Let consider A=lim_(x→0) (∫_0 ^1 (Γ(t))^x dt)^(1/x) Prove that A=∫


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Question Number 75228 by ~blr237~ last updated on 08/Dec/19

$Let consider$ $A = \lim_{x \to 0} {(\int_{0}^{1} {(Γ (t))}^{x} dt)}^{\frac{1}{x}}$ $Prove that A = \int_{0}^{1} \ln (Γ (t)) dt$ $Deduce the value of A$
Commented by mind is power last updated on 08/Dec/19
$ln{(∫_0 ^1 Γ^x (t)dt)^(1/x) }=g(x) =((ln(∫_0 ^1 Γ^x (t)dt))/x)=g(x) let h(x)=ln(∫_0 ^1 Γ^x (t)dt) g(x)=((h(x))/x)=((h(x)−h(0))/(x−0)) lim_(x→0) g(x)=h′(x)∣_(x=0) h′(x)=((∫_0 ^1 ln{Γ(t)}.Γ^x (t))/(∫_0 ^1 Γ^x (t)dt))dt x=0,give us ∫_0 ^1 ln(Γ(t))dt ∫_0 ^1 ln(Γ(t))dt=∫_0 ^1 ln(Γ(1−t))dt ⇒2A=∫_0 ^1 {ln(Γ(t))+ln(Γ(1−t))}dt =∫_0 ^1 ln(Γ(t)Γ(1−t))dt Γ(t).Γ(1−t)=(π/(sin(πt))) =∫_0 ^1 {ln(π)−ln(sin(πt))}dt =ln(π)−∫_0 ^1 ln(sin(πt))dt u=πt⇒∫_0 ^1 ln(sin(πt))dt=((∫_0 ^π ln(sin(u))du)/π) ∫_0 ^π ln(sin(u))du=2∫_0 ^(π/2) ln(sin(u))du ∫_0 ^(π/2) ln(sin(2u))du=(π/2)ln(2)+2∫_0 ^(π/2) ln(sin(u))du =∫_0 ^π ln(sin(w))(dw/2)=∫_0 ^(π/2) ln(sin(w))dw=2∫_0 ^(π/2) ln(sin(y))dy+(π/2)ln(2) ⇒∫_0 ^(π/2) ln(sin(w))dw=−(π/2)ln(2) 2A=ln(π)−2.−(1/2)ln(2) 2A=ln(2π)⇒A=((ln(2π))/2)$
$\ln {{(\int_{0}^{1} Γ^{x} (t) dt)}^{\frac{1}{x}}} = g (x)$ $= \frac{\ln (\int_{0}^{1} Γ^{x} (t) dt)}{x} = g (x)$ $let h (x) = \ln (\int_{0}^{1} Γ^{x} (t) dt)$ $g (x) = \frac{h (x)}{x} = \frac{h (x) - h (0)}{x - 0}$ $\lim_{x \to 0} g (x) = h^{'} (x) ∣_{x = 0}$ $h^{'} (x) = \frac{\int_{0}^{1} \ln {Γ (t)} . Γ^{x} (t)}{\int_{0}^{1} Γ^{x} (t) dt} dt$ $x = 0, give us \int_{0}^{1} \ln (Γ (t)) dt$ $\int_{0}^{1} \ln (Γ (t)) dt = \int_{0}^{1} \ln (Γ (1 - t)) dt$ $\Rightarrow 2 A = \int_{0}^{1} {\ln (Γ (t)) + \ln (Γ (1 - t))} dt$ $= \int_{0}^{1} \ln (Γ (t) Γ (1 - t)) dt$ $Γ (t) . Γ (1 - t) = \frac{π}{\sin (π t)}$ $= \int_{0}^{1} {\ln (π) - \ln (\sin (π t))} dt$ $= \ln (π) - \int_{0}^{1} \ln (\sin (π t)) dt$ $u = π t \Rightarrow \int_{0}^{1} \ln (\sin (π t)) dt = \frac{\int_{0}^{π} \ln (\sin (u)) du}{π}$ $\int_{0}^{π} \ln (\sin (u)) du = 2 \int_{0}^{\frac{π}{2}} \ln (\sin (u)) du$ $\int_{0}^{\frac{π}{2}} \ln (\sin (2 u)) du = \frac{π}{2} \ln (2) + 2 \int_{0}^{\frac{π}{2}} \ln (\sin (u)) du$ $= \int_{0}^{π} \ln (\sin (w)) \frac{dw}{2} = \int_{0}^{\frac{π}{2}} \ln (\sin (w)) dw = 2 \int_{0}^{\frac{π}{2}} \ln (\sin (y)) dy + \frac{π}{2} \ln (2)$ $\Rightarrow \int_{0}^{\frac{π}{2}} \ln (\sin (w)) dw = - \frac{π}{2} \ln (2)$ $2 A = \ln (π) - 2 . - \frac{1}{2} \ln (2)$ $2 A = \ln (2 π) \Rightarrow A = \frac{\ln (2 π)}{2}$
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