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- Feb 5, 2012

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**Original Title: Please Need an help on this integral!!**

Hi chamilka,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 .

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

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.