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...
Definition of the gamma function
4
$\begingroup$
calculus
definite-integrals
improper-integrals
special-functions
gamma-function
-
1Yes, 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. – 2012-03-21
-
0@GEdgar: Thanks! I shall look forward to that :) – 2012-03-21
-
1Basically 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 $\Gamma(1/2) = \int_0^\infty t^{-1/2} e^{-t}\ dt$. – 2012-03-21