::::prove that : i: Ω= ∫_0 ^( 1) (x^(s−1) /(1+x))dx=ψ(s) − ψ((s/2))−ln(2) ii: ψ(2x)=(1/2)ψ(x)+(1/2)ψ(x+(1/2))+ln(2)


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Question Number 125297 by mnjuly1970 last updated on 09/Dec/20

$:::: p r o v e t h a t :$ $i : Ω = \int_{0}^{1} \frac{x^{s - 1}}{1 + x} d x = ψ (s) - ψ (\frac{s}{2}) - l n (2)$ $i i : ψ (2 x) = \frac{1}{2} ψ (x) + \frac{1}{2} ψ (x + \frac{1}{2}) + l n (2)$
	Answered by Bird last updated on 09/Dec/20

	$d e f i n e Ψ s i r m n j$
	Commented by hatakekakashi1729gmailcom last updated on 10/Dec/20

	$d i g a m m a f u n c t i o n$
	Commented by mnjuly1970 last updated on 10/Dec/20

	$ψ (s) = - γ + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + s - 1})$ $= - γ + \int_{0}^{1} (\frac{1 - t^{s - 1}}{1 - t}) d t$ $γ := e u l e r - m a s c h e r o n i c o n s t a n t$
	Commented by mnjuly1970 last updated on 10/Dec/20

	$t h a n k y o u s i r . .$
	Answered by Dwaipayan Shikari last updated on 10/Dec/20

	$\int_{0}^{1} \frac{x^{s - 1}}{1 + x} d x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{s + n - 1} = \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + s}$
	Answered by mindispower last updated on 11/Dec/20
	$Ω=∫_0 ^1 ((x^(s−1) −x^s )/(1−x^2 ))dx x^2 =r =∫_0 ^1 ((r^((s−1)/2) −r^(s/2) )/(1−r)).(1/2)r^(−(1/2)) dr =(1/2)∫_0 ^1 ((r^((s/2)−1) −1)/(1−r))dr−(1/2)∫_0 ^1 ((r^(((s+1)/2)−1) −1)/(1−r))dr =(1/2){−Ψ((s/2))−γ}−(1/2){−Ψ(((s+1)/2))−γ} =(1/2){−Ψ((s/2))+Ψ(((s+1)/2))}...E in other hand we have Γ(x)Γ(x+(1/2))=Γ(2x).(√π).2^(1−2x) ⇒ln(Γ(x))+ln(Γ(x+(1/2)))=lnΓ(2x)+((ln(π))/2)+(1−2x)ln(2) tack (d/dx),bothe sid⇒ Ψ(x)+Ψ(x+(1/2))=2Ψ(2x)−2ln(2)....2 ⇒Ψ((s/2)+1)=2Ψ(s)−2ln(2)−Ψ((s/2)) E⇔ (1/2)(−Ψ((s/2))+2Ψ(s)−2ln(2)−Ψ((s/2))) =Ψ(s)−Ψ((s/2))+ln(2)....1$
	$Ω = \int_{0}^{1} \frac{x^{s - 1} - x^{s}}{1 - x^{2}} d x$ $x^{2} = r$ $= \int_{0}^{1} \frac{r^{\frac{s - 1}{2}} - r^{\frac{s}{2}}}{1 - r} . \frac{1}{2} r^{- \frac{1}{2}} d r$ $= \frac{1}{2} \int_{0}^{1} \frac{r^{\frac{s}{2} - 1} - 1}{1 - r} d r - \frac{1}{2} \int_{0}^{1} \frac{r^{\frac{s + 1}{2} - 1} - 1}{1 - r} d r$ $= \frac{1}{2} {- Ψ (\frac{s}{2}) - γ} - \frac{1}{2} {- Ψ (\frac{s + 1}{2}) - γ}$ $= \frac{1}{2} {- Ψ (\frac{s}{2}) + Ψ (\frac{s + 1}{2})} . . . E$ $i n o t h e r h a n d w e h a v e$ $Γ (x) Γ (x + \frac{1}{2}) = Γ (2 x) . \sqrt{π} . 2^{1 - 2 x}$ $\Rightarrow l n (Γ (x)) + l n (Γ (x + \frac{1}{2})) = l n Γ (2 x) + \frac{l n (π)}{2} + (1 - 2 x) l n (2)$ $t a c k \frac{d}{d x}, b o t h e s i d \Rightarrow$ $Ψ (x) + Ψ (x + \frac{1}{2}) = 2 Ψ (2 x) - 2 l n (2) . . . . 2$ $\Rightarrow Ψ (\frac{s}{2} + 1) = 2 Ψ (s) - 2 l n (2) - Ψ (\frac{s}{2})$ $E \Leftrightarrow$ $\frac{1}{2} (- Ψ (\frac{s}{2}) + 2 Ψ (s) - 2 l n (2) - Ψ (\frac{s}{2}))$ $= Ψ (s) - Ψ (\frac{s}{2}) + l n (2) . . . . 1$
	Commented by mnjuly1970 last updated on 12/Dec/20

	$m e r c e y s i r . .$
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