Question Number 102089 by Ar Brandon last updated on 06/Jul/20

  The Gamma function Γ(α) is defined as follows;  Γ(α)=∫_0 ^∞ y^(α−1) e^(−y) dy , α>0  a\ Show that Γ(α+1)=αΓ(α).  b\Conclude that Γ(n)=(n−1)! , n=1, 2, 3, ...  c\Determine Γ(55).

Answered by Ar Brandon last updated on 06/Jul/20

a\Γ(α+1)=∫_0 ^∞ y^α e^(−y) dy=[y^α ∫e^(−y) dy]_0 ^∞ −∫_0 ^∞ {(dy^α /dy)∙∫e^(−y) dy}dy                          =[−y^α e^(−y) ]_0 ^∞ +∫_0 ^∞ αy^(α−1) e^(−y) dy=α∫_0 ^∞ y^(α−1) e^(−y) dy                          =αΓ(α)

Answered by floor(10²Eta[1]) last updated on 06/Jul/20

Γ(55)=54!       :)