Legendre Duplication Formula

Gamma Functions of argument can be expressed in terms of Gamma Functions of smaller arguments. From the definition of the Beta Function,

$\begin{displaymath} B(m,n)={\Gamma(m)\Gamma(n)\over \Gamma(m+n)} = \int_0^1 u^{m-1}(1-u)^{n-1}\,du. \end{displaymath}$

(1)

Now, let $m=n\equiv z$ , then

$\begin{displaymath} {\Gamma(z)\Gamma(z)\over \Gamma(2z)} = \int_0^1 u^{z-1}(1-u)^{z-1}\,du \end{displaymath}$

(2)

and $u\equiv (1+x)/2$ , so

and

$\displaystyle {\Gamma(z)\Gamma(z)\over \Gamma(2z)}$	$\textstyle =$	$\displaystyle \int_0^1 \left({1+x\over 2}\right)^{z-1}\left({1-{1+x\over 2}}\right)^{z-1} ({\textstyle{1\over 2}}\,dx)$
	$\textstyle =$	$\displaystyle {1\over 2}\int_0^1 \left({1+x\over 2}\right)^{z-1}\left({1-x\over 2}\right)^{z-1}\,dx$
	$\textstyle =$	$\displaystyle {1\over 2^{1+2(z-1)}} \int_0^1 (1-x^2)^{z-1}\,dx$
	$\textstyle =$	$\displaystyle 2^{1-2z}\int_0^1 (1-x^2)^{z-1}\,dx.$	(3)

Now, use the Beta Function identity

$\begin{displaymath} B(m,n)=2\int_0^1 x^{2z-1}(1-x^2)^{z-1}\,dx \end{displaymath}$

(4)

to write the above as

$\begin{displaymath} {\Gamma(z)\Gamma(z)\over \Gamma(2z)} = 2^{1-2z}B({\textstyle... ...1-2z} {\Gamma({1\over 2})\Gamma(z)\over \Gamma(z+{1\over 2})}. \end{displaymath}$

(5)

Solving for $\Gamma(2z)$ ,

$\displaystyle \Gamma(2z)$	$\textstyle =$	$\displaystyle {\Gamma(z)\Gamma(z+{1\over 2})2^{2z-1}\over\Gamma({1\over 2})} ={\Gamma(z)\Gamma(z+{1\over 2})2^{2z-1}\over\sqrt{\pi}}$
	$\textstyle =$	$\displaystyle (2\pi)^{-1/2} 2^{2z-1/2}\Gamma(z)\Gamma(z+{\textstyle{1\over 2}}),$	(6)

since $\Gamma({\textstyle{1\over 2}})=\sqrt{\pi}$ .

References

Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 256, 1972.

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 561-562, 1985.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 424-425, 1953.