# Negative integers in Euler's Reflection Formula?

Euler's reflection formula is $\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$, where $\Gamma \left({z}\right)$ is the Gamma-function, which is only defined for non-negative numbers. To me, it looks like the reflection formula must take negatives. How can this be? Is it because $z$ must be a complex number?

• Huh? $\Gamma(z)$ is defined for all complex numbers, except negative integers. – Gerry Myerson Sep 25 '15 at 7:29

The $\Gamma$ function can be extended to a meromorpphic function on the complex plane, with poles at the negative integers. The integral formula $$\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt$$ is absolutely convergent for $z\in\mathbb{C}$ with $\Re z>0$. The formula $$\Gamma(z+1)=z\,\Gamma(z),\text{ or }\Gamma(z)=\frac{\Gamma(z+1)}{z},$$ is then used to extend it to complex numbers with $\Re z\le0$.