To evaluate integral using Beta function - Which substitution should i use?

Question

$$\int_{0}^{1} \frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}dx = \frac{B(m,n)}{(a+b)^ma^n}$$

I have to use some kind of substitution but i do not understand what i use and why ?

Thanks

Answer 1

Let's try the substitution

$$x=\frac{1-y}{1+cy}$$

so that $1-x=(1+c)\frac{y}{1+cy}$.

Then, when $x=0$, $y=1$ and when $x=1$, $y=0$. We also have $dx=-(1+c)\frac{1}{(1+cy)^2}\,dy$. Then, we can write

$$\begin{align}\int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\int_0^1 \frac{\left(\frac{1-y}{1+cy}\right)^{m-1}\left((1+c)\frac{y}{1+cy}\right)^{n-1}}{\left(a+b\frac{1-y}{1+cy}\right)^{m+n}}\,(1+c)\frac{1}{(1+cy)^2}\,dy\\\\ &=\frac{(1+c)^n}{(a+b)^{n+m}}\int_0^1\frac{y^{n-1}(1-y)^{m-1}}{\left(1-\frac{b-ac}{a+b}y\right)^{m+n}} \end{align}$$

Choosing $c=b/a$, we obtain

$$\begin{align} \int_0^1\frac{x^{m-1}(1-x)^{n-1}}{(a+bx)^{m+n}}\,dx&=\frac{1}{a^n(a+b)^m}\int_0^1 y^{n-1}(1-y)^{m-1}\,dy\\\\ &=\frac{1}{a^n(a+b)^m}B(m,n) \end{align}$$

And we are done!