∫_0 ^1 ((ln(x)ln(1−x))/( x(√(1−x))))dx


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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

	$thanx have a nice day$
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