# On definition of gamma function.

We all know that gamma function's definition is $$\Gamma \left( x \right) = \int\limits_0^\infty {s^{x - 1} e^{ - s} ds}$$ and it is divergent for $x<0$.

Yesterday, I was studying about Bessel function and i came up with a dilemma. In Bessel function for negative numbers, i.e. $J_{-\alpha}$ they use a term like $\Gamma(m-\alpha+1)$ and when $m<\alpha-1$ series is defined although gamma function is undefined. Also in class there were questions like finding $\Gamma\left(-\frac{1}{2}\right)$ where it is a definite value? $$\Gamma\left(-\frac{1}{2}\right)=-2\sqrt{\pi}$$ but how is it well-defined, isn't it diverging? And how is it a negative number?

Summarizing, how exactly is gamma function defined?

• No. It is defined for $x<0$. It just has poles on $-\Bbb N$, the negative integers. – Ali Caglayan Sep 20 '14 at 16:29
• If you are asking why it works then there are many good questions on MSE to answer that. This essentially becomes a duplicate question. – Ali Caglayan Sep 20 '14 at 16:39
• It is defined as an analytic continuation. For example for negative $t$ you can always have $\Gamma (t)=\Gamma(t+n)/(t(t+1)\ldots(t-n+1))$ for large enough $n$. – Artem Sep 20 '14 at 16:40
• @Alizter: What software did you use to plot the beautiful graphic from your now-deleted answer to this question? – Lucian Sep 21 '14 at 6:30
• @Lucian Google images – Ali Caglayan Sep 21 '14 at 10:05

You are right. The integral

$$\Gamma \left( x \right) = \int\limits_0^\infty {s^{x - 1} e^{ - s} ds},\,\,x\in\mathbb{R}\tag{1}$$

converges only if $x>0$. Therefore the definition only works for positive $x$. However, one can use the functional equation $$\Gamma(x+1)=x\Gamma(x)$$

to give a meaning to $\Gamma(x)$ also when $x$ is negative. Namely, suppose $-1<x<0$. Then $x+1>0$, so $\Gamma(x+1)$ is defined by $(1)$. Now we define $\Gamma(x)$ (for which the formula $(1)$ doesn't make sense) as

$$\Gamma(x):=\frac{\Gamma(x+1)}{x}\tag{2}$$

You can already see that this definition makes no sense for $x=0$. In the same way we can define $\Gamma(x)$ for all $x\in\mathbb{R}$ (except $0,-1,-2,\dots$) Namely, if $-2<x<-1$, then $-1<x+1<0$, so we already know what $\Gamma(x+1)$ is by $(2)$, hence

$$\Gamma(x):=\frac{\Gamma(x+1)}{x}=\frac{\Gamma(x+2)}{x(x+1)}$$

In general, if $x>-n$, then

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

So you were interested in $\Gamma(-1/2)$. We get,

$$\Gamma(-1/2)=-2\Gamma(1/2)$$

which you can now calculate using $(1)$.

• so it is defined as a recurring relation?? – Hero_Zero Sep 20 '14 at 16:49
• This is one possible definition and the easiest. The nice thing is, which you will understand if you study complex analysis, is that no matter how you extend the definition of $\Gamma(x)$, as long as it is consistent and reasonable in some sense, the resulting function will be the same. This is called analytic continuation. – J.R. Sep 20 '14 at 16:51