I tried to simplfy my equation $\int_{0}^{\infty } \frac{\Gamma(a, \frac{a}{b} x^{2/n})) }{\Gamma(a)(x+1)} {d}x$ with a manipulation of $\frac{1}{1+x} dx =dy $. The integral turns the form as follows $$ \frac{1}{\Gamma(a)} \int_{0}^{\infty } \Gamma(a, \frac{a}{b} {(e^y -1)}^{2/n})) {d}y$$
where $0 < n< 2$, $a>1$, $b>0$. Lets say $f(y) = {(e^y -1)}^{2/n} $. Then, the integral turns to $$ \frac{1}{\Gamma(a)}\int_{0}^{\infty } \Gamma(a,f(y)) dy $$ I found that $\int \Gamma(a,y) dy == y \Gamma(a,y) - \Gamma(a+1,y)$. How can I implement this to my equation $\int \Gamma(a,f(y)) dy $ ?
I will be grateful for all comments and suggestion. Thanks a lot!