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.

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.


3 Answers 3


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

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


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

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

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

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


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


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

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

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

  • 1
    $\begingroup$ :D Beautiful! +1 $\endgroup$ Commented Jan 15, 2017 at 15:48

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.


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.

  • $\begingroup$ That's the exact identity I was looking for for the odd case. $\endgroup$
    – bryanj
    Commented Apr 14, 2014 at 5:45
  • $\begingroup$ @bryanj No problem. As a matter of fact I just finished reading Ahlfors (and solving almost all exercises) a few months back. I know nothing of a simpler proof. $\endgroup$
    – user1337
    Commented Apr 14, 2014 at 5:52
  • $\begingroup$ I bet @DanielFischer knows one. $\endgroup$
    – bryanj
    Commented Apr 14, 2014 at 5:54
  • $\begingroup$ He's a beast. ${}$ $\endgroup$
    – user1337
    Commented Apr 14, 2014 at 5:54
  • 1
    $\begingroup$ @bryanj Unless you use Stirling's formula (which might be considered cheating, since that's the next section in Ahlfors' book), I don't know of a simpler or nicer proof than this. With Stirling's formula, a short computation shows that the difference of the logarithms is $O\left(\frac{1}{\operatorname{Re} z}\right)$, and since it's a constant, that constant hence must be $0$, which is simpler in so far as you don't need to know the sine product, but arguably not as nice if you do know it. $\endgroup$ Commented Apr 15, 2014 at 8:51

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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