solution of Φ=∫_0 ^1 xH_x dx =^(γ+ψ(x+1)) ∫_0 ^( 1) x(γ+ψ(x+1))dx =^(ψ(x+1)=(1/x)+ψ(x)) ∫_0 ^( 1) x(γ+(1/x)+ψ(x))dx =(γ/2)+1+ ∫_0 ^( 1) x.(d/dx)(ln(Γ(x))) =(γ/2)+1+[xln(Γ(x))]_0 ^1 −∫_0 ^( 1) ln(Γ(x))dx we know (why?) ∫_0 ^( 1) ln(Γ(x))dx=ln((√(2π)) ) and lim_(x→0^+ ) (xln(Γ(x)))=0 (why??) finally Φ =∫_0 ^( 1) xH_x dx=(γ/2)+1−ln((√(2π))) ✓ m.n.july 1970....


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Question Number 112114 by mnjuly1970 last updated on 06/Sep/20

$s o l u t i o n o f$ $Φ = \int_{0}^{1} {x H}_{x} d x \overset{γ + ψ (x + 1)}{=} \int_{0}^{1} x (γ + ψ (x + 1)) d x$ $\overset{ψ (x + 1) = \frac{1}{x} + ψ (x)}{=} \int_{0}^{1} x (γ + \frac{1}{x} + ψ (x)) d x$ $= \frac{γ}{2} + 1 + \int_{0}^{1} x . \frac{d}{d x} (l n (Γ (x)))$ $= \frac{γ}{2} + 1 + {[x l n (Γ (x))]}_{0}^{1} - \int_{0}^{1} l n (Γ (x)) d x$ $w e k n o w (w h y ?) \int_{0}^{1} l n (Γ (x)) d x = l n (\sqrt{2 π})$ $a n d {l i m}_{x \to 0^{+}} (x l n (Γ (x))) = 0 (w h y ? ?)$ $f i n a l l y Φ = \int_{0}^{1} {x H}_{x} d x = \frac{γ}{2} + 1 - l n (\sqrt{2 π}) ✓$ $m . n . j u l y 1970 . . . .$
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