Chamilka's Question from Math Help Forum.

  • Context: MHB 
  • Thread starter Thread starter Sudharaka
  • Start date Start date
  • Tags Tags
    Forum
Click For Summary
SUMMARY

The integral \(\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx\) can be evaluated using the Binomial series expansion for \((1-x^b)^{d-1}\). The result is expressed as \(\sum_{i=0}^{\infty}(-1)^{i}\binom{d-1}{i}B(a+b+bi,\,c-a+1)\), where \(B\) denotes the Beta function. This evaluation requires the conditions \(Re(a+b+bi)>0\) and \(Re(c-a+1)>0\) to hold true. The discussion highlights the importance of series expansion techniques in solving complex integrals.

PREREQUISITES
  • Understanding of Beta functions and their properties
  • Familiarity with Binomial series expansions
  • Knowledge of complex analysis, particularly regarding convergence conditions
  • Basic calculus skills, specifically integration techniques
NEXT STEPS
  • Study the properties of the Beta function in detail
  • Learn about series expansion techniques in calculus
  • Explore convergence criteria for series in complex analysis
  • Investigate the Gaussian hypergeometric function and its applications
USEFUL FOR

Mathematicians, students studying advanced calculus, researchers in mathematical analysis, and anyone interested in integral evaluation techniques.

Sudharaka
Gold Member
MHB
Messages
1,558
Reaction score
1
Original Title: Please Need an help on this integral!

chamilka said:
Hi everyone!

I have to integrate the following function.

\[\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx\]

Here a,b,c and d are constants. I have to integrate the above function with respect to x in the region from zero to infinity. This may seem like a Beta integral function, but there is a slight change that there are three terms x, (1-x) and (1-x^b).

Also I have found in a journal article that the answer for the integral is given as below

\[\sum_{i=0}^{\infty}(-1)^{i}\binom{d-1}{i}B(a+b+bi,\,c-a+1)\]

But the evaluation methods are not given. They might used a series expansion, but nothing is given there.

Please help me on this problem.

Thank you .

Hi chamilka, :)

\[\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx\]

Using the Binomial series of \((1-x^b)^{d-1}\) we get,

\begin{eqnarray}

\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx&=&\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}\sum_{i=0}^{\infty} \; {d-1\choose i}\;(-x^b)^{i}\,dx\\

&=&\int_{0}^{1}\left(\sum_{i=0}^{\infty}(-1)^{i}{d-1\choose i}x^{a+b+bi-1}(1-x)^{c-a}\right)\,dx

\end{eqnarray}

The series, \(\displaystyle\sum_{i=0}^{\infty}(-1)^{i}{d-1\choose i}x^{a+b+bi-1}(1-x)^{c-a}\) is a power series and hence could be integrated term by term. Therefore,

\begin{eqnarray}

\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx&=&\sum_{i=0}^{\infty}\left(\int_{0}^{1}(-1)^{i}{d-1\choose i}x^{a+b+bi-1}(1-x)^{c-a}\,dx\right)\\

&=&\sum_{i=0}^{\infty}(-1)^{i}{d-1\choose i}\left(\int_{0}^{1}x^{(a+b+bi)-1}(1-x)^{(c-a+1)-1}\,dx\right)\\

\end{eqnarray}

By the definition of the Beta function, if \(Re(a+b+bi)>0\mbox{ and }Re(c-a+1)>0\) we get,

\[\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx=\sum_{i=0}^{\infty}(-1)^{i}{d-1\choose i}B(a+b+bi,\,c-a+1)\]

Kind Regards,
Sudharaka.
 
Physics news on Phys.org
Thank you Sudharaka. Here I got a more clear proof for my question. This is very great!

Btw I have posted a new post here It's regarding Beta functions. If you can please help me there too.. Thanks again.

Also quoted here:
chamilka said:
Hi everyone!
I got two versions of one particular function and now I need to show those two versions are equivalent.
For that I need to show the follwing, View attachment 223

Is it possible to show this by using the properties of Beta functions, Gaussian hypergeometric function etc?

Thanks in advance!
 
Question answered it seems and there is a new thread for the other question so I'll close this thread now.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K