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I want to find a simple expression for the following integral :

$ F(x)=\displaystyle \int_0^x \dfrac{ay^{a\alpha -1}\left[ 1-(1-c)y^a \right]^{\beta -1}}{B(\alpha,\beta)\left[1+cy^a\right]^{\alpha +\beta}}dy$

If I put $(1-c)y^a=t$, then I get

$ F(x)=\dfrac{1}{(1-c)^{\alpha}B(\alpha,\beta)}\displaystyle \int_0^{(1-c)x^a}t^{\alpha-1}(1-t)^{\gamma - \alpha -1}\left[ 1-\dfrac{c}{1-c}t \right]^{-\gamma}dt$

where $\gamma=\alpha +\beta$.

My aim is to put the integral in the RHS in the form of a hypergeometric function as it looks similar to the same :

$ {}_2F_1 (a,b,c;z)=\dfrac{1}{B(b,c-b)}\displaystyle \int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx$

This is where I am stuck. Any ideas ?

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Edit

I change my answer to a hint: You could start by doing the substitution $$ u=\frac{(1-c)(1-t)}{1-c-ct}. $$ It will transform the integral $$ \int t^{\alpha-1}(1-t)^{\gamma-\alpha-1}\Bigl[1-\frac{c}{1-c}t\Bigr]^{-\gamma}\,dt $$ into (modulo a multiplicative constant, I hope I did the calculations correctly) $$ \int (1-u)^{a-1}u^{\alpha+\gamma-1}\,du. $$ This will lead you to an incomplete beta function, and then you can use equation (3) at this site to transform it into the hypergeometric function you want.

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  • $\begingroup$ Kindly elaborate a bit $\endgroup$
    – Debashish
    Sep 9 '15 at 7:51
  • $\begingroup$ Thanks a lot. It works fine. $\endgroup$
    – Debashish
    Sep 10 '15 at 6:30

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