I know that the Gamma function with argument $(-\frac{1}{ 2})$ -- in other words $\Gamma(-\frac{1}{2})$ is equal to $-2\pi^{1/2}$. However, the definition of $\Gamma(k)=\int_0^\infty t^{k-1}e^{-t}dt$ but how can $\Gamma(-\frac{1}{2})$ be obtained from the definition? WA says it does not converge... 

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    $\begingroup$ Yes, the integral diverges at $k=-1/2$. For values $\Gamma(z)$ with real part of $z \le 0$, you need analytic continuation. If you study complex analysis, you will learn about that. $\endgroup$ – GEdgar Mar 21 '12 at 21:04
  • $\begingroup$ @GEdgar: Thanks! I shall look forward to that :) $\endgroup$ – pomme Mar 21 '12 at 21:06
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    $\begingroup$ Basically the point is that the functional equation $z \Gamma(z) = \Gamma(z+1)$ lets you define $\Gamma(z)$ anywhere, even in the region where the integral doesn't converge, as long as you don't run into a division by $0$. So $\Gamma(-1/2) = -2 \Gamma(1/2)$ where $\Gamma(1/2) = \int_0^\infty t^{-1/2} e^{-t}\ dt$. $\endgroup$ – Robert Israel Mar 21 '12 at 21:12

Your doubt makes sense if for $k=-1/2$ you try using the definition of the gamma function by the integral you have written, because it diverges at $k=-1/2$ as you stated. I will try to clarify if as follows. This integral representation of the gamma function (I use $x$ instead of $k$)

$$\Gamma (x)=\int_{0}^{\infty }t^{x-1}e^{-t}dt\tag{0}$$

holds in the reals if and only if $x>0$. Using integration by parts we can show that $$ \Gamma (x+1)=x\Gamma (x).\tag{1}$$

For $x<0$ we can define $\Gamma (x)$ for all negative values of $x<0$ except $-1,-2,-3,\cdots $ not by the integral $(0)$, rather by means of the functional equation $(1)$ in the form $$ \Gamma (x)=\frac{\Gamma (x+1)}{x}.\tag{2}$$ Then $x+1>0$ $\Gamma (x+1)$ is convergent. In your example $x=-1/2$, so $x+1=1/2$ and $\Gamma (1+x)=\Gamma (1/2)$. Hence we obtain $$\Gamma (-1/2)=\frac{\Gamma (1/2)}{-1/2}=-2\Gamma (1/2)=-2\sqrt{\pi },\tag{3}$$ where we use the known value of the integral $\Gamma(1/2)=\sqrt{\pi }$, which can be evaluated, e.g. from the equality

$$\int_{0}^{\infty }\int_{0}^{\infty }e^{-x^{2}-y^{2}}dxdy=\frac{\pi }{4} =\left( \int_{0}^{\infty }e^{-x^{2}}dx\right) ^{2}.\tag{4}$$ Similarly, for $ x=-3/2$ by $(2)$ we find $\Gamma (-3/2)=-\frac{2}{3}\Gamma (-1/2)$, using $(3)$ twice.

This process is called analytic continuation, but its true understanding requires knowledge of complex analysis.

enter image description here

$$\text{Plot of }y=\Gamma(x)\quad -5<x<5$$


The functional equation $\Gamma(z+1)=z\Gamma(z)$ allows you to define $\Gamma(z)$ for all $z$ with real part greater than $-1$, other than $z=0$: just set $\Gamma(z) = \Gamma(z+1)/z$, and the integral definition of $\Gamma(z+1)$ does converge. The value $\Gamma(\frac12)=\sqrt\pi$ is well known and can be derived from Euler's reflection formula.

(Repeated use of the functional equation shows that $\Gamma$ can be defined for all complex numbers other than nonpositive integers.)


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