prove the duplication formula
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Gamma functions of argument 2z can be expressed in terms of gamma functions of smaller arguments. From the definition of the beta function,
B(m,n)=(Gamma(m)Gamma(n))/(Gamma(m+n))=int_0^1u^(m-1)(1-u)^(n-1)du.
(1)
Now, let m=n=z, then
(Gamma(z)Gamma(z))/(Gamma(2z))=int_0^1u^(z-1)(1-u)^(z-1)du
(2)
and u=(1+x)/2, so du=dx/2 and
(Gamma(z)Gamma(z))/(Gamma(2z)) = int_(-1)^1((1+x)/2)^(z-1)(1-(1+x)/2)^(z-1)(1/2dx)
(3)
= 1/2int_(-1)^1((1+x)/2)^(z-1)((1-x)/2)^(z-1)dx
(4)
= 1/(2^(1+2(z-1)))int_(-1)^1(1-x^2)^(z-1)dx
(5)
= 2^(1-2z)[2int_0^1(1-x^2)^(z-1)dx].
(6)
Now, use the beta function identity
B(m,n)=2int_0^1x^(2m-1)(1-x^2)^(n-1)dx
(7)
to write the above as
(Gamma(z)Gamma(z))/(Gamma(2z))=2^(1-2z)B(1/2,z)=2^(1-2z)(Gamma(1/2)Gamma(z))/(Gamma(z+1/2)).
(8)
Solving for Gamma(2z) and using Gamma(1/2)=sqrt(pi) then gives
Gamma(2z) = (2pi)^(-1/2)2^(2z-1/2)Gamma(z)Gamma(z+1/2)
(9)
= (2^(2z-1)Gamma(z)Gamma(z+1/2))/(sqrt(pi)).
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