Is there a relatively elementary proof that the gamma function has no zero?

Question

I have never studied functional analysis. So I am unable to understand terms such as meromorphic, holomorphic and etc.

So far, I have showed Gauss', Euler's , Weierstrass' definition of the Gamma function are identical.

I have proved that $\Gamma(z)\Gamma(w)=\Gamma(z+w)B(z,w)$ too where $Re(z),Re(w)>0$.

I think the proof in the link below is elementary if i could understand, but i think proof here does not make sense.. If so, please explain me how exactly..

(Link : http://www.proofwiki.org/wiki/Zeroes_of_Gamma_Function)

Answer 1

By Euler's Reflection Formula , we have that

$$\Gamma(1-z)\Gamma(z)=\frac\pi{\sin\pi z}$$

This clearly shows the function cannot vanish (unless it'd vanish at $\;z\;$ and also it'd have a pole in $\;1-z\;$ with the same residue, which doesn't happen as the poles of the function are only at the non-positive integers...)

Answer 2

Let $z\in\mathbb{C}$ such that $Re(z)>0$

Then $\Gamma(z)=\frac{1}{z} \prod_{n=1}^\infty \frac{(1+1/n)^z}{1+z/n}$ (Euler Form)

Suppose $\Gamma(z)=0$

Then, $\prod_{n=1}^\infty \frac{(1+1/n)^{Re(z)}}{|n+z|/n} = 0$

Since the Euler form is true for $Re(z)$ too, $\prod_{n=1}^\infty \frac{n+Re(z)}{|n+z|}=0$

(Note that this limit exists)

Let $M$ be a positive integer such that $|Im(z)|<M$.

Then, $\forall n\in\mathbb{Z}^+, |z+n|\leq |z|+n \leq Re(z)+|Im(z)|+n < Re(z)+M+n$

Hence, $1\leq \prod_{n=1}^\infty \frac{Re(z)+M+n}{|z+n|}$

Hence, $0<\prod_{n=1}^\infty \frac{Re(z)+n}{|z+n|}$

Contradiction

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Edit:

Note that $0<\Gamma(Re(z))=\frac{1}{Re(z)} \prod_{n=1}^\infty \frac{(1+1/n)^{Re(z)}}{1+Re(z)/n}$

From the Euler second integral, it's direct to check that the limit on the right-hand side is nonzero.

Hence, $0< \prod_{n=1}^\infty \frac{(1+1/n)^{Re(z)}}{1+Re(z)/n}$

Multiply the inverse of this limit to the limit in the 4th line, which yields $\prod_{n=1}^\infty \frac{n+Re(z)}{|n+z|}=0$

Answer 3

An infinite product $\prod_{j=1}^{\infty} (1+a_{j})$ of complex numbers $a_{j}\ne -1$ converges to a non-zero complex number if $\sum_{j=1}^{\infty}|a_{j}| < \infty$. This applies to your case because, if $z \ne -1,-2,\ldots$, $$ \frac{\left(1+\frac{1}{n}\right)^{z}}{\left(1+\frac{z}{n}\right)}=1+\frac{z(z-1)}{2n^{2}}+O\left(\frac{1}{n^{3}}\right). $$