Convolution theorem problem

Question

Use the convolution theorem to show that $$\int_{0}^1 u^m(1-u)^ndu = \frac{m!n!}{(m+n+1)!}$$ I don't know where to start on this, do I have you use commutative property of convolution?

Answer 1

Let $I(m,n)$ be LHS and $J(m,n)$ be RHS of equality. We have $I(m,0)=J(m,0)=\frac1{m+1}$ Integrating by parts gives $I(m,n)=\frac{n}{m+1}I(m+1,n-1).$

It is evident that $J(m,n)=\frac{n}{m+1}J(m+1,n-1).$ So $I$ and $J$ have the same recursion formula and the same anitial conditions $I(m,0)=J(m,0)$. Therefore they must be equal.

Answer 2

With the substitution $u=e^{-x}$ the problem boils down to finding $$ I(m,n) = \int_{0}^{+\infty}(1-e^{-x})^m e^{-(n+1)x}\,dx $$ that for a fixed $m\in\mathbb{N}$, due to the binomial theorem, equals $$ g(n)=\sum_{k=0}^{m}\binom{m}{k}\frac{(-1)^k}{n+1+k}=\frac{m!}{(m+n+1)(m+n)\cdots(1+n)} $$ where the last identity follows from the fact that $\frac{m!}{(m+n+1)(m+n)\cdots(1+n)}$ and $\sum_{k=0}^{m}\binom{m}{k}\frac{(-1)^k}{n+1+k}$ share the same simple poles and the same residues.