prove-that-i-0-1-x-s-1-1-x-dx-s-s-2-ln-2-ii-2x-1-2-x-1-2-x-1-2-ln-2-

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 = \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})} \dots 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) \dots . 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) \dots . 1

Commented by mnjuly1970 last updated on 12/Dec/20

m e r c e y s i r . .