Corollary

Tag 08JC, permanent. Cite it as [Bourbaki, Tag 08JC].

Functions of a Real Variable, chapter VII, § 2. THE GAMMA FUNCTION IN THE COMPLEX DOMAIN

For every real $t$ one has

|\Gamma(it)| = \sqrt{\frac{\pi}{t \sinh \pi t}} \quad (t \neq 0)

|\Gamma(\frac{1}{2} + it)| = \sqrt{\frac{\pi}{\cosh \pi t}}.

Indeed, one deduces from (8) that $\Gamma(it)\Gamma(-it) = \frac{i\pi}{t \sin \pi it} = \frac{\pi}{t \sinh \pi t}$ , and one has $\Gamma(-it) = \overline{\Gamma(it)}$ ; similarly, (8) gives

\Gamma\left(\frac{1}{2} + it\right)\Gamma\left(\frac{1}{2} - it\right) = \frac{\pi}{\sin (\frac{\pi}{2} + \pi it)} = \frac{\pi}{\cos \pi it} = \frac{\pi}{\cosh \pi t},

and one has

\Gamma\left(\frac{1}{2} - it\right) = \overline{\Gamma\left(\frac{1}{2} + it\right)}.

Now let $p$ be any integer $> 0$ and consider the product

f(z) = \Gamma\left(\frac{z+1}{p}\right) \Gamma\left(\frac{z+2}{p}\right) \ldots \Gamma\left(\frac{z+p}{p}\right).

By (3) (VII, p. 315), for every $z \neq -n \ (n \in \mathbf{N})$ , $f(z)$ is the limit of the product

\frac{n^{(z+1)/p} n!}{\left(\frac{z+1}{p}\right) \left(\frac{z+1}{p} + 1\right) \ldots \left(\frac{z+1}{p} + n\right)}

\frac{n^{(z+2)/p} n!}{\left(\frac{z+2}{p}\right) \left(\frac{z+2}{p} + 1\right) \ldots \left(\frac{z+2}{p} + n\right)} \cdots

\ldots \frac{n^{(z+p)/p} n!}{\left(\frac{z+p}{p}\right) \left(\frac{z+p}{p} + 1\right) \ldots \left(\frac{z+p}{p} + n\right)}

= \frac{n^{z+(p+1)/2} p^{(n+1)p} (n!)^p}{(z+1)(z+2) \ldots (z+(n+1)p)}

and in particular $f(0)$ is the limit of the product

\frac{n^{(p+1)/2} p^{(n+1)p} (n!)^p}{((n+1)p)!}

from which it follows that $f(z)/f(0)$ is the limit of

\frac{n^z ((n+1)p)!}{(z+1)(z+2)\ldots(z+(n+1)p)} = z\ p^{-z} \left( \frac{n}{n+1} \right)^z \cdot \frac{((n+1)p)^z ((n+1)p)!}{z(z+1)(z+2)\ldots(z+(n+1)p)}

which, by (3) (VII, p. 315), gives

f(z) = f(0)\ z\ p^{-z}\ \Gamma(z).

But one can write

f(0) = \prod_{k=1}^{p-1} \Gamma \left( \frac{k}{p} \right) = \prod_{k=1}^{p-1} \Gamma \left( 1 - \frac{k}{p} \right) = \sqrt{\prod_{k=1}^{p-1} \Gamma \left( \frac{k}{p} \right) \Gamma \left( 1 - \frac{k}{p} \right)}

since

f(0) > 0

; the complements’ relation then gives

f(0) = \sqrt{\pi^{p-1} / \prod_{k=1}^{p-1} \sin \frac{k\pi}{p}}

and since the product on the right-hand side is equal to

p/2^{p-1}

(VI, p. 284, cor. 1) one sees finally that:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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