I want to show that $\Gamma(z)$ has no zeros. My idea is to use the formula $$\Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin(\pi z)}$$ which holds for all $z\in\mathbb{C}$.

If $\Gamma(z)=0$, then the left-hand side is $0$ and the right-hand side isn't, so impossible. But I'm worried that it wouldn't work if $\Gamma(1-z)=\pm\infty$. How to resolve this case?

  • 7
    $\begingroup$ $\Gamma(1-z) = \infty$ implies $1-z = -n$ for some $n=0,1,2,...$ and so $\Gamma(z) = \Gamma(n+1) = n! \ne 0$ $\endgroup$
    – Cocopuffs
    Dec 14, 2013 at 7:36

1 Answer 1


A slight variation of the comment-answer by Cocopuffs: if $\Gamma(z)=0$, then from $\Gamma(z)=(z-1)\Gamma(z-1)$ we find that either $z=1$ or $\Gamma(z-1)=0$. But $\Gamma(1)=\int_0^\infty e^{-t}\,dt=1$. Hence, any zero of $\Gamma$ propagates to the left: it creates the sequence of zeros $z-n$, $n=1,2,3,\dots$. According to the functional equation cited in the question, this creates a sequence of poles propagating to the right. But $\Gamma(z)=\int_0^\infty t^{z-1} e^{-t}\,dt$ is evidently holomorphic when $\operatorname{Re}z>0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .