Gamma Function Question

Question

I am relatively new to this idea. What values can use for x for the gamma function? For example would all complex numbers work?

Answer 1

If you use the integral representation

$$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$$

You will find that it only converges if the real part of $x$ is greater than zero. That is, for $x=a+bi$, we must have $a>0$, which is due to comparison with the p-series around $t=0$.

To extend to all complex values, use

$$\Gamma(x-1)=\frac{\Gamma(x)}{x-1}$$

with the exception that $x\notin\{0,-1,-2,\dots\}$, since clearly we will run into things like

$$\Gamma(0)=\frac{\Gamma(1)}0=\text{undefined}$$

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For representations of the Gamma function that converge for all $x\notin\{0,-1,-2,\dots\}$, see the wikipedia.

Answer 2

As said by @Simple Art, it can be defined as $$\Gamma(z)=\int_0^\infty t^{z-1}\mathrm e^{-z}\,\mathrm d\mkern 1mu t$$ for all $z$ with $\;\operatorname{Re} z>0$.

Then one can show this function has an analytic continuation on $\mathbf C\smallsetminus\{0,-1,\dots,-n,\dots\}$ which satisfies the functional relation: $$\Gamma(z+1)=z\Gamma(z).$$

Note: It also can be defined as $$\Gamma(z)=\lim_{n\to\infty}\frac{n!\,n^z}{z(z+1)\dotsm(z+n)}\qquad\qquad\text{ (Euler's definition)}$$ and as the infinite product: $$\Gamma(z)=\frac{\mathrm e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\frac{\mathrm e^{\tfrac zn}}{1+\dfrac zn}\qquad\qquad\text{(Weierstraß' definition)}$$ where $\gamma$ is Euler-Mascheroni's constant.