e f u l . . .$
	Commented by Ar Brandon last updated on 03/Jan/21

	$My pleasure, Sir$
	Answered by Dwaipayan Shikari last updated on 03/Jan/21

	$\int_{1}^{\infty} \frac{{(x^{2} - 1)}^{\frac{3}{2}}}{x} + \frac{{(x^{2} - 1)}^{- \frac{1}{2}}}{x} d x x^{2} - 1 = j$ $= \frac{1}{2} \int_{0}^{\infty} \frac{j^{\frac{3}{2}}}{j + 1} + \frac{1}{2} \int_{0}^{\infty} \frac{j^{\frac{- 1}{2}}}{j + 1} d j \frac{j}{j + 1} = ζ \Rightarrow \frac{1}{{(j + 1)}^{2}} = \frac{d ζ}{d j}$ $= \frac{1}{2} \int_{0}^{1} {(\frac{ζ}{1 - ζ})}^{\frac{3}{2}} {(1 - ζ)}^{- 1} + {(\frac{ζ}{1 - ζ})}^{- \frac{1}{2}} {(1 - ζ)}^{- 1} d ζ$ $= \frac{1}{2} . \frac{Γ (- \frac{3}{2}) Γ (\frac{5}{2})}{Γ (1)} + \frac{1}{2} . Γ^{2} (\frac{1}{2}) = \frac{3 \sqrt{π}}{8} Γ (- \frac{3}{2}) + \frac{π}{2}$ $I n t e g r a l i s D i v e r g e n t . . .$
	Commented by Ar Brandon last updated on 03/Jan/21

	$cool bro$ 😃
	Commented by Dwaipayan Shikari last updated on 03/Jan/21

	$\int_{0}^{\infty} \frac{j^{n}}{j + 1} d j i s c o n v e r g e n t w h e n 0 < n < - 1$
	Commented by Dwaipayan Shikari last updated on 03/Jan/21

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