show that ∫_0 ^1 lnΓ(x)dx=ln(√(2π))


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Question Number 111620 by mathdave last updated on 04/Sep/20

$s h o w t h a t$ $\int_{0}^{1} \ln Γ (x) d x = \ln \sqrt{2 π}$
	Answered by mathmax by abdo last updated on 04/Sep/20
	$we have Γ(x).Γ(1−x) =(π/(sin(πx))) (compliment formulae) with o<x<1 ⇒ln(Γ(x)+lnΓ(1−x) =ln(π)−ln(sin(πx)) ⇒ ∫_0 ^1 lnΓ(x)dx+∫_0 ^1 lnΓ(1−x)dx =ln(π)−∫_0 ^1 ln(sin(πx)dx but ∫_0 ^1 lnΓ(1−x)dx =_(1−x=t) ∫_0 ^1 lnΓ(t)dt ⇒ 2∫_0 ^1 lnΓ(x)dx =ln(π)−∫_0 ^1 ln(sin(πx)dx we have ∫_0 ^1 ln(sin(πx))dx =_(πx=u) (1/π)∫_0 ^π ln(sinu) du =(1/π){ ∫_0 ^(π/2) ln(sinu)du +∫_(π/2) ^π ln(sinu)du (→u =(π/2)+t)} =(1/π){−(π/2)ln(2) +∫_0 ^(π/2) ln(cost)dt} =(1/π){−(π/2)ln(2)−(π/2)ln(2)} =−ln2 ⇒2∫_0 ^1 lnΓ(x)dx =ln(π)+ln(2) =ln(2π) ⇒ ∫_0 ^1 lnΓ(x)dx =(1/2)ln(2π) =ln((√(2π))).$
	$we have Γ (x) . Γ (1 - x) = \frac{π}{\sin (π x)} (compliment formulae)$ $with o < x < 1 \Rightarrow \ln (Γ (x) + \ln Γ (1 - x) = \ln (π) - \ln (\sin (π x)) \Rightarrow$ $\int_{0}^{1} \ln Γ (x) dx + \int_{0}^{1} \ln Γ (1 - x) dx = \ln (π) - \int_{0}^{1} \ln (\sin (π x) dx but$ $\int_{0}^{1} \ln Γ (1 - x) dx =_{1 - x = t} \int_{0}^{1} \ln Γ (t) dt \Rightarrow$ $2 \int_{0}^{1} \ln Γ (x) dx = \ln (π) - \int_{0}^{1} \ln (\sin (π x) dx we have$ $\int_{0}^{1} \ln (\sin (π x)) dx =_{π x = u} \frac{1}{π} \int_{0}^{π} \ln (sinu) du$ $= \frac{1}{π} {\int_{0}^{\frac{π}{2}} \ln (sinu) du + \int_{\frac{π}{2}}^{π} \ln (sinu) du (\to u = \frac{π}{2} + t)}$ $= \frac{1}{π} {- \frac{π}{2} \ln (2) + \int_{0}^{\frac{π}{2}} \ln (cost) dt} = \frac{1}{π} {- \frac{π}{2} \ln (2) - \frac{π}{2} \ln (2)}$ $= - \ln 2 \Rightarrow 2 \int_{0}^{1} \ln Γ (x) dx = \ln (π) + \ln (2) = \ln (2 π) \Rightarrow$ $\int_{0}^{1} \ln Γ (x) dx = \frac{1}{2} \ln (2 π) = \ln (\sqrt{2 π}) .$
	Commented by mnjuly1970 last updated on 06/Sep/20

	$t h a n k y o u s o m u c h^{'} {s i r}^{'} f o r$ $y o u r e f f o r t . . .$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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