Legendre's Duplication Formula

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:

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

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

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

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

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

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

Substituting $n = 2$ yields:

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

$\blacksquare$

Also denoted as

Some sources report this as:

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

Source of Name

This entry was named for Adrien-Marie Legendre.
The passage is a detailed mathematical proof of Legendre's Duplication Formula for the gamma function, including two different proofs and some additional notes.
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system
Legendre's Duplication Formula

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:

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

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

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

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

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

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

Substituting $n = 2$ yields:

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

$\blacksquare$

Also denoted as

Some sources report this as:

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

Source of Name

This entry was named for Adrien-Marie Legendre.
user
The passage is a detailed mathematical proof of Legendre's Duplication Formula for the gamma function, including two different proofs and some additional notes.

TYPE: NO_QUESTION