
#1
Dec2005, 10:35 AM

P: 19

Hiya,
I'm having trouble finding a simple proof for gamma of 1/2 = root pi? Any suggestions 



#2
Dec2005, 11:23 AM

Admin
P: 21,637

One approach would be to show
[itex]\Gamma(x)\Gamma(1x)[/itex] = [itex]\frac{\pi}{sin(\pi x)}[/itex] then let x = 1/2. 



#3
Dec2005, 02:11 PM

HW Helper
P: 1,024

If you know the Betafunction too, and the relation with the Gammafunction:
[tex]B\left( {u,v} \right) = \frac{{\Gamma \left( u \right)\Gamma \left( v \right)}}{{\Gamma \left( {u + v} \right)}}[/tex] Then it's easy to use the definition of the Betafunction to compute B(1/2,1/2) which gives [itex]\pi[/itex], so: [tex]B\left( {\frac{1}{2},\frac{1}{2}} \right) = \pi = \frac{{\Gamma \left( {\frac{1}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma \left( 1 \right)}} = \Gamma \left( {\frac{1}{2}} \right)^2 \Leftrightarrow \Gamma \left( {\frac{1}{2}} \right) = \sqrt \pi [/tex] 



#4
Dec2005, 03:27 PM

P: 130

Gamma Function, Gamma 1/2=root pi
Oh I use to tell people that (1/2)!^2=pi and then they'd ask me to show them why so I kept this one in memory. (of course this doesn't prove that (1/2)!^2 = pi since factorial isn't really defined on nonnegative nonnatural numbers).
Start with the Gamma Function: [tex]\Gamma (x)= \int \limits_0^\infty \exp (t) t^{x1} dt [/tex] [tex]\Gamma (\frac{1}{2})= \int \limits_0^\infty \exp (t) t^{\frac{1}{2}} dt[/tex] Then make the following substitution: [tex]u=t^{\frac{1}{2}} [/tex] [tex]du=\frac{1}{2}t^{\frac{1}{2}}dt \Rightarrow t^{\frac{1}{2}}dt=2du[/tex] And the original equation becomes: [tex]\Gamma (\frac{1}{2})= 2 \int \limits_0^\infty \exp (u^2) du[/tex] The next part is the 'trick.' The trick is to then square gamma so you have two integrals with two different variables of integration (say, x and y). Because of their form, you should be able to combine them into one integral and then change it into another twovariable coordinate system where pi's are used. Becareful when changing limits (hint: when you integrate over two variables from 0 to infinity, you are effectively integrating over the first quadrant. Therefore, you should change your limits to your new coordinate system to make sure you are also integrating over the first quadrant) (EDIT from before: sorry I forgot this was the homework help section; As you can see, I've curtailed my answer ) 



#5
Dec2005, 04:30 PM

P: 789

Yeah, you'll need to switch to polar coordinates and the evaluating the double integral is pretty straightfoward. You can Google "Gaussian Integral" to see the technique used.



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