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Question Number 75228 by ~blr237~ last updated on 08/Dec/19
Let consider A=lim_(x→0) (∫_0 ^1 (Γ(t))^x dt)^(1/x) Prove that A=∫_0 ^1 ln(Γ(t))dt Deduce the value of A
$$\:\mathrm{Let}\:\mathrm{consider}\: \\ $$$$\mathrm{A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} \left(\Gamma\left(\mathrm{t}\right)\right)^{\mathrm{x}} \mathrm{dt}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\:\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{t}\right)\right)\mathrm{dt}\:\: \\ $$$$\mathrm{Deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{A} \\ $$
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)
$$\mathrm{ln}\left\{\left(\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma^{\mathrm{x}} \left(\mathrm{t}\right)\mathrm{dt}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \right\}=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$=\frac{\mathrm{ln}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma^{\mathrm{x}} \left(\mathrm{t}\right)\mathrm{dt}\right)}{\mathrm{x}}=\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{let}\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{ln}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma^{\mathrm{x}} \left(\mathrm{t}\right)\mathrm{dt}\right) \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{h}\left(\mathrm{x}\right)}{\mathrm{x}}=\frac{\mathrm{h}\left(\mathrm{x}\right)−\mathrm{h}\left(\mathrm{0}\right)}{\mathrm{x}−\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{h}'\left(\mathrm{x}\right)\mid_{\mathrm{x}=\mathrm{0}} \\ $$$$\mathrm{h}'\left(\mathrm{x}\right)=\frac{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left\{\Gamma\left(\mathrm{t}\right)\right\}.\Gamma^{\mathrm{x}} \left(\mathrm{t}\right)}{\int_{\mathrm{0}} ^{\mathrm{1}} \Gamma^{\mathrm{x}} \left(\mathrm{t}\right)\mathrm{dt}}\mathrm{dt} \\ $$$$\mathrm{x}=\mathrm{0},\mathrm{give}\:\mathrm{us}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{t}\right)\right)\mathrm{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{1}−\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$\Rightarrow\mathrm{2A}=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{ln}\left(\Gamma\left(\mathrm{t}\right)\right)+\mathrm{ln}\left(\Gamma\left(\mathrm{1}−\mathrm{t}\right)\right)\right\}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{t}\right)\Gamma\left(\mathrm{1}−\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$\Gamma\left(\mathrm{t}\right).\Gamma\left(\mathrm{1}−\mathrm{t}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{t}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{ln}\left(\pi\right)−\mathrm{ln}\left(\mathrm{sin}\left(\pi\mathrm{t}\right)\right)\right\}\mathrm{dt} \\ $$$$=\mathrm{ln}\left(\pi\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{sin}\left(\pi\mathrm{t}\right)\right)\mathrm{dt} \\ $$$$\mathrm{u}=\pi\mathrm{t}\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{sin}\left(\pi\mathrm{t}\right)\right)\mathrm{dt}=\frac{\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}}{\pi} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{2u}\right)\right)\mathrm{du}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{u}\right)\right)\mathrm{du} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{w}\right)\right)\frac{\mathrm{dw}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{w}\right)\right)\mathrm{dw}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{y}\right)\right)\mathrm{dy}+\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}\left(\mathrm{w}\right)\right)\mathrm{dw}=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{2A}=\mathrm{ln}\left(\pi\right)−\mathrm{2}.−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{2A}=\mathrm{ln}\left(\mathrm{2}\pi\right)\Rightarrow\mathrm{A}=\frac{\mathrm{ln}\left(\mathrm{2}\pi\right)}{\mathrm{2}} \\ $$

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