Question

Prove that ga ma pa all 2 p - gone Gama p - 1

Answer 1

Step-by-step explanation:

Prove that Γ(p)×Γ(1−p)=πsin(pπ),∀p∈(0,1)

integration

Prove that

Γ(p)×Γ(1−p)=πsin(pπ),∀p∈(0,1)

With

Γ(p)=∫∞0xp−1e−xdx

My tried:

We have

B(p,q)=∫10xp−1(1−x)q−1dx=Γ(p)×Γ(q)Γ(p+q)

Hence

B(p,1−p)=Γ(p)×Γ(1−p)Γ(1)=Γ(p)×Γ(1−p)=∫10xp−1(1−x)−pdx