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I want to express the Integral $$\int_{z}^{1} x^{\alpha-2}(1-x)^{\beta-1}dx, \quad 0 < z < 1, \text{ and } \alpha,\beta > 0$$ as a more simple function of the constants $z,\alpha$ and $\beta$. Due to the truncation of the integral and the exponent on $x$ it differs from the usual Beta function.

Thank you in advance.

EDIT1: I tried using the incomplete beta function, i.e. \begin{align*} \int_{z}^{1} x^{\alpha-2}(1-x)^{\beta-1}dx &= \int_{0}^{1} x^{\alpha-2}(1-x)^{\beta-1}dx - \int_{0}^{z} x^{\alpha-2}(1-x)^{\beta-1}dx \\ &= B(\alpha-1,\beta)- B(z;\alpha-1,\beta)\end{align*}

However the solution requires me to pick $\alpha > 1$ and not $\alpha > 0$ else the integrals would not be convergent.

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    $\begingroup$ this integral can expressed by the Gamma and Beta function $\endgroup$ – Dr. Sonnhard Graubner Mar 20 '17 at 12:29
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    $\begingroup$ look at Incomplete beta function mathworld.wolfram.com/IncompleteBetaFunction.html $\endgroup$ – Zaid Alyafeai Mar 20 '17 at 12:55
  • $\begingroup$ You should not use the beta function. Just use the substitution $y =1-x $ and then use the Incomplete beta function. $\endgroup$ – Zaid Alyafeai Mar 20 '17 at 16:14
  • $\begingroup$ (+1) because you showed your attempt. I don't know why would someone downvote this! $\endgroup$ – Zaid Alyafeai Mar 20 '17 at 16:26

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