MHB Chamilka's Question from Math Help Forum.

  • Thread starter Thread starter Sudharaka
  • Start date Start date
  • Tags Tags
    Forum
AI Thread Summary
Chamilka seeks assistance with the integral \[\int_{0}^{1}x^{a+b-1}(1-x)^{c-a}(1-x^b)^{d-1}\,dx\], suspecting it relates to the Beta function but noting its complexity due to multiple terms. A response explains that using the Binomial series expansion allows the integral to be evaluated term by term, leading to the result \[\sum_{i=0}^{\infty}(-1)^{i}\binom{d-1}{i}B(a+b+bi,\,c-a+1)\]. The conditions for convergence are also clarified, ensuring the integral is valid under specific parameters. Chamilka expresses gratitude for the clear proof and indicates a new question regarding Beta functions in a separate thread. The discussion concludes with the initial question resolved and a new topic introduced.
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.
 
Mathematics 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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top