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Integrate the following $$\int_0^\infty \alpha\,\beta\, c\, k\, x^s\, x^{c-1} (1+x^c)^{k-1} \left[(1+x^c)^k-1\right]^{-\beta-1} \left[1+\gamma \left((1+x^c)^k-1\right)^{-\beta}\right]^{-(\alpha/\gamma)-1} \,dx$$

where $\alpha$, $\beta$, $\gamma$, $c$ and $k$ are positive real numbers and $s$ is positive integer.

This is how I have tried to solve it enter image description here

substituting enter image description here and putting beta and gamma equal to "1" and taking its derivative we have enter image description here

expanding the first term using binomial expansion we have enter image description here =enter image description here

substituting t-1=1/y we getenter image description here again substituting 1+y=1/v , we get enter image description here

the terms in the integral form beta function therefore

enter image description here

enter image description here is the solution I come up with. Can anyone hint another solution?

Note: When I put beta and gamma not equal to one I need to expand the term with "s" power twice using binomial expansion.

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  • $\begingroup$ If the asterisk is the usual product, it can be omitted... Was the formula copy-pasted from a math program? $\endgroup$ – ajotatxe Jan 28 '15 at 12:28
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    $\begingroup$ Under no circumstance use asterisks for multiplication (except in computer programming, where you don't have any other choice, of course). In mathematics, it most commonly denotes convolution, which is another type of beast entirely. If you need a multiplication sign, use \cdot ($\cdot$) instead. $\endgroup$ – Harald Hanche-Olsen Jan 28 '15 at 12:32
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    $\begingroup$ Surely, this cries out for subsituting $u=(1+x^c)^k-1$, does it not? $\endgroup$ – Harald Hanche-Olsen Jan 28 '15 at 12:34
  • $\begingroup$ Or even $u=((1+x^c)^k-1)^{-\beta}$. But I guess troubles come from the factor $x^s$. $\endgroup$ – Sary Jan 30 '15 at 12:39
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    $\begingroup$ When $\alpha/\gamma$ is an integer, the integrand is a rational fraction that should be manageable by residues. All poles can be found analytically I guess. In the non-integer case, the last factor generates a nasty branch point. $\endgroup$ – Yves Daoust Feb 5 '15 at 8:52
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This is how I have tried to solve it enter image description here

substituting enter image description here and putting beta and gamma equal to "1" and taking its derivative we have enter image description here

expanding the first term using binomial expansion we have enter image description here =enter image description here

substituting t-1=1/y we getenter image description here again substituting 1+y=1/v , we get enter image description here

the terms in the integral form beta function therefore

enter image description here is the solution I come up with. Can anyone hint another solution?

Note: When I put beta and gamma not equal to one I need to expand the term with "s" power twice using binomial expansion.

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  • $\begingroup$ You should post this in the body of the question. $\endgroup$ – Alex R. Feb 1 '15 at 7:19
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    $\begingroup$ You shouldn't have this in both your question and answer. Plus you have yet to give your bounty, so why have you already selected this as your answer? $\endgroup$ – Squirtle Feb 6 '15 at 3:03
  • $\begingroup$ @Squirtle since it seemed highly unlikely that anyone was giving a solution I accepted the answer that I posted. $\endgroup$ – SA-255525 Feb 6 '15 at 12:08
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    $\begingroup$ It's still not a good reason; if anything, this only prevented an answer from coming....why bother with a question if the question is already answered? This only hurts you. $\endgroup$ – Squirtle Feb 6 '15 at 22:52
  • $\begingroup$ Plus, this is not an answer since it soon assumes that beta = gamma = 1 although the question does not. $\endgroup$ – Did Feb 8 '15 at 17:44

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