# Homework Help: Gamma Function

1. Oct 9, 2011

### Ted123

If the Gamma function $$\Gamma (z) = \int_0^{\infty} t^{z-1} e^{-t}\;dt$$ only converges for $\text{Re}(z)>0$ then why is, for example, $\Gamma (-1+i)$ defined when clearly $\text{Re} (-1+i)<0$ ?

2. Oct 9, 2011

### lineintegral1

You are correct, the integral formula does not converge if the real part is less than one. This is why it is useful to use the analytic continuation of the gamma function. In other words, we can a way to extend the domain of the gamma function. Take a look at the following for more information (this is somewhat beyond my area of expertise, at the moment),

http://en.wikipedia.org/wiki/Gamma_function#The_gamma_function_in_the_complex_plane

3. Oct 9, 2011

### Ted123

The book that I've got says that the Gauss formula for complex parameters a,b,c: $$\displaystyle _2 F_1 (a,b;c;1) = \frac{\Gamma (c) \Gamma (c-a-b)}{\Gamma (c-a) \Gamma (c-b)}$$ is valid for $\text{Re}(c-a-b)>0,\;c\neq 0,-1,-2,-3,...$.

But if a=1, b=-0.6 and c=-0.5 for example then all the gamma functions appear to be defined (or aren't they?) yet $\text{Re}(c-a-b) = -0.9 \not > 0$ $$\displaystyle \frac{\Gamma (-0.5) \Gamma (-0.9)}{\Gamma (-1.5) \Gamma (0.1)}$$ seems to be all defined to me?

Are the conditions $\text{Re}(c-a-b)>0,\;c\neq 0,-1,-2,-3,...$ sufficient for the Gauss formula to be valid?

Last edited: Oct 9, 2011