Legendre's Duplication Formula

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Theorem

Let $\Gamma$ denote the gamma function.

Then:

$\forall z \notin \left\{{-\dfrac n 2: n \in \N}\right\}: \Gamma \left({z}\right) \Gamma \left (z + \dfrac 1 2 \right) = 2^{1 - 2 z} \sqrt \pi \ \Gamma \left({2 z}\right)$

where $\N$ denotes the natural numbers.


Proof 1

From the definition of the Beta function:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \Beta \left({z_1, z_2}\right)\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \frac {\Gamma \left({z_1}\right) \Gamma \left({z_2}\right)} {\Gamma \left({z_1 + z_2}\right)}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \int_0^1 u^{z_1 - 1} \left({1 - u}\right)^{z_2 - 1} \ \mathrm d u\) \(\displaystyle \) \(\displaystyle \)          Equivalence of Definitions of Beta Function          


Letting $z_1 = z_2 = z$ gives:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \frac {\Gamma \left({z}\right) \Gamma \left({z}\right)} {\Gamma \left({2 z}\right)}\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \int_0^1 u^{z - 1} \left({1 - u}\right)^{z - 1} \ \mathrm d u\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \frac 1 2 \int_{-1}^1 \left({\frac {1 + x} 2 }\right)^{z - 1} \left({\frac {1 - x} 2}\right)^{z - 1} \ \mathrm d x\) \(\displaystyle \) \(\displaystyle \)          Substitute $u = \dfrac {1 + x} 2$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \frac 1 {2^{2 z - 1} } \int_{-1}^1 \left({1 - x^2}\right)^{z - 1} \ \mathrm d x\) \(\displaystyle \) \(\displaystyle \)                    
\((1):\)      \(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle 2^{2 z - 1} \Gamma \left({z}\right) \Gamma \left({z}\right)\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle 2 \Gamma \left({2 z}\right) \int_0^1 \left({1 - x^2}\right)^{z - 1} \ \mathrm d x\) \(\displaystyle \) \(\displaystyle \)          by Definite Integral of Even Function          


Now substituting $u = x^2$ into the Beta function:

$\displaystyle \Beta \left({z_1, z_2}\right) = \int_0^1 x^{2z_1 - 2} \left({1 - x^2}\right)^{z_2 - 1} 2x \ \mathrm d x$

Letting $z_1 = \dfrac 1 2$ and $z_2 = z$ gives:

$(2): \quad \displaystyle \Beta \left({\frac 1 2, z}\right) = 2 \int_0^1 \left({1 - x^2}\right)^{z - 1} \ \mathrm d x$


Combining results $(1)$ and $(2)$:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle 2^{2z - 1} \Gamma \left({z}\right) \Gamma \left({z}\right)\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \Gamma \left({2z}\right) \Beta \left({\frac 1 2, z}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \Gamma \left({2z}\right) \frac {\Gamma \left({\frac 1 2}\right) \Gamma \left({z}\right)} {\Gamma \left({\frac 1 2 + z}\right)}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \Gamma \left({z}\right) \Gamma \left({z + \dfrac 1 2}\right)\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle 2^{1 - 2z} \Gamma \left({\frac 1 2}\right) \Gamma \left({2z}\right)\) \(\displaystyle \) \(\displaystyle \)                    

From Gamma Function of One Half:

$\Gamma \left({\dfrac 1 2}\right) = \sqrt \pi$

It follows that:

$\Gamma \left({z}\right) \Gamma \left({z + \dfrac 1 2}\right) = 2^{1 - 2z} \sqrt \pi \, \Gamma \left({2 z}\right)$

$\blacksquare$


Proof 2

From Gauss Multiplication Formula:

$\displaystyle \prod_{k \mathop = 0}^{n - 1} \Gamma \left({z + \frac k n}\right) = \left({2 \pi}\right)^{\left({n - 1}\right) / 2} n^{1/2 - n z} \Gamma \left({n z}\right)$

Substituting $n=2$ yields:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \Gamma \left({z}\right) \Gamma \left({z + \frac 1 2}\right)\) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle \left({2 \pi}\right)^{1 / 2} 2^{1/2 - 2 z} \Gamma \left({2 z}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\) \(\displaystyle \) \(\displaystyle 2^{1 - 2 z} \sqrt \pi \, \Gamma \left({2 z}\right)\) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


Also denoted as

Some sources report this as:

$\forall z \notin \left\{{-\dfrac n 2: n \in \N}\right\}: 2^{2 z - 1} \Gamma \left({z}\right) \Gamma \left (z + \dfrac 1 2 \right) = \sqrt \pi \ \Gamma \left({2 z}\right)$


Source of Name

This entry was named for Adrien-Marie Legendre.


Sources