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?

  • 6
    $\begingroup$ $\Gamma(1-z) = \infty$ implies $1-z = -n$ for some $n=0,1,2,...$ and so $\Gamma(z) = \Gamma(n+1) = n! \ne 0$ $\endgroup$ – Cocopuffs Dec 14 '13 at 7:36

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 $\operatorname{Re}z>0$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.