Let $0<\theta\leq 1$ and $a, b>0$. Then for $x>0$, we have $0< e^{-(ax+\frac{b}{2}x^2)}<1$. Making Binomial series expansion, we have $$\left[1-e^{-(ax+\frac{b}{2}x^2)}\right]^{\theta-1}=\sum_{i=0}^{\infty}(-1)^{i}\binom{\theta-1}{i} e^{-i(ax+\frac{b}{2}x^2)}.$$ Why
$$\int_{0}^{\infty} x^{k}(ax+b)\sum_{i=0}^{\infty}(-1)^{i}\binom{\theta-1}{i} e^{-i(ax+\frac{b}{2}x^2)}dx$$
$$=\sum_{i=0}^{\infty}(-1)^{i}\binom{\theta-1}{i}\int_{0}^{\infty} x^{k}(ax+b) e^{-i(ax+\frac{b}{2}x^2)}dx$$
for $k=1,2,...$.
