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?
 A: 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 $(1)$.
