Check this value of $\int_{0}^{x}\frac{t^m}{(x-t)^\alpha}dt$

Question

I want to prove that:

$$\int_{0}^{x}\frac{t^m}{(x-t)^\alpha}dt=\frac{\Gamma(1-\alpha)\Gamma(m+1)}{\Gamma(m-\alpha+2)}x^{m-\alpha+1}$$ where $m$ is a positive integer and $\alpha \in [0,1]$.

I try to prove this formula with $\beta(x,y)=\dfrac{\beta(x)\beta(y)}{\beta(x+y)}$ and definition of $\Gamma$ function but i can't and confused!

Answer 1

Doing what @vnd suggested, you have $$\int_{0}^{x}\dfrac{t^m}{(x-t)^\alpha}dt=x^{-\alpha +m+1}\int_{0}^{1}u^m (1-u)^{-\alpha }\,du $$ and $$\int u^m (1-u)^{-\alpha }\,du= B_u(m+1,1-\alpha )$$ where appears the incomplete beta function.

Now, consider the bounds and the relation to the gamma function.