How does one solve this Sum?

Question

How does one compute,for integer $a,b\ge0$,$$\sum_{i=0}^b (-1)^{(b-i)} \dfrac{1}{a+b-i}\dbinom{b}{i}$$

Answer 1

$\displaystyle(1-x)^{b}=\sum_{i=0}^{i=b}\dbinom{b}{i}(1)^{i}(-x)^{b-i}=\sum_{i=0}^{i=b}\dbinom{b}{i}(-1)^{b-i}x^{b-i}$

Multiplying by $x^{a-1}$ we have,

$\displaystyle x^{a-1}(1-x)^{b}=\sum_{i=0}^{i=b}\dbinom{b}{i}(-1)^{b-i}x^{(a+b-1)-i}$

integrating both sides from $0$ to $1$ we have,

$\displaystyle \int_{0}^{1}x^{a-1}(1-x)^{b}=\sum_{i=0}^b (-1)^{(b-i)} \dfrac{1}{a+b-i}\dbinom{b}{i}$

We know that ,

$\displaystyle\int_{0}^{1}x^{a-1}(1-x)^{b}=\frac{\Gamma (a)\Gamma (b+1)}{\Gamma (a+b+1)}=\frac{(a-1)!b!}{(a+b)!}$