Solution of Integral $\int_0^\infty \frac{x^{r/\beta}e^{-\alpha x}}{\left(1-(1-\beta)e^{-x}\right)^{\alpha+1}} \, dx$ I need to solve the following integral
$$\int_0^\infty \frac{x^{r/\beta}e^{-\alpha x}}{\left(1-(1-\beta)e^{-x}\right)^{\alpha+1}} \, dx$$
where $r,\alpha,\beta>0$.
kindly help me to solve this integral.
 A: As 
$$\frac{1}{(1-x)^s}=\sum_{j=0}^\infty{s+j-1\choose j}x^j        ----(1)$$
for $|x|<1$ (reference is here). Since $e^{-x}$ lies between 0 and 1 for interval $(0,\infty)$. So we have two cases for $0<\beta<2$ and $\beta>2$.
For $0<\beta<2$, using above result we can write
$$\sum_{j=0}^\infty{\alpha+j\choose j}(1-\beta)^j\int_0^\infty x^{r/\beta} e^{-x(\alpha +j)}dx$$
$$\sum_{j=0}^\infty{\alpha+j\choose j}\frac{(1-\beta)^j}{(\alpha+j)^{r/\beta+1}}\int_0^\infty x^{r/\beta} e^{-x}dx$$
$$\sum_{j=0}^\infty{\alpha+j\choose j}\frac{(1-\beta)^j}{(\alpha+j)^{r/\beta+1}}\Gamma \left( r/\beta +1 \right)$$
and now for $\beta>2$, rewriting the integral as
$$\int_0^\infty \frac{x^{r/\beta} e^{-\alpha x}}{\beta(1-(e^{-x}-1)\frac{(\beta-1)}{\beta})^{\alpha+1}}dx$$
using (1) we can write
$$\frac{1}{\beta}\sum_{j=0}^\infty{\alpha+j\choose j}(\frac{\beta-1}{\beta})^{j}\int_0^\infty x^{r/\beta} e^{-\alpha x}(e^{-x}-1)^{j} dx$$ 
$$\frac{1}{\beta}\sum_{j=0}^\infty{\alpha+j\choose j}(\frac{\beta-1}{\beta})^{j}\sum_{k=0}^j (-1)^{j+k}\int_0^\infty x^{r/\beta} e^{-(\alpha+k) x} dx$$ 
$$\frac{1}{\beta}\sum_{j=0}^\infty{\alpha+j\choose j}(\frac{\beta-1}{\beta})^{j}\sum_{k=0}^j \frac{(-1)^{j+k}}{(\alpha+k)^{r/\beta +1}}\Gamma \left( r/\beta +1 \right)$$
hence proved. 
