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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)

Did
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Number 9
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  • The proof is fairly easy to understand if you accept the Euler form of $\Gamma(z)$. What is it that you're having trouble with? Also, holomorphic and meromorphic are terms generally used within complex analysis. – doppz Feb 24 '14 at 03:18
  • @D.Clark Clark I have neither learned complex analysis. I have learned real analysis that's it. How do i exactly show that $\Gamma(z)\neq 0$ from the Euler's second integral? – Number 9 Feb 24 '14 at 03:32
  • Hmmm, I didn't know you haven't yet studied complex analysis before I wrote my answer... – DonAntonio Feb 24 '14 at 03:38
  • @DonAntonio Dear Don. Does the proof in the link make sense? – Number 9 Feb 24 '14 at 03:40
  • Well, the Euler form, which is the one used in the Bohr-Mollerup theorem (which is a pearl in itself and does not require complex analysis to understand it, only advanced real analysis) is pretty clear there, though the argument is not that clear to me (and less at these hours). Yet I think it is, or can be put in a form as to be, basically true. – DonAntonio Feb 24 '14 at 03:44
  • @DonAntonio I know the Bohr-Mollerup theorem, but isn't it only true for the gamma function with real domain? – Number 9 Feb 24 '14 at 03:47
  • Yes @Number9...but then "analytic continuation" kicks in and things get swell for complex variable...but this perhaps is out of your actual reach. – DonAntonio Feb 24 '14 at 03:52
  • @DonAntonio Would you please check the answer i wrote? is it correct? – Number 9 Feb 24 '14 at 04:10
  • See Gamma function has no zeros. – user127096 Feb 24 '14 at 04:39
  • @127.0.9.6 I saw that post before i post my question. I couldn't understand it though.. what is holomorphic..? – Number 9 Feb 24 '14 at 04:40
  • There is a Wikipedia article about that. I think that most people learn what a holomorphic function is before they begin to worry about whether $\Gamma$ has zeros in the complex plain... – user127096 Feb 24 '14 at 04:45

3 Answers3

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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...)

DonAntonio
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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 $z \ne -1,-2,\ldots$, $$ \frac{\left(1+\frac{1}{n}\right)^{z}}{\left(1+\frac{z}{n}\right)}=1+\frac{z(z-1)}{2n^{2}}+O\left(\frac{1}{n^{3}}\right). $$

Disintegrating By Parts
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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

===

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

Number 9
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  • Where does the Re(z) exponent in the 4th line come from? – DonAntonio Feb 24 '14 at 04:13
  • @DonAntonio Absolute value of $|\Gamma(z)z|$ on the second line. – Number 9 Feb 24 '14 at 04:18
  • Since the limit in the second line converges to 0, it converges to 0 absolutely. – Number 9 Feb 24 '14 at 04:21
  • I see what I asked, I cannot though follow the other parts' logic now: why "converges to zero" means "converges to zero abs."? Why that limit exists (6th line)? From where the 5th line? – DonAntonio Feb 24 '14 at 04:25
  • @DonAntonio I'm really glad that you take time to help me. Thank you. I have added an explanation below. – Number 9 Feb 24 '14 at 04:35
  • Since if a sequence ${a_n}$ converges to 0, $\forall \epsilon>0, \exists N\in\mathbb{Z}^+$ such that $n\geq N \Rightarrow |a_n|<\epsilon$. Hence, ${|a_n|}$ converges to 0 too. – Number 9 Feb 24 '14 at 04:39