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How can I prove the Gamma function has no zeros in its holomorphy domain $\Bbb C\setminus\Bbb Z_{\le0}$ using only its integral definition $\Gamma(z)=\int_0^{+\infty}t^{z-1}e^{-t}\,dt$ valid when $\Re z>0$ and the functional equation $\Gamma(z+1)=z\Gamma(z)$?

From the integral definition we can find easily the holomorphic extension; thus it would be enough to prove that $\Gamma\neq0$ in $\{\Re z>0\}$, using thus the integral form. But I can't prove neither this.

Can someone help me?

EDIT: This question is not a duplicate because all the solution given use more "advanced" tools. Here I'm asking to prove that Gamma has no zeros using ONLY its integral representation

Joe
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    Possible duplicate of Is there a relatively elementary proof that the gamma function has no zero? – DonAntonio Apr 14 '16 at 17:38
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    That's not a duplicate because all the solution given there used more "advanced" tools. Here I'm asking to prove that Gamma has no zeros using ONLY its integral representation – Joe Apr 14 '16 at 17:41
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    And why do you believe such a proof using "only the integral representation" of the gamma functions exists? And "more advanced" may depend on the beholder – DonAntonio Apr 14 '16 at 17:42
  • Use the integral definition and functional equation to prove an infinite product representation or the reflection formula, and then ... – Robert Israel Apr 14 '16 at 17:44
  • Yes, I agree, and I know other formulas are only different point to start. I wrote "advanced" between inverted commas exactly for this reason.

    But now I need to show that Gamma has no zero using its integral representation. If it's not possible just tell me this.

     – Joe Apr 14 '16 at 17:46
  • Also Robert's suggestion uses the reflection formula or the product formula. If you want only the integral to appear it may prove to be pretty tough, if possible at all. – DonAntonio Apr 14 '16 at 18:00
  • Using only the integral representation requires showing that we cannot have simultaneously $$\int_0^\infty t^{x-1}e^{-t}\cos(y\log(t)),dt=0$$and $$\int_0^\infty t^{x-1}e^{-t}\sin(y\log(t)),dt=0$$for any $(x,y)$. That task looks challenging. – Mark Viola Apr 14 '16 at 18:03

2 Answers2

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Assume that $\Gamma(\alpha) = 0$ for some $\Re(\alpha) > 0$. Then for any $s \geq 0$, the substitution $ t = (1+s)x$ gives

$$ 0 = \frac{\Gamma(\alpha)}{(1+s)^{\alpha}} = \int_{0}^{\infty} x^{\alpha-1} e^{-x} e^{-sx} \, \mathrm{d}x. \tag{1}$$

  1. This is already enough to give a contradiction since the right-hand side is the Laplace transform of $x \mapsto x^{\alpha-1} e^{-x}$ and hence cannot be identically zero.

  2. If we avoid the use of Laplace transform, still we can derive a contradiction. Let $0 < \sigma < \Re(\alpha)$. For this $\sigma$, we know that $\Gamma(\sigma) > 0$. Then multiply both sides of $\text{(1)}$ by $s^{\sigma-1}/\Gamma(\sigma)$ and integrate w.r.t. $s$ on $[0, \infty)$. By the Fubini's theorem, this yields \begin{align*} 0 & = \int_{0}^{\infty} x^{\alpha-1} e^{-x} \left( \frac{1}{\Gamma(\sigma)} \int_{0}^{\infty} s^{\sigma-1} e^{-sx} \, \mathrm{d}y \right) \, \mathrm{d}x \\ &= \int_{0}^{\infty} x^{\alpha-\sigma-1} e^{-x} \, \mathrm{d}x \\ &= \Gamma(\alpha-\sigma). \end{align*} This shows that $\Gamma(z) = 0$ along the line segment joining $i\Im(\alpha)$ and $\alpha$. Then the identity theorem tells us that $\Gamma(z)$ is identically zero for $\Re(z) > 0$, which is impossible.

Sangchul Lee
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$$ \eqalign{ & \Gamma (z) = \int_{t\, = \,0}^{\,\infty } {e^{\, - z} t^{\,z - 1} dt} = \int_{t\, = \,0}^{\;1} {e^{\, - z} t^{\,z - 1} dt} + \int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt} = \cr & = \sum\nolimits_{t\, = \,0}^{\;\infty } {{{\left( { - 1} \right)^{\, - k} } \over {k!(z + k)}}} + \int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt} \cr} $$ where $\int_{t\, = \,1}^{\;\infty } {e^{\, - z} t^{\,z - 1} dt} $ is an entire function. This a well known development, you can find further details in many publications on Gamma function (e.g.: in "The theory of analytic functions, a brief course - A. I. Markushevich ")

G Cab
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