Legendre's Duplication Formula

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Theorem

Let $\Gamma$ denote the gamma function.

Then:

$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$

where $\N$ denotes the natural numbers.


Proof 1

From the definition of the Beta function:

\(\ds \map \Beta {z_1, z_2}\) \(=\) \(\ds \frac {\map \Gamma {z_1} \map \Gamma {z_2} } {\map \Gamma {z_1 + z_2} }\)
\(\ds \) \(=\) \(\ds \int_0^1 u^{z_1 - 1} \paren {1 - u}^{z_2 - 1} \rd u\) Equivalence of Definitions of Beta Function


Letting $z_1 = z_2 = z$ gives:

\(\ds \frac {\map \Gamma z \map \Gamma z} {\map \Gamma {2 z} }\) \(=\) \(\ds \int_0^1 u^{z - 1} \paren {1 - u}^{z - 1} \rd u\)
\(\ds \) \(=\) \(\ds \frac 1 2 \int_{-1}^1 \paren {\frac {1 + x} 2 }^{z - 1} \paren {\frac {1 - x} 2}^{z - 1} \rd x\) Substitute $u = \dfrac {1 + x} 2$
\(\ds \) \(=\) \(\ds \frac 1 {2^{2 z - 1} } \int_{-1}^1 \paren {1 - x^2}^{z - 1} \rd x\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 2^{2 z - 1} \map \Gamma z \map \Gamma z\) \(=\) \(\ds 2 \map \Gamma {2 z} \int_0^1 \paren {1 - x^2}^{z - 1} \rd x\) by Definite Integral of Even Function


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

$\ds \map \Beta {z_1, z_2} = \int_0^1 x^{2 z_1 - 2} \paren {1 - x^2}^{z_2 - 1} 2 x \rd x$

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

$(2): \quad \ds \map \Beta {\frac 1 2, z} = 2 \int_0^1 \paren {1 - x^2}^{z - 1} \rd x$


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

\(\ds 2^{2 z - 1} \map \Gamma z \map \Gamma z\) \(=\) \(\ds \map \Gamma {2 z} \map \Beta {\frac 1 2, z}\)
\(\ds \) \(=\) \(\ds \map \Gamma {2 z} \frac {\map \Gamma {\frac 1 2} \map \Gamma z} {\map \Gamma {\frac 1 2 + z} }\)
\(\ds \leadsto \ \ \) \(\ds \map \Gamma z \map \Gamma {z + \dfrac 1 2}\) \(=\) \(\ds 2^{1 - 2 z} \map \Gamma {\frac 1 2} \map \Gamma {2 z}\)

From Gamma Function of One Half:

$\map \Gamma {\dfrac 1 2} = \sqrt \pi$

It follows that:

$\map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$

$\blacksquare$


Proof 2

From Gauss Multiplication Formula:

$\ds \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$

Substituting $n = 2$ yields:

\(\ds \map \Gamma z \map \Gamma {z + \frac 1 2}\) \(=\) \(\ds \paren {2 \pi}^{1 / 2} 2^{1/2 - 2 z} \map \Gamma {2 z}\)
\(\ds \) \(=\) \(\ds 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}\)

$\blacksquare$


Also denoted as

Some sources report this as:

$\forall z \notin \set{-\dfrac n 2: n \in \N}: 2^{2 z - 1} \map \Gamma z \, \map \Gamma {z + \dfrac 1 2} = \sqrt \pi \, \map \Gamma {2 z}$


Also see


Source of Name

This entry was named for Adrien-Marie Legendre.


Sources

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