$$\newcommand{acoloneqq}[0]{\ &\hspace-2pt\coloneqq}
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定理
ガンマ関数の倍数公式
$\displaystyle \Gamma(nz) =
(2\pi)^\frac{1-n}2n^{nz-\frac12}\prod_{k=0}^{n-1} \Gamma\fqty(z+\frac kn)$
補題
$\displaystyle \prod_{k=1}^{n-1} \sin\frac{k\pi}n = 2^{1-n}n$
$\beginend{align}{
\sahen &=
\prod_{k=1}^{n-1} \qty(i\sinh\fqty(-\frac{k\pi i}n)) =
\qty(\frac i2)^{n-1}\prod_{k=1}^{n-1} \qty(e^{-\frac{k\pi i}n}-e^\frac{k\pi i}n) \\&=
\qty(\frac i2)^{n-1}e^{-\frac{\pi i}n\frac{n(n-1)}2}\prod_{k=1}^{n-1} \qty(1-e^\frac{2k\pi i}n) \\&=
\qty(\frac i2)^{n-1}e^{-\frac{(n-1)\pi i}2}\lim_{a\to1} \frac1{a-1}\prod_{k=0}^{n-1} \qty(a-e^\frac{2k\pi i}n) \\&=
\qty(\frac i2)^{n-1}i^{-n+1}\lim_{a\to1} \frac{a^n-1}{a-1} =
\uhen
}$
ディガンマ関数の倍数公式
$\displaystyle n\psi(nz) = n\ln n + \sum_{k=0}^{n-1} \psi\fqty(z+\frac kn)$
証明
証明
ガンマ関数の倍数公式
補題3を両辺積分して、
$\beginend{alignat}{2
\ln\Gamma(nz) &=
n\ln n\cdot z + \sum_{k=0}^{n-1} \ln\Gamma\fqty(z+\frac kn) + C \\
\Gamma(nz) &= Dn^{nz}\prod_{k=0}^{n-1} \Gamma\fqty(z+\frac kn) \quad\fqty(D\coloneqq e^C) \\
\asupplement{-3.5em}{$z=\tfrac1n$を代入して、} \\
D &=
\frac{\Gamma\fqty(\frac nn)}{n\prod_{k=0}^{n-1} \Gamma\fqty(\frac1n+\frac kn)} =
\frac1{n\prod_{k=1}^{n-1} \Gamma\fqty(\frac kn)} \\&=
\frac1{n\sqrt{\prod_{k=1}^{n-1} \Gamma\fqty(\frac kn)\Gamma\fqty(1-\frac kn)}} \\&=
\frac1n\sqrt{\prod_{k=1}^{n-1} \frac{\sin\frac{k\pi}n}\pi}
&&\hspace-8em\because\textsf{相反公式} \\&=
\frac{\sqrt{\pi^{1-n}2^{1-n}n}}n
&&\hspace-8em\because\textsf{補題2} \\&=
\frac{(2\pi)^\frac{1-n}2}{\sqrt n} \\
\asupplement{-3.5em}{よって、} \\
\Gamma(nz) &=
\frac{(2\pi)^\frac{1-n}2}{\sqrt n}n^{nz}\prod_{k=0}^{n-1} \Gamma\fqty(z+\frac kn) \\&=
(2\pi)^\frac{1-n}2n^{nz-\frac12}\prod_{k=0}^{n-1} \Gamma\fqty(z+\frac kn)
}$
