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

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To solve the gamma function \Gamma(5/4), it is essential to understand that the gamma function generalizes the factorial and can be expressed through an integral representation. The discussion highlights the use of the integral \Gamma(x) = ∫(0,∞) t^(x-1)e^(-t) dt for x > 0, and how it can be simplified using the beta function. The relationship \beta(5/4, 3/2) = \Gamma(5/4)Γ(3/2)/Γ(4) is explored, but \Gamma(5/4) itself does not have a simple expression, as \Gamma(1/4) is known to be transcendental. Additionally, Euler's reflection formula is mentioned, which relates gamma functions at complementary arguments, providing insights into the nature of \Gamma(0) and its complexities. Understanding these relationships is crucial for evaluating gamma functions effectively.
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I'am not sure how to solve this gamma function \Gamma(5/4). any help ?

sorry if this is in the wrong section.
 
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The Gamma function is the generalization of factoral. The simplest representation is an integral:

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

It can be extended by analytic continuation.
 
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)
 
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\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.
 
thanks for your help
 
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
 
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how about \Gamma(0) I read somewhere that its a complex infinity but don't understand what that means
 
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.
 
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