integral involving Incomplete Gamma Function and exponential

Question

I was trying to solve the following integral;

$\int_{0}^{\infty} \Gamma (s,x) (1- e^{by})^{K-1} e^{by} dy$.

The lower limit of the incomplete gamma function $x = \frac{a(1+y)-1}{c}$, which contains $y$.

I rewrote the gamma function as the integral; $\int_{0}^{\infty} \int_{x}^{\infty} t^{s-1} e^{-t} e^{by} (1- e^{by})^{K-1} dt dy$.

and tried to change order of integration such as;

$\int_{0}^{\infty} \int_{0}^{H} t^{s-1} e^{-t} e^{by}(1- e^{by})^{K-1} dy dt$.

where $H = \frac{ct+1}{a}-1$.

After taking the integral w.r.t. $y$ , I ended up something like the following, I am not sure whether I did mistakes until here.

$\int_{0}^{\infty} (-\frac{1}{bK}) t^{s-1} e^{-t} (1-e^{Hb})^{K}dt$.

I could not proceed any further. Could someone give me tips to solve this? Thank you!

Answer 1

$$I=\int_0^\infty\Gamma(s,x)(1-e^{by})^{K-1}e^{by}dy=\int_0^\infty\int_x^\infty t^{s-1}e^{-t}(1-e^{by})^{K-1}e^{by}dy$$ You could use the fact that: $$(1-e^{by})^{K-1}=\sum_{k=0}^{K-1}{{K-1}\choose{k}}(-1)^ke^{byk}$$ and if we assume we can interchange the summation and the integral we get: $$I=\sum_{k=0}^{K-1}(-1)^k{{K-1}\choose{k}}\int_0^\infty\int_x^\infty t^{s-1}e^{b(k+1)y-t}dtdy$$