# Definition of the gamma function for non-integer negative values

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.

• The integral definition is not violated if we assume (in addition) $\Gamma(x)=\frac{\Gamma(x)}{x}$ is valid for all non-integer negative values. – user368558 Mar 26 '17 at 17:14
• @YifanWei ITYM $\Gamma(x+1)/x$ to the right of =. – Jens Mar 27 '17 at 7:22
• @Jens you're right, sorry for the typo... – user368558 Mar 27 '17 at 17:22
• You can look at the concepts developed by van der Corput i.e Neutrix Calculus. Brian Fisher in 1987 and 2012 and in other publications defined the Gamma function for negative values of x using the notions of neutrix calculus. Am still trying to grasp the concepts. – Gregory Abe-i-kpeng May 20 at 19:20

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.

• That a very nice explanation. Let me expand upon it. Nobody says $x$ has to be real. – Cye Waldman Mar 26 '17 at 17:25
• @Ptheguy As a wise mathematician once said to me, "We can't argue with definitions.". We can define pretty much anything we want, as long as we are consistent. The definition in question is consistent, as it matches the result, given by the integration by parts formula, when $x$ is positive. – Stefano Mar 26 '17 at 17:34
• @Ptheguy because the original definition is "what this function is for $x>0$". It doesn't say "for $x<0$ the function mustn't be defined", it rather doesn't say anything at all for this range of $x$. Then we extend the domain of the function by imposing a property, which already holds for $x>0$, on the function on the whole complex plane (except nonpositive integers). We can then prove that this gives us the analytic continuation of original function. – Ruslan Mar 26 '17 at 21:04
• This is almost the same thing as when you say "by $a^n$, we mean the product of $n$ factors equal to $a$ if $n > 0$; $1$ if $n = 0$; and $1/a^{-n}$ if $n < 0$." The first part of the gives the natural definition of $a^n$ when $n$ is positive. This definition wouldn't make any sense for $n \leq 0$. So you extend the definition of the symbol $a^n$ to these values of $n$ by giving a different definition in those cases. And this extension is in fact the only way $a^n$ can be defined for those values of $n$ so that the relation $a^m a^n = a^{m+n}$ continues to hold. Analogously, in your case... – user49640 Mar 27 '17 at 5:07
• you extend the definition of $\Gamma(x)$ to negative values of $x$ in the only way that works if you want the relation $\Gamma(x+1) = x \Gamma(x)$ to continue to hold. – user49640 Mar 27 '17 at 5:08

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 $-1<x$.

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 $|z|< 1$

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.

• I understand the simple poles. My question is how can this new definition that is originally derived from a definition assuming $x>0$ apply to $x<0$. Shouldn't the previous assumptions carry on? – Ptheguy Mar 26 '17 at 17:25
• The reflection formula allows you to define it. – John Kontol Mar 26 '17 at 17:30