can anyone help me with a tough integral?

[Silent Acorns]

I've been trying to solve the following integral:

Given:

f(u)=B*(u^n)*(1-u)^(B-1) where n and B can be any positive real number

find the integral from 0 to 1 of f(u)du.

If I limit myself to integer values of B and n I noticed that the solutions seem to follow the following pattern:

Integral = n!B!/(n+B)!

This is certainly true for B=1,2, 3, and 4 (and even B = 0). Unfortunately, as B gets large, expanding f(u), integrating it, and making all the cancelations gets real tedious, real fast.

I was wondering two things:

1) can anyone show that the pattern I found works for all positive integer values of n and B?
2) If so, do you think I could generalize to all positive real numbers by using the gamma function?

[Friar Bellows]

I think you could use mathematical induction to prove it true for all integers B > 0. But then I don't think you could generalise this to real numbers.

[Tetlepanquetzatzin]

Originally posted by Silent Acorns
1) can anyone show that the pattern I found works for all positive integer values of n and B?
The integral depends on n and B so let's denote it I(n,B). It is easy to show using partial integration that, for n > -1 and B > 1,

I(n,B) = B/(n+1) I(n+1,B-1).

Your hypothesis is at least consistent with this recursion relation.
2) If so, do you think I could generalize to all positive real numbers by using the gamma function?
I have tested your generalization

I(n,B) = gamma(1+n) gamma(1+B) / gamma(1+n+B)

for various values of n and B and it works. Thus, I consider the equation to be established "empirically". I think it holds for all n and B, for which I(n,B) converges, i.e. for all n > -1 and B > 0.

[Tat]

What you seem to have hold of there is the Beta function:

For real positive x and y:
beta(x,y)
=Intergal{0 to 1}(u^(x-1)*(1-u)^(y-1)du)
=Gamma(x)*Gamma(y)/Gamma(x+y)

Proof of the second equality can be found in Adavanced Calculus by Taylor and Mann, and no doubt elsewhere.

There is some symbol pushing to get your expression to look like the beta function above. The trick I used was to use parts once, and rewrite so that you have a integral in the form of the beta function, and an integral which looks like the original. Solve that algebraically for the integral.

It is relativly strightforward, but I took the liberty of attaching (hopefully), the details above.

[lpetrich]

This problem,

B(a,b) = Integral(u^(a-1)*(1-u)^(b-1),u,(0,1))

can be solved by a very nice trick.

Multiply it by Integral(exp(-x)*x^(a+b-1),x,(0,inf))

and one gets a double integral:

Integral(exp(-x)*x^(a+b-1)*u^(a-1)*(1-u)^(b-1),(x,y),((0,1),(0,inf)))

The form suggests defining new variables:

y1 = u*x
y2 = (1-u)*x

The "d" parts are related by

dy1*dy2 = J((y1,y2),(u,x))*du*dx
= x*du*dx

where J is the Jacobian: J(y,x) = det(d(y)/d(x))

where d(y)/d(x) is expanded in components:

dy1/dx1 dy1/dx2
dy2/dx1 dy2/dx2

etc.

Thus,

B(a,b)*Integral(exp(-x)*x^(a+b-1),x,(0,inf)) =

Integral(exp(-x)*x^(a+b-2)*u^(a-1)*v(b-1),(y1,y2),((0,inf),(0,inf)))

or after changing over to y1 and y2,

Integral(exp(-y1)*y1^(a-1)*exp(-y2)*y2(b-1),(y1,y2),((0,inf),(0,inf)))

These are all gamma-function integrals, and

B(a,b)*G(a+b) = G(a)*G(b)

or B(a,b) = G(a)*G(b)/G(a+b)

which is what we were trying to prove.

[Silent Acorns]

Thanks to everyone, especially Tat, for finding the answer to my question. It's going to come in real handy.

[DigitalChicken]

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The only place I can read math problems!

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DC