Chamilka's Question from Math Help Forum.

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In summary, Chamilka's Question from Math Help Forum is a specific and potentially challenging math problem posted on the website. To solve it, one must carefully analyze and understand the question, identify given information, and apply appropriate mathematical techniques. Asking for help on the forum is allowed, but it is important to show effort and provide relevant information. Some tips for approaching the question include breaking it into smaller parts, using visual aids, and seeking different perspectives.
  • #1
Sudharaka
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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.
 
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  • #2
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!
 
  • #3
Question answered it seems and there is a new thread for the other question so I'll close this thread now.
 

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1. What is Chamilka's Question from Math Help Forum?

Chamilka's Question from Math Help Forum is a specific math problem or question that has been posed on the Math Help Forum website.

2. What makes Chamilka's Question unique or challenging?

Chamilka's Question may be unique or challenging because it requires a deep understanding of mathematical principles or involves multiple steps and concepts.

3. How can I solve Chamilka's Question?

To solve Chamilka's Question, it is important to carefully read and understand the question, identify any given information or constraints, and use appropriate mathematical techniques or formulas to arrive at a solution.

4. Can I ask for help with Chamilka's Question on the Math Help Forum?

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5. Are there any tips for approaching Chamilka's Question from Math Help Forum?

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