Welcome to our community

Be a part of something great, join today!

Chamilka's Question from Math Help Forum.

Not open for further replies.


Well-known member
MHB Math Helper
Feb 5, 2012
Original Title: Please Need an help on this integral!!

chamilka said:
Hi everyone!!

I have to integrate the following function.


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


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, :)


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


\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


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,


\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)\\


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,


New member
Jul 3, 2012
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:

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!!


Staff member
Jan 26, 2012
Question answered it seems and there is a new thread for the other question so I'll close this thread now.
Not open for further replies.