Reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$? So I am trying to find the reduction formula for $\int_0^1 x^m(1-x)^{n-1}\,\mathrm{d}x$.
My attempt: Taking $x^m$ as the second function and integrating by parts, I got,
$I_n=\frac{(n-1)}{m+1}I_{n-2,m+1}$. But the given answer is $\frac{(n-1)!m!}{(m+n)}$. How do I go from my answer to the one given in the textbook? I tried manipulating my answer, but it isn't the same as the one in the textbook. Thanks in advance.
 A: I will assume that the integral in question is from $0$ to $1$. The reduction formula you obtained is correct, except for the fact that you should write $I_{n-1,m+1}$ instead of $I_{n-2,m+1}$. The answer in the textbook intends obviously to show the value of the integral (after the reduction is repeatedly done till $I_{0,m+n} $).
However there is an error in the textbook formula (if you correctly cited it). The correct one is
$$\frac{(n-1)!m!}{(m+n)\color{red}!}.
$$
A: Let me take $$J_{m,n}=\displaystyle\int_0^1x^m(1-x)^n\,dx.$$ This is really your $I_{m, n+1},$ but this renaming make it symmetric and easy to work with. For example, the substitution $x=1-t$ gives us $$J_{m,n}=J_{n, m}\tag1$$ which will transform as $I_{m, n+1}=I_{n, m+1}$ in to old notation. Note that $J_{m,0}=\dfrac{1}{m+1}$ for all $m\in\mathbb{N}$ (including zero). Letting $u=x^m$ and $dv=(1-x)^ndx$ we get $$J_{m,n}=\dfrac{m}{n+1}J_{m-1, n+1}\tag2$$ via integration by parts. Also because of $(1)$ identity, we have $$J_{m,n}=\dfrac{n}{m+1}J_{m+1, n-1}\tag3$$ for free. In fact, this is exactly what you get after integrating by parts with $u=(1-x)^n$ and $dv=x^mdx.$ Lets apply this third identity repeatedly as: $$J_{m,n}=\dfrac{n}{m+1}J_{m+1, n-1}=\dfrac{n}{m+1}\dfrac{n-1}{m+2}J_{m+2, n-2}$$ to derive $$J_{m,n}=\dfrac{n}{m+1}\dfrac{n-1}{m+2}\cdots\dfrac{1}{m+n}J_{m+n, 0}=\color{Green}{\dfrac{m!\,n!}{(m+n+1)!}}.$$ Now it is not difficult to rewrite this in your old notation.
