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Consider the definition of the Gamma function

$$ \Gamma(s) = \int_{0}^{\infty}\left[x^{s-1}e^{-x} \right] dx $$

Clearly: $x^{s-1}$ may have multiple defined values for $s$ if $s-1$ is rational or even infinitely many if $s-1$ is irrational.

Does that mean that when we look at the Gamma function we are really only observing a specific branch cut of it?

iadvd
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Sidharth Ghoshal
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    "Clearly", when s is real (and positive), this is the usual power function with real values hence there is no branch cut there, only a banal real function. – Did Mar 14 '15 at 18:37
  • If you look at, for example, http://en.wikipedia.org/wiki/Gamma_function, you will see that the Gamma function can be extended to the complex plane without any branch cuts. The only difficulty will be with poles which are at the non-positive integers. – Stephen Montgomery-Smith Mar 14 '15 at 18:43

1 Answers1

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No, $\Gamma$ has no branch cuts. That definition can only be taken for $s$ whose real part is positive, as the integral diverges otherwise. Noting that we define $x^s=e^{s\log(x)}$, we are tempted to worry about the fact that $\log$ is multi-valued - but this proves to be a non-issue. Why? Because the principal value of $\log$ is continuous in the left half-plane - there are no branch cuts there. Given that the gamma function, from the left half plane, can be continued (analytically) to the whole plane using the identity $\Gamma(t+1)=t\Gamma(t)$, which introduces no branch cuts, it's clear that the function as a whole must have no branch cuts.

You might be tempted to worry about what happens if we take $x^s=e^{s(\log(x)+2n i \pi)}$ for some integer $n$ - and the answer is that we get another meromorphic function, also satisfying $f(t+1)=tf(t)$ and $f(1)=1$. It would, in fact, satisfy $f(s)=e^{(s-1)2n i \pi}\Gamma(s)$ - but it is a distinct function.

Milo Brandt
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    So it seems that there are "other gamma like functions" just rotated around the complex plane. The one corresponding $e^0$ in the latter product is generally accepted as the main one – Sidharth Ghoshal Mar 15 '15 at 07:07