find-0-1-1-x-ln-2-x-dx-

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} \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}) \dots

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