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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:

$\Gamma (1+\frac {1}{2})$

$\Gamma (-2+\frac {1}{2})$

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You can use the functional equation $ \Gamma(z+1)=z \, \Gamma(z). $ – user58512 Jan 29 '13 at 13:27
@user58512 this is my question infact i do not know how to use the functional equation. – Neo Jan 29 '13 at 13:29
I fail to see any equations at all in the post. – mrf Jan 29 '13 at 13:30
up vote 2 down vote accepted

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

$$ \Gamma\left(\frac{1}{2}-m\right) = \frac{\Gamma\left(\frac{1}{2}\right)}{(-1)^m\left(\frac{1}{2}\right)_{m}} = \frac{\sqrt{\pi}}{(-1)^m\left(\frac{1}{2}\right)_{m}}. $$

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

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From here:

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.

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Using the functional equation $\Gamma(z+1)=z\Gamma(z)$ would be much easier, no? – 1015 Jan 29 '13 at 14:52
@julien, you mean to show the identity for $\Gamma(z)\Gamma(z+0.5)$, I don't see how you can use the functional equation to show this identity. – MathematicalPhysicist Jan 30 '13 at 16:10
Well, $\Gamma(3/2)=(1/2)\Gamma(1/2)$ and so on by induction. Likewise $\Gamma(-1/2)=-2\Gamma(1/2)$ etc... – 1015 Jan 30 '13 at 16:52

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