Legendre's Duplication Formula/Proof 1
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Theorem
- $\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
where $\N$ denotes the natural numbers.
Proof
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:
- $\map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
$\blacksquare$