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Let $$\Gamma (z)= \int_{0}^{\infty} e^{-t}t^{z-1}dz$$ for $\Re z\gt 0$ Can be extended as a meromorphic function to the entire complex plane with simple poles at non positive integers.

How can I show this statement?

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  • $\begingroup$ The simplest way is by taking the product representation of $\Gamma$, and showing that both definitions agree say on $(0,1)$. $\endgroup$ – Daniel Fischer Jan 6 '14 at 0:01
  • $\begingroup$ Can you write write you said more explicitly please? @DanielFischer thank you. $\endgroup$ – user315 Jan 6 '14 at 0:06
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    $\begingroup$ I'm on my way to bed, sorry. If nobody did until then, I'll expand tomorrow. $\endgroup$ – Daniel Fischer Jan 6 '14 at 0:11
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The key is the following relation: $$ \Gamma(z) =\frac{\Gamma(z+n)}{ z(z+1)...(z+n-1)}$$ for all $z$ such that ${\Re}z>0$ and noticing that $\Gamma(z+n)$ is holomorphic for all $z$ such that ${\Re}z>-n$.

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  • $\begingroup$ Please can you expand your answer more? Thank you :) $\endgroup$ – user315 Jan 6 '14 at 0:50
  • $\begingroup$ Since you have asked this question, I assume you are familiar with the formula $\Gamma(x+1)=x\Gamma(x)$. Since you are self-learning, try to use induction to show that that formula implies $\Gamma(z+n)=(z+n-1)(z+n-2)\cdots(z+1)z\Gamma(z)$. Divide both sides by $(z+n-1)(z+n-2)\cdots(z+1)z$. $\endgroup$ – robjohn Jan 6 '14 at 1:11
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    $\begingroup$ @B11b See robjohn comment. Also, the right hand side function is holomorphic except for $0,-1,...,-n+1$, and it coincides with $\Gamma$ (on $\Gamma$'s domain). Hence its value does not depend on $n$ (due to the identity principle). Taking sufficiently large $n$, we can compute the value of the holomorphic continuation for all complex numbers except $0,-1,-2,...$. $\endgroup$ – ir7 Jan 6 '14 at 2:39
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The long-time standard idea is to integrate by parts, giving $\Gamma(z)=\Gamma(z+1)/z$. This extends $\Gamma(z)$ to the left by unit-width strips, inductively.

Historically, Euler's (and others') discussion of this extension of "factorial" often used the ("Hadamard") product expansion of the Gamma function, but it is not the simplest thing to prove that the integral has the appropriate product expansion, etc. True, it is interesting to understand this, probably best in the light of (post-Weierstrass) Hadamard products, rather than the (to my taste weird, immemorable) explicit manipulations of Whittaker-and-Watson. But the quickest more-or-less airtight proof is just integration by parts.

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Hint: use the functional equation $\Gamma(z+1) = z\Gamma(z)$ to repeatedly extend the Gamma function to a larger and larger domain...

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