How Do You Solve Gamma Functions Like \Gamma(5/4)?

[mjk1]

I'am not sure how to solve this gamma function \Gamma(5/4). any help ?

sorry if this is in the wrong section.

 

[mathman]

The Gamma function is the generalization of factoral. The simplest representation is an integral:

Gamma(x)=int(0,oo)t^x-1e^-tdt, for x>0.

It can be extended by analytic continuation.

 

[mjk1]

yeah the original integral was \int[0,4]Y^{3/2}(16-Y^{2})^{1/2} dy

which i simplified to

64\int(0,1) t^{(5/4)-1}(1-t)^{(3/2)-1} dt

using Gamma(x)=int(0,oo)tx-1e-tdt, =>> \beta(5/4, 3/2)hence i am trying to solve \beta(5/4, 3/2)
= \Gamma(5/4)\Gamma(3/2) / \Gamma((5/2)+(3/2))

i am trying to solve this to get an answer in terms of a number.

i don't know how to solve \Gamma(5/4)

 

[Santa1]

\Gamma\left(\frac{5}{4}\right) = \Gamma\left(1+\frac{1}{4}\right) = \frac{1}{4}\Gamma\left(\frac{1}{4}\right)

\Gamma\left(\frac{1}{4}\right) has no known basic expression, but is known to be transcendental.

 

[mjk1]

thanks for your help

 

[robert Ihnot]

I read that (1/q)! has no representation for integer q>2, except decimal form. But, if you are concerned with gamma(1/4), you can see it to 1,000,000 decimals at:

http://www.dd.chalmers.se/~frejohl/math/gamma14_1_000_000.txt

Working with Euler's reflection formula, I get

(1/4)!(3/4)! = \frac{3\pi\sqrt2}{16}=.833041

 

[mjk1]

how about \Gamma(0) I read somewhere that its a complex infinity but don't understand what that means

 

[robert Ihnot]

mjk1: how about (0)

Well, when you have a form like\int_{0}^{\infty} \frac{e^-x}{x} dx, you have trouble.

If you check out the Euler's reflection formula, we have


\Gamma(Z)\Gamma(1-Z) = \frac{\pi}{sin\pi(x)} for Z=0.