Definition of the gamma function for non-integer negative values

Question

The gamma function is defined as $$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$$ for $x>0$.

Through integration by parts, it can be shown that for $x>0$, $$\Gamma(x)=\frac{1}{x}\Gamma(x+1).$$

Now, my textbook says we can use this definition to define $\Gamma(x)$ for non-integer negative values. I don't understand why. The latter definition was derived by assuming $x>0$. So shouldn't the whole definition not be valid for any $x$ value less than zero?

P.S. I have read other mathematical sources and most of them explain things in mathematical terms that are beyond my level. It would be appreciated if things could be kept in relatively simple terms.

Answer 1

The definition you gave is valid only for $x >0$, has you have pointed out. However, you can extend $\Gamma$ to negative non integer values by defining

$$\Gamma(x) := \frac{1}{x}\Gamma(x+1) $$

whenever $x <0$, $x \notin \mathbb Z$. For example you get

$$\Gamma\left(-\frac{1}{2}\right) = -2 \Gamma\left(\frac{1}{2}\right) $$

and $\Gamma\left(\frac{1}{2}\right)$ is given by the integral you have presented.

Answer 2

Note that the gamma function has simple poles at all the negative integers, so it cannot be defined there. If you do not understand this, you may want to find a book on introductory complex analysis. As far as the analytic continuation of the gamma function goes, it should be obvious that the functional equation :

$$\Gamma(x) := \frac{1}{x}\Gamma(x+1)$$

does the job fairly easily.

Answer 3

The integral $\int_0^\infty t^{x-1}e^{-t}\,dt$ is a "representation" of the Gamma function for $x>0$. That is, for $x>0$

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

But the Gamma Function exists for all complex values of $x$ provided $x$ is not $0$ or a negative integer.

The idea of "representing" a function on a subset of the domain of definition is introduced in elementary calculus courses. For example, we can represent the function $f(x)=\log(1+x)$ as the series

$$f(x)=\log(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n} \tag 1$$

which is valid for $-1<x\le 1$. However, $f(x)$ exists for all $-1<x$.

Answer 4

To add on the other answers :

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This is one of the 1st example of analytic continuation. It is clear that $$\Gamma(z) = \frac{\Gamma(z+1)}{z}=\frac{\Gamma(z+2)}{z(z+1)}=\frac{\Gamma(z+n+1)}{z(z+1)\ldots(z+n)}$$ makes $\Gamma(z)$ well-defined for $z \in \mathbb{C}, -z \not \in \mathbb{N}$.

But it is not so obvious (without a lot of theorems in complex analysis) that this continuation is the only one being analytic, in the same way that $\frac{1}{1-z}$ is the only one analytic continuation of $\sum_{n=0}^\infty z^n$ beyond $|z|< 1$