Undergrad Formula for integration of natural coordinates over an element

[Arjan82]

TL;DR

For FEM analysis use is made of elements and their natural coordinates. For integration my book provides a neat formula, but unfortunately without source.

In a textbook I own a formula is given for the integration of natural coordinates over an element. In this case it is a 1 dimensional element (i.e. a line segment) with coordinates $x_i$ and $x_j$. The coordinate $x$ over the element is written as:
$$
x = L_1(x) x_i + L_2(x) x_j
$$

with
$$
L_1(x) = \frac{x_j - x}{L}\text{, }
L_2(x) = \frac{x - x_i}{L}
$$
with $L= x_j - x_i$

It is stated, unfortunately without source, that:
$$
\int_L L_1^\alpha(x) L_2^\beta(x) dx = \frac{\alpha!\beta!}{(\alpha + \beta +1)!}L
$$
It then continues to show an example where $x_i = 2$ and $x_j = 6$ so that $L_1(x) = (6-x)/4$ and $L_2(x) = (x-2)/4$ and then computes the integral:
$$
\int_2^6 L_1^2(x) L_2(x) dx = 0.333...
$$
Using the long-hand method and the formula stated above. Which obviously give the same result. But where does this formula come from? How is it derived? I want to know the limitations of this formula (which I assume there are).

 

[pasmith]

For \operatorname{Re}(\alpha) > -1, \operatorname{Re}(\beta) > -1 and x_j > x_i, substitute u = (x - x_i)/L to obtain
<br /> \frac{1}{L^{\alpha + \beta}}\int_{x_i}^{x_j} (x_j - x)^\alpha(x - x_i)^\beta\,dx <br /> =L\int_0^1 (1-u)^\alpha u^\beta\,du. The integral on the right is the definition of the Beta function B(\alpha + 1, \beta + 1). It can be shown from this and the integral representation <br /> \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\,dt valid for \operatorname{Re}(z) &gt; 0 that B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p + q)}. Lastly, if n \geq 0 is an integer then \Gamma(n + 1) = n!.

 

[Arjan82]

Thanks! That's enough hints for me to figure it out :)

 

[pasmith]

Alternatively, if \alpha and \beta are positive integers then set <br /> I(\alpha,\beta) = \int_0^1 (1-u)^\alpha u^\beta\,du and integrate by parts to obtain <br /> I(\alpha, \beta) = \frac{\beta}{\alpha + 1}I(\alpha+1,\beta-1) so that <br /> I(\alpha,\beta) = \frac{\alpha! \beta!}{(\alpha + \beta)!}I(\alpha + \beta,0) = \frac{\alpha! \beta!}{(\alpha + \beta + 1)!}.