0-1-ln-x-ln-1-x-x-1-x-dx-

Question Number 202731 by MrGHK last updated on 02/Jan/24

\int_{0}^{1} \frac{l n (x) l n (1 - x)}{x \sqrt{1 - x}} d x

Answered by witcher3 last updated on 02/Jan/24

\int_{0}^{1} t^{a - 1} {(1 - t)}^{b - 1} dt = β (a, b)

\partial_{a} . \partial_{b} β (a, b) = \int_{0}^{1} \ln (t) t^{a - 1} \ln (1 - t) t^{b - 1} dt = f (a, b)

\underset{x \to 0}{\lim f} (x, \frac{1}{2}) = \int_{0}^{1} \frac{\ln (x)}{x} . \frac{\ln (1 - x)}{\sqrt{1 - x}} dx

\partial_{b} β (a, b) = β (a, b) (Ψ (b) - Ψ (a + b))

\partial_{a} \partial_{b} β (a, b) = β (a, b) (Ψ (a) - Ψ (a + b)) (Ψ (b) - Ψ (a + b))

- Ψ^{'} (a + b) β (a, b)

I = \lim_{x \to 0} β (x, \frac{1}{2}) (Ψ (x) - Ψ (x + \frac{1}{2})) (Ψ (\frac{1}{2}) - Ψ (x + \frac{1}{2})) - Ψ^{'} (x + \frac{1}{2}))

= (Ψ (x) (Ψ (\frac{1}{2}) - Ψ (x + \frac{1}{2})) - Ψ (x + \frac{1}{2}) (Ψ (\frac{1}{2}) - Ψ (x + \frac{1}{2}) - Ψ^{'} (x + \frac{1}{2})) β (x, \frac{1}{2}) =

Ψ (x) = Ψ (1 + x) - \frac{1}{x}

Ψ (\frac{1}{2}) - Ψ (x + \frac{1}{2}) = Ψ (\frac{1}{2}) - Ψ (\frac{1}{2}) - x Ψ^{'} (\frac{1}{2}) - \frac{x^{2}}{2} Ψ^{″} (\frac{1}{2}) + o (x^{2})

= x (- Ψ^{'} (\frac{1}{2}) - \frac{x}{2} Ψ^{″} (\frac{1}{2}) + o (x))

x . (Ψ (1 + x) - \frac{1}{x} - Ψ (x + \frac{1}{2})) (- Ψ^{'} (\frac{1}{2}) - \frac{x}{2} Ψ^{″} (\frac{1}{2}) + o (x)) = A

= x Ψ (1 + x) - 1 - x Ψ (x + \frac{1}{2}) = - 1 - x (Ψ (\frac{1}{2}) - Ψ (1)) + o (x)

A = Ψ^{'} (\frac{1}{2}) + x (- \frac{Ψ^{″} (\frac{1}{2})}{2} + (Ψ (\frac{1}{2}) - Ψ (1)) Ψ^{'} (\frac{1}{2})) + o (x)

Ψ^{'} (\frac{1}{2} + x) = Ψ^{'} (\frac{1}{2}) + Ψ^{″} (\frac{1}{2}) x + o (x)

= (Ψ (x) - Ψ (x + \frac{1}{2})) (Ψ (\frac{1}{2}) - Ψ (\frac{1}{2} + x)) - Ψ^{'} (x + \frac{1}{2})

= x (- \frac{Ψ^{″} (\frac{1}{2})}{2} + (Ψ (\frac{1}{2}) - Ψ (1)) Ψ^{'} (\frac{1}{2})) + o (x)

β (x, \frac{1}{2}) \sim Γ (x) = \frac{Γ (1 + x)}{x}

I = \lim_{x \to 0} β (x, \frac{1}{2}) . x (- \frac{Ψ^{″} (\frac{1}{2})}{2} + (Ψ (\frac{1}{2}) - Ψ (1)) Ψ^{'} (\frac{1}{2}))

β (x, \frac{1}{2}) \sim Γ (x) = \frac{Γ (1 + x)}{x}

\Leftrightarrow \lim_{x \to 0} \frac{(- \frac{Ψ^{″} (\frac{1}{2})}{2} + (Ψ (\frac{1}{2}) - Ψ (1)) Ψ^{'} (\frac{1}{2}) x Γ (1 + x)}{x}

= (- \frac{Ψ^{″} (\frac{1}{2})}{2} + (Ψ (\frac{1}{2}) - Ψ (1)) Ψ^{'} (\frac{1}{2})) = \int_{0}^{1} \frac{\ln (x) \ln (1 - x)}{x \sqrt{1 - x}} dx

Ψ (1) = - γ, Ψ (\frac{1}{2}) = - γ - \log (4); Ψ^{'} (\frac{1}{2}) = \frac{π^{2}}{2};

Ψ^{″} (\frac{1}{2}) = - 14 ζ (3)

\int_{0}^{1} \frac{\ln (x) \ln (1 - x)}{x \sqrt{1 - x}} dx = (7 ζ (3) - π^{2} \ln (2)) \sim 1.57

Commented by MrGHK last updated on 03/Jan/24

nice solution sir

Commented by witcher3 last updated on 08/Jan/24

thanx have a nice day