I am interested in computing the following integral
$$ I=\int_{-1}^{1}\mathrm{d} u~_2F_1\left(-k,\alpha+k-\frac{1}{2};\alpha, 1-u^2\right)(1-u^2)^{\alpha-1}u^{2m} $$
where $_2F_1$ is the standard hypergeometric function, $k,m$ are positive integers and $\alpha>0$. I want to prove that it's equal to $0$ for $m>k$.
I wrote the hypergeometric function as a polynomial, $$ ~_2F_1\left(-k,\alpha+k-\frac{1}{2};\alpha, t\right)=\sum_{l=0}^{k}\binom{k}{l}(-1)^l \frac{\Gamma(\alpha-1/2+k+l)}{\Gamma(\alpha-1/2+k)}\frac{\Gamma(\alpha)}{\Gamma(\alpha+l)}t^l $$
and got the expression
$$ I=\frac{\Gamma(\alpha)\Gamma(1/2+m)}{\Gamma(\beta+k)}\sum_{l=0}^{k}\binom{k}{l}(-1)^l \frac{\Gamma(\beta+k+l)}{\Gamma(\beta+1+l+m)} $$ where $\beta:=\alpha-1/2$.
I am now left with trying to prove that the expression $$ A = \sum_{l=0}^{k}\binom{k}{l}(-1)^l \frac{\Gamma(\beta+k+l)}{\Gamma(\beta+1+l+m)} = \sum_{l=0}^{k}\binom{k}{l}(-1)^l \frac{\prod_{j=0}^{k-1}(\beta+l+j)}{\prod_{i=0}^{m}(\beta+l+i)} $$ equals $0$ when $m>k$. In this case it can also be written as $$ A = \sum_{l=0}^{k}\binom{k}{l}(-1)^l \frac{1}{(\beta+l+k)\ldots(\beta+l+m)} $$
So far I've tried representing $A$ as the $m-k$th integral between $0$ and $1$ of the polynomial
$$ P(X) = X^{\beta+k}(1-X)^k = \sum_{l=0}^k \binom{k}{l}(-1)^l X^{\beta+l+k} $$ and it's clear that the right hand side of this expression equals $A$ when integrated $m-k$ times. It isn't however clear why this is $0$.
I'm not sure I've tried the right method for this. Perhaps this is a known integral for hypergeometric functions, or a known sum for Gamma functions?