Let-consider-A-lim-x-0-0-1-t-x-dt-1-x-Prove-that-A-0-1-ln-t-dt-Deduce-the-value-of-A-

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 {{(\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}