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Legendre Duplication Formula

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)\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
{\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 $du=dx/2$ 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
B(m,n)=2\int_0^1 x^{2z-1}(1-x^2)^{z-1}\,dx
\end{displaymath} (4)

to write the above as
{\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 $\Gamma({\textstyle{1\over 2}})=\sqrt{\pi}$.

See also Gamma Function, Gauss Multiplication Formula


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.

© 1996-9 Eric W. Weisstein