prove-that-s-1-s-pi-sinspi-

Question Number 122045 by mohammad17 last updated on 13/Nov/20

p r o v e t h a t Γ s Γ 1 - s = \frac{π}{s i n s π}

Commented by Dwaipayan Shikari last updated on 14/Nov/20

Γ (s) = \frac{e^{- γ s}}{s} \prod^{\infty} e^{\frac{s}{n}} {(1 + \frac{s}{n})}^{- 1}

Γ (1 - s) = \frac{e^{- γ + γ s}}{1 - s} \prod^{\infty} e^{\frac{1 - s}{n}} {(1 + \frac{1 - s}{n})}^{- 1}

Γ (s) Γ (1 - s) = \frac{e^{- γ}}{s (1 - s)} \prod^{\infty} e^{\frac{1}{n}} {(1 + \frac{s}{n})}^{- 1} {(1 - \frac{s}{n} + \frac{1}{n})}^{- 1}

\frac{π s}{s i n π s} = \prod^{\infty} {(1 - \frac{s}{n})}^{- 1} {(1 + \frac{s}{n})}^{- 1}

Γ (s) Γ (1 - s) = \frac{e^{- γ + \sum^{\infty} \frac{1}{n}}}{s (1 - s)} . \frac{π s}{s i n π s} . \prod^{\infty} (1 - \frac{s}{n}) \prod^{\infty} {(1 - \frac{s}{n} + \frac{1}{n})}^{- 1}

Γ (s) Γ (1 - s) = \frac{e^{l o g (n)}}{(1 - s)} . \frac{π}{s i n π s} ((1 - s) (\frac{2 - s}{2}) \dots) \frac{1.2.3}{(2 - s) (3 - s) (4 - s)}

Γ (s) Γ (1 - s) = \lim_{n \to \infty} \frac{π n}{s i n π s} . \frac{1}{n - s + 1}

Γ (s) Γ (1 - s) = \frac{π}{s i n π s}

Commented by mnjuly1970 last updated on 14/Nov/20

b r a v o m r d w a i p a y a n . .

v e r y n i c e a s a l w a y s \dots

Commented by Dwaipayan Shikari last updated on 14/Nov/20

With pleasure��

Answered by mnjuly1970 last updated on 14/Nov/20

p r o o f b y u s i n g d o u b l e

i n t e g r a l .

Γ (s) = \int_{0}^{\infty} x^{s - 1} e^{- x} d x \dots (i)

Γ (1 - s) = \int_{0}^{\infty} y^{- s} e^{- y} d y \dots (i i)

f r o m (i) a n d (i i) :

Γ (s) Γ (1 - s) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{x} {(\frac{x}{y})}^{s} e^{- (x + y)} d x d y

{_{\frac{x}{y} = v}^{x + y = u} \Rightarrow {_{x = \frac{u v}{1 + v}}^{y = \frac{u}{1 + v}}

d x d y =∣ \frac{\partial (x, y)}{\partial (u, v)} ∣ d u d v w h e r e

\frac{\partial (x, y)}{\partial (u, v)} = J (u, v) = | \begin{matrix} \frac{\partial x}{\partial u} \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} \frac{\partial y}{\partial v} \end{matrix} |

∴ J (u, v) = x_{u} y_{v} - x_{v} y_{u} = (\frac{v}{1 + v}) (\frac{- u}{{(1 + v)}^{2}}) - (\frac{u}{{(1 + v)}^{2}}) (\frac{1}{1 + v})

J (u, v) = \frac{- u}{{(1 + v)}^{2}} ✓

Γ (s) Γ (1 - s) = \int_{0}^{\infty} \int_{0}^{\infty} \frac{1 + v}{u v} v^{s} e^{- u} ∣ \frac{- u}{{(1 + v)}^{2}} ∣ d u d v

= \int_{0}^{\infty} \int_{0}^{\infty} \frac{v^{s - 1}}{1 + v} e^{- u} d u d v = \int_{0}^{\infty} \frac{v^{s - 1}}{1 + v} d v \underset{a n a l y s i s}{\overset{c o m p l e x}{=}} \frac{π}{s i n (π s)} ✓

\dots m . n \dots

Commented by Dwaipayan Shikari last updated on 14/Nov/20

G r e a t!

Commented by mnjuly1970 last updated on 14/Nov/20

t h a n k y o u . .