Editing Gamma function (section)

=== Residues ===
The behavior for non-positive <math>z</math> is more intricate. Euler's integral does not converge for {{nowrap|<math>\Re(z) \le 0</math>,}} but the function it defines in the positive complex half-plane has a unique [[analytic continuation]] to the negative half-plane. One way to find that analytic continuation is to use Euler's integral for positive arguments and extend the domain to negative numbers by repeated application of the recurrence formula,<ref name="Davis" />
<math display="block">\Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n)},</math>
choosing <math>n</math> such that <math>z + n</math> is positive. The product in the denominator is zero when <math>z</math> equals any of the integers <math>0, -1, -2, \ldots</math>. Thus, the gamma function must be undefined at those points to avoid [[division by zero]]; it is a [[meromorphic function]] with [[simple pole]]s at the non-positive integers.<ref name="Davis" />

For a function <math>f</math> of a complex variable <math>z</math>, at a [[simple pole]] <math>c</math>, the [[Residue (complex analysis)|residue]] of <math>f</math> is given by:
<math display="block">\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).</math>

For the simple pole <math>z = -n</math>, the recurrence formula can be rewritten as:
<math display="block">(z+n) \Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n-1)}.</math>
The numerator at <math>z = -n,</math> is
<math display="block">\Gamma(z+n+1) = \Gamma(1) = 1</math>
and the denominator 
<math display="block">z(z+1)\cdots(z+n-1) = -n(1-n)\cdots(n-1-n) = (-1)^n n!.</math>
So the residues of the gamma function at those points are:<ref name="Mathworld">{{MathWorld|urlname=GammaFunction |title=Gamma Function}}</ref>
<math display="block">\operatorname{Res}(\Gamma,-n)=\frac{(-1)^n}{n!}.</math>The gamma function is non-zero everywhere along the real line, although it comes arbitrarily close to zero as {{math|''z'' → −∞}}. There is in fact no complex number <math>z</math> for which <math>\Gamma (z) = 0</math>, and hence the [[reciprocal gamma function]] <math display="inline">\frac {1}{\Gamma (z)}</math> is an [[entire function]], with zeros at <math>z = 0, -1, -2, \ldots</math>.<ref name="Davis" />