...nice calculus.. evaluate : 2∫_1 ^( ∞) ((({x}−(1/2))/x))dx−∫_0 ^( 1) ln(Γ(x))dx=??? {x}: fractional part...


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Question Number 124887 by mnjuly1970 last updated on 06/Dec/20
$...nice calculus.. evaluate : 2∫_1 ^( ∞) ((({x}−(1/2))/x))dx−∫_0 ^( 1) ln(Γ(x))dx=??? {x}: fractional part...$
$. . . n i c e c a l c u l u s . .$ $e v a l u a t e :$ $2 \int_{1}^{\infty} (\frac{{x} - \frac{1}{2}}{x}) d x - \int_{0}^{1} l n (Γ (x)) d x = ? ? ?$ ${x} : f r a c t i o n a l p a r t . . .$
	Answered by Dwaipayan Shikari last updated on 06/Dec/20
	$∫_1 ^∞ ((({x}−(1/2))/x))dx =(1/2)log(2π)−1 (Proved previously) ∫_0 ^1 log(Γ(x))=(1/2)(∫_0 ^1 log(π)−log(sinπx))=(1/2)log(2π) 2∫_1 ^∞ ((({x}−(1/2))/x))dx −∫_0 ^1 log(Γ(x))dx=(1/2)log(2π)−2$
	$\int_{1}^{\infty} (\frac{{x} - \frac{1}{2}}{x}) d x = \frac{1}{2} l o g (2 π) - 1 (P r o v e d p r e v i o u s l y)$ $\int_{0}^{1} l o g (Γ (x)) = \frac{1}{2} (\int_{0}^{1} l o g (π) - l o g (s i n π x)) = \frac{1}{2} l o g (2 π)$ $2 \int_{1}^{\infty} (\frac{{x} - \frac{1}{2}}{x}) d x - \int_{0}^{1} l o g (Γ (x)) d x = \frac{1}{2} l o g (2 π) - 2$
	Commented by mnjuly1970 last updated on 06/Dec/20

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