prove-that-1-2-it-2pi-e-pit-e-pit-and-1-it-2pit-e-pit-e-pit-

Question Number 67540 by mathmax by abdo last updated on 28/Aug/19

p r o v e t h a t ∣ Γ (\frac{1}{2} + i t) ∣ = \sqrt{\frac{2 π}{e^{π t} + e^{- π t}}}

a n d ∣ Γ (1 + i t) ∣ = \sqrt{\frac{2 π t}{e^{π t} - e^{- π t}}}

Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19

I^{'} m n o t s u r e i f w e a l w a y s h a v e \bar{Γ} (z) = Γ (\bar{z})

B u t i f i s t r u e

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

= \frac{π}{s i n (\frac{π}{2} + i π t)} u s i n g c o m p l e m e n t s f o r m u l a s Γ (s) Γ (1 - s) = \frac{π}{s i n (π s)}

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

∣ Γ (1 + i t) ∣^{2} = Γ (1 + i t) Γ (1 - i t) = i t Γ (i t) Γ (1 - i t) = \frac{i π t}{s i n (i π t)} = \frac{i π t}{\frac{e^{- π t} - e^{π t}}{2 i}} = \frac{2 π t}{e^{π t} - e^{- π t}}

Commented by mathmax by abdo last updated on 29/Aug/19

t h a n k y o u s i r .