- #1

- 8

- 0

sorry if this is in the wrong section.

- Thread starter mjk1
- Start date

- #1

- 8

- 0

sorry if this is in the wrong section.

- #2

mathman

Science Advisor

- 7,888

- 458

Gamma(x)=int(0,oo)t

It can be extended by analytic continuation.

- #3

- 8

- 0

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

which i simplified to

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

using Gamma(x)=int(0,oo)tx-1e-tdt, =>> [tex]\beta[/tex](5/4, 3/2)

hence i am trying to solve [tex]\beta[/tex](5/4, 3/2)

= [tex]\Gamma[/tex](5/4)[tex]\Gamma[/tex](3/2) / [tex]\Gamma[/tex]((5/2)+(3/2))

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

i dont know how to solve [tex]\Gamma[/tex](5/4)

which i simplified to

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

using Gamma(x)=int(0,oo)tx-1e-tdt, =>> [tex]\beta[/tex](5/4, 3/2)

hence i am trying to solve [tex]\beta[/tex](5/4, 3/2)

= [tex]\Gamma[/tex](5/4)[tex]\Gamma[/tex](3/2) / [tex]\Gamma[/tex]((5/2)+(3/2))

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

i dont know how to solve [tex]\Gamma[/tex](5/4)

Last edited:

- #4

- 107

- 0

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

- #5

- 8

- 0

thanks for your help

- #6

- 1,056

- 0

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)! = [tex]\frac{3\pi\sqrt2}{16}=.833041[/tex]

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

Working with Euler's reflection formula, I get

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

Last edited by a moderator:

- #7

- 8

- 0

- #8

- 1,056

- 0

mjk1: how about (0)

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

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

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

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

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

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

Last edited:

- Replies
- 16

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 745

- Last Post

- Replies
- 8

- Views
- 3K

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 18

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 852

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 2K