# How to evaluate this$\int_{0}^{c} y^{\alpha-1}(1-y)^{\beta-1}dy$?

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.

## 1 Answer

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.

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