Ω := ∫_0 ^( (π/2)) (√(sin(x))) ln(sin( x ))dx=? m.n..


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Question Number 156028 by mnjuly1970 last updated on 07/Oct/21

$Ω := \int_{0}^{\frac{π}{2}} \sqrt{s i n (x)} \ln (s i n (x)) d x = ?$ $m . n . .$
	Answered by Ar Brandon last updated on 07/Oct/21
	$f(α)=∫_0 ^(π/2) sin^α xdx=(1/2)β(((α+1)/2), (1/2))=((Γ(((α+1)/2))(√π))/(2Γ((α/2)+1))) f ′(α)=∫_0 ^(π/2) sin^α xln(sinx)dx=((√π)/2){((Γ((α/2)+1)Γ′(((α+1)/2))−Γ(((α+1)/2))Γ′((α/2)+1))/(Γ^2 ((α/2)+1)))} f ′((1/2))=∫_0 ^(π/2) (√(sinx))ln(sinx)dx=((√π)/2){((Γ((5/4))Γ′((3/4))−Γ((3/4))Γ′((5/4)))/(Γ^2 ((5/4))))} =((√π)/2){(((1/4)Γ((1/4))Γ((3/4))[ψ((3/4))−ψ((5/4))])/((1/(16))Γ^2 ((1/4))))} ...$
	$f (α) = \int_{0}^{\frac{π}{2}} \sin^{α} x d x = \frac{1}{2} β (\frac{α + 1}{2}, \frac{1}{2}) = \frac{Γ (\frac{α + 1}{2}) \sqrt{π}}{2 Γ (\frac{α}{2} + 1)}$ $f^{'} (α) = \int_{0}^{\frac{π}{2}} \sin^{α} x \ln (\sin x) d x = \frac{\sqrt{π}}{2} {\frac{Γ (\frac{α}{2} + 1) Γ^{'} (\frac{α + 1}{2}) - Γ (\frac{α + 1}{2}) Γ^{'} (\frac{α}{2} + 1)}{Γ^{2} (\frac{α}{2} + 1)}}$ $f^{'} (\frac{1}{2}) = \int_{0}^{\frac{π}{2}} \sqrt{\sin x} \ln (\sin x) d x = \frac{\sqrt{π}}{2} {\frac{Γ (\frac{5}{4}) Γ^{'} (\frac{3}{4}) - Γ (\frac{3}{4}) Γ^{'} (\frac{5}{4})}{Γ^{2} (\frac{5}{4})}}$ $= \frac{\sqrt{π}}{2} {\frac{\frac{1}{4} Γ (\frac{1}{4}) Γ (\frac{3}{4}) [ψ (\frac{3}{4}) - ψ (\frac{5}{4})]}{\frac{1}{16} Γ^{2} (\frac{1}{4})}}$ $. . .$
	Commented by Ar Brandon last updated on 07/Oct/21

	$Γ (x) Γ (1 - x) = \frac{π}{\sin (π x)}$ $ψ (s + 1) = \frac{1}{s} + ψ (s)$ $ψ (x) - ψ (1 - x) = - π \cot (π x), x = \frac{1}{4}$
	Commented by mnjuly1970 last updated on 07/Oct/21

	$g r a t e f u l m r b r a n d o n . .$
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