How do I evaluate the following integral? $$ \int_{0}^{c} y^{\alpha-1} (1-y)^{\beta-1}dy $$ where $1\geq c>0$.

Thank you in advance.


As already pointed out by Somos within the comment section this is the so-called Incomplete Beta Function denoted as

$$B(z;a,b)=\int_0^z t^{a-1}(1-t)^{b-1}dt$$

Further note that for $z=1$ this equals the Beta Function $($as mentioned by Travis$)$

$$B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}dt$$

The Beta Function $-$ and the Incomplete Beta Function aswell $-$ are special functions and cannot be represented in terms of elementary functions $($such as polynomials, logarithms, etc.$)$ in a finite combination. So the integral

$$\int_0^c y^{\alpha-1}(1-y)^{\beta-1}dy$$

cannot be "evaluated" in the classical sense as long as the numbers $\alpha,\beta,c$ are choosen arbitrarily. You may look up the two links for further information especially concerning their different ways of representation.

  • 1
    $\begingroup$ @andrew No problem! Make sure to mark it as accepted if your are satisfied with the given answer :) $\endgroup$ – mrtaurho Dec 12 '18 at 21:30

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