Let $\alpha,\beta \in \mathbb{R}$. I was wondering if there is a systematic way to solve integrals of the following form:

\begin{equation} \int^1_{0} x^\alpha(1-x)^\beta dx \end{equation}

I have seen similar kind of integrals many times. Any help/hints would be really appreciated. Thanks!


Using the definition of the Beta Function: $$B(x,y)=\int_0^1t^{x-1}(1-t)^{1-y}dt$$ By matching coefficients,we would conclude that: $$\int_0^1x^\alpha(1-x)^\beta dx=B(\alpha+1,1-\beta)$$


Begin with the definition of the Gamma function:

$$\Gamma(\alpha+1) = \int_0^{\infty} du\; u^{\alpha} e^{-u} $$


$$\Gamma(\beta+1) = \int_0^{\infty} dv\; v^{\beta} e^{-v} $$

Multiply these together...

$$\begin{align}\Gamma(\alpha+1) \Gamma(\beta+1) &= \int_0^{\infty} du\; \int_0^{\infty} dv\; u^{\alpha} v^{\beta} e^{-u-v}\\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; \int_{-p}^p dq \, (p+q)^{\alpha} (p-q)^{\beta} e^{-p} \\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; p^{\alpha+\beta+1} e^{-p} \, \int_{-1}^1 dr \, (1+r)^{\alpha} (1-r)^{\beta}\\ &= \Gamma(\alpha+\beta+2) \frac1{2^{\alpha+\beta+1}} \int_0^2 dr \, r^{\alpha} (2-r)^{\beta} \\ &= \Gamma(\alpha+\beta+2) \int_0^1 dx \, x^{\alpha} (1-x)^{\beta}\end{align}$$

In the second line, I used the transformation $p=u+v$, $q=u-v$, which has Jacobian $1/2$. Thus,

$$\int_0^1 dx \, x^{\alpha} (1-x)^{\beta} = \frac{\Gamma(\alpha+1) \Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.