We first consider the integral:
ò 1 o x u-1 (1 – x) v-1 dx = b (u,v)
For Re u > 0, Re v > 0
Which is known as the Beta function.
This function is symmetric so can also be written:
ò 1 o y v-1 (1 – y) u-1 dx = b (v, u)
This can be expressed in trigonometric form by writing:
x = sin 2 q
Thence:
ò p/2 o sin 2u-2 q cos 2v-2 q 2 sin q cos q dq = b (u,v)
Or:
b (u,v) = 2 ò p/2 o sin 2u-1 q cos 2v-1 q dq
The earlier expressions for b (u,v) and b (v, u) represent analytic functions in each of the complex variables u and v. One can then introduce a new variable of integration: w = x / (1 - x) so the original integral becomes:
b (u,v) = ò ¥ o w u-1 dw / (1 + w) u+v
For Re u > 0, Re v > 0
Using suitable substitutions, we find that:
(1 + w) u+v =
1/ G(u + v) ò ¥ o exp (-1 + w) xu+v-1 dx
Where G(u + v) is the gamma function
Substituting the preceding expression (for (1 + w) u+v )
into: b (u,v) = ò ¥ o w u-1 dw / (1 + w) u+v
We get:
b (u,v) = G(u) / G(u + v) ò ¥ o e - x xv-1 dx = G(u) G(v) / G(u + v)
A useful identity.
Now, to fix ideas, consider the Beta function b(3, 4):
b(3, 4) = G (3) G (4)/ G (3 + 4) = (2) (6)/ G (7) = 12 / G (7)
Recall: G (a) = (a - 1)!
So: G (3) = (3 - 1)! = 2! = 2 x 1 = 2
G (4) = (4 - 1)! = 3! = 3 x 2 x 1 = 6
Then b(3, 4) = 12/ (7 - 1)! = 12/ (6)!
= 12/ (6 x 5 x 4 x 3 x 2 x 1) = 12/ 720 = 1/ 60
Problems for Math Wizards:
1) The error function erf x
= (2 /Ö p ) ò ¥ o exp ( -x 2) dx
The gamma function:
G (½ ) = ò ¥ o x -1/2 e - x dx
Write the relationship between erf x and G (½)
2) Given: ò 1 o x n dx /Ö ( 1 - x 2)
Find an integral expression for the Beta function: ½ b(n + ½, ½)
Hint: Let x = sin q
3) Astrophysicist Brian Greene in an episode of 'The Elegant Universe', wrote out a form of the Beta function used in string theory as:
b(p, q) = G (p) G (q)/ G (p + q)
Where: p = [1- a(s)] and q = [1 - a(t)]
Are string theory parameters
If: a(s) = ln e/ 20 and a(t) = p ln e/ 2
Find the applicable Beta function.
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