Math Help Forum - View Single Post - Integration of Incomplete Beta Functions and Hypergeometric Functions

shawsend · #3 November 6th, 2009, 06:05 PM

Here's what I believe are the integrals for others in case you're interested:

$I_1=\int_0^1 \frac{(ab)^2 v^{r+2b+s}(1-v)}{r+b+1}dv$

$I_2=\int_0^1 \frac{abcd v^{r+b+s+1}(1-v)^d}{r+b+1}dv$

$I_3=\int_0^{1} abcd v^{s+b-1}(1-v) \text{B}(v;r+2,d)dv$

$I_4=\int_0^{1} (cd)^2 v^s(1-v)^d \text{B}(v;r+2,d)dv$

where $\text{B}(v;s,t)$ is the incomplete Beta function

I went through the first one and apparently r,b s are integers since you're taking factorials and I get the same answer as you. You write further down in solving for the second one:

"However this is not the 5th term of I that I was asked to show. If you look closely, you will see that in the denominator in the Gamma function of my working there is a 3 instead of a 2. I just can't see where I'm going wrong."

I get a 3 also and so does Mathematica:

Code:

Integrate[((a*b*c*d)/(r + b + 1))*
   v^(r + b + s + 1)*(1 - v)^d, {v, 0, 1}]

(1/(1 + b + r))*(a*b*c*d*
   If[Re[d] > -1 && Re[b + r + s] > -2, 
    (Gamma[1 + d]*Gamma[2 + b + r + s])/
     Gamma[3 + b + d + r + s], 
    Integrate[(1 - v)^d*
      v^(1 + b + r + s), {v, 0, 1}, 
     Assumptions -> Re[d] <= -1 || 
       Re[b + r + s] <= -2]])

Therefore, I believe your initial answer is incorrect and the 2 should be changed to a 3. And as far as integrating the beta function, what about setting it up as a double integral and switching the order of integration?

$I_3=\int_0^{1} abcd v^{s+b-1}(1-v) \text{B}(v;r+2,d)dv=$ $abcd\int_0^1 v^{s+b-1}(1-v)\int_0^v u^{r+1}(1-u)^{d-1}dudv$