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Question Number 145602 by mnjuly1970 last updated on 06/Jul/21

$. . . . . Advanced . . . . . . . . . Calculus . . . . .$ $Q :: Find the value of ::$ $\begin{array}{cc} i :: ϕ := \int_{0}^{1} Ln (Γ (2 + x)) d x = ? \\ i i :: Ω := \underset{n = 1}{\sum^{\infty}} \frac{1}{n (2 n + 3)} = ? \end{array}$ $. . . . . m . n . j u l y . 1970 . . . . . ◼$
	Answered by ajfour last updated on 06/Jul/21
	$(ii) S=Σ_(n=1) ^∞ (1/(n(2n+3))) ((3S)/2)=Σ_(n=1) ^∞ ((2n+3−2n)/(2n(2n+3))) =Σ_(n=1) ^∞ ((1/(2n))−(1/(2n+3))) ln (1−x)=−(x+(x^2 /2)+(x^3 /3)+..) Σ_(n=1) ^∞ (1/(2n))=−(1/2)lim_(x→1) ln (1−x) ln (1+x)−ln (1−x) =2(x+(x^3 /3)+(x^5 /5)+...) lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] =Σ_(n=1) ^∞ (1/(2n+3)) ((3S)/2)=−(1/2)lim ln (1−x)− lim_(x→1) [(1/2){ln (1+x)−ln (1−x)} −x−(x^3 /3)] ((3S)/2)= (4/3)−((ln 2)/2) S=(8/9)−((ln 2)/3)$
	$(i i)$ $S = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (2 n + 3)}$ $\frac{3 S}{2} = \underset{n = 1}{\sum^{\infty}} \frac{2 n + 3 - 2 n}{2 n (2 n + 3)}$ $= \underset{n = 1}{\sum^{\infty}} (\frac{1}{2 n} - \frac{1}{2 n + 3})$ $\ln (1 - x) = - (x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + . .)$ $\underset{n = 1}{\sum^{\infty}} \frac{1}{2 n} = - \frac{1}{2} \lim_{x \to 1} \ln (1 - x)$ $\ln (1 + x) - \ln (1 - x)$ $= 2 (x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + . . .)$ $\lim_{x \to 1} [\frac{1}{2} {\ln (1 + x) - \ln (1 - x)}$ $- x - \frac{x^{3}}{3}] = \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n + 3}$ $\frac{3 S}{2} = - \frac{1}{2} \lim \ln (1 - x) -$ $\lim_{x \to 1} [\frac{1}{2} {\ln (1 + x) - \ln (1 - x)}$ $- x - \frac{x^{3}}{3}]$ $\frac{3 S}{2} = \frac{4}{3} - \frac{\ln 2}{2}$ $S = \frac{8}{9} - \frac{\ln 2}{3}$
	Commented by mnjuly1970 last updated on 06/Jul/21

	$b r a v o m r a j f o r y o u r w o r k$ $i s a d m i r a b l e . . .$
	Answered by Olaf_Thorendsen last updated on 06/Jul/21

	$Ω = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (2 n + 3)} = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + 1) (n + \frac{5}{2})}$ $Ω = \frac{1}{2} . \frac{ψ (\frac{5}{2}) - ψ (1)}{\frac{5}{2} - 1} = \frac{1}{3} (ψ (\frac{5}{2}) + γ)$ $ψ (\frac{5}{2}) = ψ (1 + \frac{3}{2}) = ψ (\frac{3}{2}) + \frac{1}{3 / 2} = ψ (\frac{3}{2}) + \frac{2}{3}$ $ψ (\frac{3}{2}) = ψ (1 + \frac{1}{2}) = ψ (\frac{1}{2}) + \frac{1}{1 / 2} = - 2 \ln (2) - γ + 2$ $Ω = \frac{1}{3} (- 2 \ln (2) - γ + 2 + \frac{2}{3} + γ) = \frac{2}{3} (\frac{4}{3} - \ln (2))$
	Commented by mnjuly1970 last updated on 06/Jul/21

	$g r a t e f u l m r o l a f$ $v e r y n i c e a s a l w a y s . . m e r c e y$
	Answered by qaz last updated on 06/Jul/21

	$Γ (x) Γ (1 - x) = \frac{π}{\sin π x}$
	Answered by Kamel last updated on 06/Jul/21

	$i) u s e t = 1 - x a n d : Γ (t) Γ (1 - t) = \frac{π}{s i n (π t)}$ $i i . . Ω = \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n (2 n + 3)} = \frac{2}{3} \underset{n = 1}{\sum^{+ \infty}} (\frac{1}{2 n} - \frac{1}{2 n + 3}) = \frac{2}{3} (\underset{n = 1}{\sum^{+ \infty}} (\frac{1}{2 n} - \frac{1}{2 n + 1}) + \underset{n = 1}{\sum^{+ \infty}} (\frac{1}{2 n + 1} - \frac{1}{2 (n + 1) + 1}))$ $= \frac{2}{3} (\underset{n = 1}{\sum^{+ \infty}} \frac{{(- 1)}^{n}}{n} + 1 + \frac{1}{3}) = \frac{2}{3} (\frac{4}{3} - L n (2)) = \frac{8}{9} - \frac{2 L n (2)}{3}$
	Commented by mnjuly1970 last updated on 06/Jul/21

	$t a s h a k o r m r k a m e l$
	Answered by ArielVyny last updated on 06/Jul/21

	$\emptyset = \int_{0}^{1} l n (Γ (2 + x)) d x$ $= \int_{0}^{1} l n ((x + 1)!) d x$ $x + 1 = t \to d t = d x$ $\emptyset = \int_{1}^{2} l n (t!) d t = \int_{1}^{2} l n [{(\frac{t}{e})}^{t} \sqrt{2 π t}]$ $\int_{1}^{2} l n {(\frac{t}{e})}^{t} + \int_{1}^{2} l n (\sqrt{2 π t}) d t$ $\int_{1}^{2} l n (e^{t l n (\frac{t}{e})}) + \int_{1}^{2} l n (\sqrt{2 π}) d t + \frac{1}{2} \int_{1}^{2} l n (t) d t$ $\int_{1}^{2} t l n (\frac{t}{e}) + l n (\sqrt{2 π}) + \frac{1}{2} {[x l n (x) - x]}_{1}^{2}$ $\int_{1}^{2} t l n (t) - \int_{1}^{2} t d t + l n (\sqrt{2 π}) + \frac{1}{2} (2 l n (2) + 1)$ $\int_{1}^{2} t l n (t) d t - \frac{3}{2} + \frac{1}{2} l n (2 π) + \frac{2 l n 2}{2} + \frac{1}{2}$ $\int_{1}^{2} t l n (t) d t$ $d u = l n (t) v = t$ $u = t l n (t) - t d v = 1$ $\int_{1}^{2} t l n (t) = [t^{2} l n (t) - t] - \int t l n t d t + \int t d t$ $2 \int_{1}^{2} t l n (t) d t = {[t^{2} l n (t) - t + \frac{1}{2} t^{2}]}_{1}^{2} = 4 l n (2) - 2 + \frac{1}{2} 4 + 1 - \frac{1}{2}$ $\int_{1}^{2} t l n (t) d t = 2 l n (2) - 1 + 1 + \frac{1}{2} - \frac{1}{4}$ $= 2 l n (2) + \frac{1}{4}$ $\emptyset = 2 l n (2) + \frac{1}{4} - \frac{3}{2} + \frac{l n (2 π)}{2} + \frac{2 l n (2)}{2} + \frac{1}{2}$ $\emptyset = 2 l n (2) + \frac{1}{4} - 1 + \frac{l n (8 π)}{2}$ $\emptyset = - \frac{3}{4} + \frac{2 l n (4)}{2} + \frac{l n (8 π)}{2}$ $\emptyset = - \frac{3}{4} + \frac{1}{2} l n (8^{3} π)$
	Commented by mathmax by abdo last updated on 06/Jul/21

	$t! \sim {(\frac{t}{e})}^{t} \sqrt{2 π t} is valable pour t assez grand (t \to \infty)$
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