0-1-2-ix-2-dx-pi-2-

Question Number 120553 by snipers237 last updated on 01/Nov/20

\int_{0}^{\infty} ∣ Γ (\frac{1}{2} - i x) ∣^{2} d x = \frac{π}{2}

Answered by mnjuly1970 last updated on 01/Nov/20

s o l u t i o n :

w e k n o w t h a t : \bar{Γ} (z) = Γ (\bar{z}) (w h y ?)

d e d u c t i b l e b y u s i n g : Γ (z) = \frac{1}{z} e^{- γ z} \prod_{k ⩾ 0} (\frac{e^{\frac{z}{k}}}{1 + \frac{z}{k}})

∣ Γ (\frac{1}{2} - i x) ∣^{2} = Γ (\frac{1}{2} - i x) Γ (\frac{1}{2} + i x)

= Γ (\frac{1}{2} - i x) Γ (1 - (\frac{1}{2} - i x))

\overset{e u l e r r e f l e c t i o n f o r m u l s}{=} \frac{π}{s i n (π (\frac{1}{2} - i x))}

= \frac{π}{c o s (π i x)} = \frac{2 π}{e^{- π x} + e^{π x}}

= \frac{2 π e^{π x}}{1 + {(e^{π x})}^{2}} ✓

Ω = 2 \int_{0}^{\infty} \frac{π e^{π x}}{1 + {(e^{π x})}^{2}} d x \overset{e^{π x} = y}{=} 2 \int_{1}^{\infty} \frac{d y}{1 + y^{2}}

= 2 (\frac{π}{2}) - 2 (\frac{π}{4}) = \frac{π}{2} ✓ ✓

\dots . m . n . j u l y . 1970 \dots .

Commented by snipers237 last updated on 01/Nov/20

Γ (z) i s n o t a l w a y s e q u a l t o Γ (\bar{z})

b u t \bar{Γ} (z) = Γ (\bar{z}) {i t}^{'} s t r u e

Γ (z) = \int_{0}^{\infty} t^{z - 1} e^{- t} d t = \int_{0}^{\infty} e^{(r e (z) - 1 + i I m (z)) l n t - t} d t

Γ (z) = \int_{0}^{\infty} e^{r e (z) - 1 - t} [c o s (I m (z) l n t) + i s i n (I m (z) l n t)] d t

L F Y T C

Commented by mnjuly1970 last updated on 01/Nov/20

y o u a r e r i g h t

i c o r r e c t e d i t .

t h a n k y o u .