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He says:

Prove the formula of Gauss:

$$

(2\pi)^\frac{n-1}{2} \Gamma(z) = n^{z – \frac{1}{2}}\Gamma(z/n)\Gamma(\frac{z+1}{n})\cdots\Gamma(\frac{z+n-1}{n})

$$

This is an exercise out of Ahlfors.

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By taking the logarithmic derivative, it’s easy to show the left & right hand sides are the the same up to a multiplicative constant.

After that I’m lost. It’s easy using another identity when $n$ is *even* to use induction. But when $n$ is odd I am lost.

It’s obvious when $n$ is a power of 2.

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After you have established that the RHS and the LHS differ by a multiplicative constant, all you are left to do it plug in $z=1$. If you pair up the factors in the RHS as $$\Gamma \left( \frac{1+k}{n} \right) \leftrightarrow \Gamma \left( \frac{n-k-1}{n} \right) ,$$

and apply the reflection formula $\Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z}$, things will be easier IMO.

Edit:

Once you apply the reflection formula, you will have to deal with a product of sines. Please see this question in order to handle it.

Another common approach is to derive it from the limit definition of the gamma function. (See below.)

The multiplication formula can be written in the form

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

Using the limit definition of the gamma function, we have

$$ \Gamma \left(z +\frac{k}{n} \right) = \lim_{m \to \infty} \frac{m! \ m^{z+\frac{k}{n}-1}}{(z+\frac{k}{n})(z+\frac{k}{n}+1) \cdots (z+\frac{k}{n} + m -1)}$$

Then using Stirling’s formula, we get

$$ \begin{align} \Gamma \left(z+\frac{k}{n} \right) &= \lim_{m \to \infty} \frac{\sqrt{2 \pi m} (\frac{m}{e})^m m^{z+\frac{k}{n}-1}}{(z+\frac{k}{n})(z+\frac{k}{n}+1) \cdots (z+\frac{k}{n} + m -1)} \\ &=\lim_{m \to \infty} \frac{\sqrt{2 \pi} (\frac{mn}{e})^m m^{z+\frac{k}{n}-1/2}}{(nz+k)(nz+k+n) \cdots (nz+k + mn -n)} \end{align}$$

So

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

$$ = n^{nz-1/2}\lim_{m \to \infty}\frac{(\sqrt{2 \pi})^{n} (\frac{mn}{e})^{mn} m^{nz-n/2} m^{\frac{1}{n} \sum_{k=1}^{n-1} k}}{(nz)(nz+1)\cdots (nz+n-1)(nz+n) \cdots (nz+mn-n)\cdots(nz+mn-1)} $$

$$ = \lim_{m \to \infty} \frac{(\sqrt{2 \pi})^{n} (\frac{mn}{e})^{mn} (mn)^{nz-1/2}}{(nz)(nz+1)\cdots (nz+n-1)(nz+n) \cdots (nz+mn-n) \cdots(nz+mn-1)} $$

Replacing $mn$ with $m$ shouldn’t change the value of the limit (I think).

Therefore,

$$ \begin{align} n^{nz-1/2} \prod_{k=0}^{n-1} \Gamma \left(z+\frac{k}{n} \right) &=\lim_{m \to \infty} \frac{(\sqrt{2 \pi})^{n} (\frac{m}{e})^{m} m^{nz-1/2}}{(nz)(nz+1)\cdots(nz+m-1)} \\ &=\lim_{m \to \infty} \frac{(\sqrt{2 \pi})^{n-1}m! \ m^{nz-1}}{(nz)(nz+1)\cdots(nz+m-1)} \\ &= (\sqrt{2\pi})^{n-1}\Gamma(nz) \end{align}$$

**EDIT**:

Wikipedia states that the limit definition is

$$ \Gamma(t) = \lim_{n \to \infty} \frac{n! \ n^{t}}{t(t+1) \cdots (t+n)}$$

But notice that

$$ \Gamma(t-1) = \frac{\Gamma(t)}{t-1} = \lim_{n \to \infty} \frac{n! \ n^{t-1}}{(t-1)t \ldots (t+n-1)}$$

$$\implies \Gamma(t) = \lim_{n \to \infty} \frac{n! \ n^{t-1}}{t(t+1) \ldots (t+n-1)}$$

Let we consider

$$ f(z) = \frac{\Gamma(2z)\,\Gamma(1/2)}{\Gamma(z)\,\Gamma(z+1/2)}.$$

Since $\Gamma(z)$ never vanishes, $f(z)$ is a continuous function on its domain. The singularity of the $\Gamma$ function are simple poles at the negative integers: in particular, the structure of the denominator and numerator of $f(z)$ implies that $f$ has no singularity and no zero on the real line. Since $\Gamma(z+1)=z\,\Gamma(z)$, we also have:

$$ \frac{f(z+1)}{f(z)} = \frac{(2z+1)(2z)}{z(z+1/2)} = 4 $$

hence it follows that $f(z)=C\cdot 4^z$. By computing $f(z)$ at $z=1$ we get the explicit value of $C$, hence Legendre’s duplication formula through a real-analytic version of Herglotz’ trick.

You may perform just the same trick to prove the full multiplication formula in the real case.

An efficient alternative is to consider $\frac{d}{dx}\log(\cdot )$ of both terms. Since

$$ \frac{d}{dx}\log\Gamma(x) = \psi(x) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+x-1}\right) $$

the duplication/multiplication formula for the $\Gamma$ function can be derived from the duplication/multiplication formula for the $\psi$ function, that is simple to prove through elementary series manipulations.

As a third alternative, Legendre duplication formula can be proved by computing

$$ \int_{0}^{+\infty}\frac{d\theta}{(1+\cosh\theta)^n} $$

in two different ways, as done by me and Marco Cantarini here.

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