find ∫_0 ^1 (√(1−(√x)))ln^2 (x)dx


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Question Number 206754 by mathzup last updated on 23/Apr/24

$f i n d \int_{0}^{1} \sqrt{1 - \sqrt{x}} {l n}^{2} (x) d x$
	Answered by Berbere last updated on 24/Apr/24
	$x=u^2 ⇒dx=2udu ∫_0 ^1 (√(1−(√x)))ln^2 (x)dx.=A=∫_0 ^1 (1−u)^(1/2) (2ln(u))^2 .2udu =8∫_0 ^1 (1−u)^(1/2) ln^2 (u).udu f(a)=∫_0 ^1 (1−u)^((3/2)−1) u^(a−1) =β(a,(3/2))du;f′′(a)=∫_0 ^1 (1−u)^(1/2) ln^2 (u).u^(a−1) du f′(a)=β(a,(3/2))(Ψ(a)−Ψ(a+(3/2))) f′′(a)=β(a,(3/2)){(Ψ(a)−Ψ(a+(3/2)))+(Ψ′(a)−Ψ′(a+(3/2))} A=8f′′(2) =8(β(2,(3/2))){(Ψ(2)−Ψ((3/2)+2))^2 +(Ψ′(2)−Ψ′(2+(3/2)))) =8.((Γ((3/2))Γ(2))/(Γ(2+(3/2)))){(1+Ψ(1)−{(2/5)+(2/3)+2+Ψ((1/2))})^2 +(ζ(2)−1((4/(25))+(4/9)+4+Ψ′((1/2))) Ψ(1)=−γ;Ψ((1/2))=−2ln(2)−γ ζ(2)=(π^2 /6) Ψ((1/2))=Σ_(n≥0) (1/(((1/2)+n)^2 ))=Σ_(n≥0) (4/((2n+1)^2 ))=4(1−(1/4))ζ(2) Γ(2+(3/2))=(1+(3/2))((3/2))Γ((3/2))...$
	$x = u^{2} \Rightarrow d x = 2 u d u$ $\int_{0}^{1} \sqrt{1 - \sqrt{x}} {l n}^{2} (x) d x . = A = \int_{0}^{1} {(1 - u)}^{\frac{1}{2}} {(2 l n (u))}^{2} . 2 u d u$ $= 8 \int_{0}^{1} {(1 - u)}^{\frac{1}{2}} {l n}^{2} (u) . u d u$ $f (a) = \int_{0}^{1} {(1 - u)}^{\frac{3}{2} - 1} u^{a - 1} = β (a, \frac{3}{2}) d u; f^{″} (a) = \int_{0}^{1} {(1 - u)}^{\frac{1}{2}} {l n}^{2} (u) . u^{a - 1} d u$ $f^{'} (a) = β (a, \frac{3}{2}) (Ψ (a) - Ψ (a + \frac{3}{2}))$ $f^{″} (a) = β (a, \frac{3}{2}) {(Ψ (a) - Ψ (a + \frac{3}{2})) + (Ψ^{'} (a) - Ψ^{'} (a + \frac{3}{2})}$ $A = 8 f^{″} (2)$ $= 8 (β (2, \frac{3}{2})) {{(Ψ (2) - Ψ (\frac{3}{2} + 2))}^{2} + (Ψ^{'} (2) - Ψ^{'} (2 + \frac{3}{2})))$ $= 8 . \frac{Γ (\frac{3}{2}) Γ (2)}{Γ (2 + \frac{3}{2})} {{(1 + Ψ (1) - {\frac{2}{5} + \frac{2}{3} + 2 + Ψ (\frac{1}{2})})}^{2} + (ζ (2) - 1 (\frac{4}{25} + \frac{4}{9} + 4 + Ψ^{'} (\frac{1}{2}))$ $Ψ (1) = - γ; Ψ (\frac{1}{2}) = - 2 l n (2) - γ$ $ζ (2) = \frac{π^{2}}{6}$ $Ψ (\frac{1}{2}) = \sum_{n ⩾ 0} \frac{1}{{(\frac{1}{2} + n)}^{2}} = \sum_{n ⩾ 0} \frac{4}{{(2 n + 1)}^{2}} = 4 (1 - \frac{1}{4}) ζ (2)$ $Γ (2 + \frac{3}{2}) = (1 + \frac{3}{2}) (\frac{3}{2}) Γ (\frac{3}{2}) . . .$
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