how to solve this gamma function i know that $\Gamma (\frac {1}{2})=\sqrt \pi$ But I do not understand how to solve these equations
$$\Gamma (m+\frac {1}{2})$$
$$\Gamma (-m+\frac {1}{2})$$
are there any general relation to solve them
for example:
$\Gamma (1+\frac {1}{2})$
$\Gamma (-2+\frac {1}{2})$
 A: By the functional equation $$\Gamma(z+1)=z \, \Gamma(z)$$ (which is easily proved for the integral definition of $\Gamma$ by parts, and may be used to analytically continue the function to the negative numbers) we find $$\Gamma(1 + \frac{1}{2}) = \frac{1}{2} \Gamma(\frac{1}{2})$$ and $$\Gamma (-2+\frac {1}{2}) = \frac{1}{-1+\frac {1}{2}}\Gamma (-1+\frac {1}{2}) = \frac{2}{-1+\frac {1}{2}}\Gamma (\frac{1}{2})$$
A: You can use the Pochhammer symbol
$$ \left(a\right)_{m} = \frac{\Gamma(m+a)}{\Gamma(a)},\quad \left(a\right)_{m}=(a)(a+1)\dots(a+m-1) $$
to write $\Gamma\left(m+\frac{1}{2}\right)$ as
$$\Gamma\left(m+\frac{1}{2}\right) = \Gamma\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)_{m} =\sqrt{\pi}\left(\frac{1}{2}\right)_{m}.$$
Added: If you put some effort you can write $\Gamma\left(\frac{1}{2}-m\right)$ in terms of the Pochhammer symbol as
$$ \Gamma\left(\frac{1}{2}-m\right) = \frac{\Gamma\left(\frac{1}{2}\right)}{(-1)^m\left(\frac{1}{2}\right)_{m}} = \frac{\sqrt{\pi}}{(-1)^m\left(\frac{1}{2}\right)_{m}}. $$
A: From here:
http://en.wikipedia.org/wiki/Gamma_function
you have the next relation:
$$\Gamma(z)\Gamma(z+0.5)=2^{1-2z}\sqrt{\pi} \Gamma(2z)$$
Plug $z=m$ for an integer $m$.
If you want a proof of the above relation is better to use the product definition of the gamma function.
