prove Ω=∫_0 ^( 1) (( x − x^( 2) )/((1+x )ln(x))) dx = ln((4/π) ) −−−−−


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Question Number 165152 by mnjuly1970 last updated on 26/Jan/22

$p r o v e$ $Ω = \int_{0}^{1} \frac{x - x^{2}}{(1 + x) l n (x)} d x = l n (\frac{4}{π})$ $- - - - -$
	Answered by mindispower last updated on 26/Jan/22
	$f(a)=∫_0 ^1 ((x−x^(a+1) )/((1+x)ln(x)))dx,f(0)=0 Ω must bee <0 ln((4/π))>0 f′(a)=∫_0 ^1 ((−x^(1+a) )/(1+x)) =∫_0 ^1 ((x^(2+a) −x^(1+a) )/(1−x^2 ))dx =(1/2)∫_0 ^1 ((t^((a+1)/2) −t^(a/2) )/(1−t)) .dt recall Ψ(z+1)=−γ+∫_0 ^1 ((1−x^z )/(1−x))dx f^′ (a)=(1/2){Ψ((a/2)+1)−Ψ(((a+3)/2))} f(a)=ln(((Γ(((a+2)/2)))/(Γ(((a+3)/2)))))+c f(0)=0⇒c=−ln((2/( (√π))))=ln(((√π)/2)) Ω=f(1)=ln((((√π)/2)/1))=ln(((√π)/2))+ln(((√π)/2))=ln((π/4)) Ω=ln((π/4))$
	$f (a) = \int_{0}^{1} \frac{x - x^{a + 1}}{(1 + x) l n (x)} d x, f (0) = 0$ $Ω m u s t b e e < 0 l n (\frac{4}{π}) > 0$ $f^{'} (a) = \int_{0}^{1} \frac{- x^{1 + a}}{1 + x}$ $= \int_{0}^{1} \frac{x^{2 + a} - x^{1 + a}}{1 - x^{2}} d x$ $= \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{a + 1}{2}} - t^{\frac{a}{2}}}{1 - t} . d t$ $r e c a l l Ψ (z + 1) = - γ + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} d x$ $f^{^{'}} (a) = \frac{1}{2} {Ψ (\frac{a}{2} + 1) - Ψ (\frac{a + 3}{2})}$ $f (a) = l n (\frac{Γ (\frac{a + 2}{2})}{Γ (\frac{a + 3}{2})}) + c$ $f (0) = 0 \Rightarrow c = - l n (\frac{2}{\sqrt{π}}) = l n (\frac{\sqrt{π}}{2})$ $Ω = f (1) = l n (\frac{\frac{\sqrt{π}}{2}}{1}) = l n (\frac{\sqrt{π}}{2}) + l n (\frac{\sqrt{π}}{2}) = l n (\frac{π}{4})$ $Ω = l n (\frac{π}{4})$
	Commented by mnjuly1970 last updated on 27/Jan/22

	$t h a n k s a l o t s i r p o w e r . . g r a t e f u l$
	Commented by mindispower last updated on 27/Jan/22

	$w t h e P l e a s u r H a v e a n i c e D a y$
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