An integral involves Gamma function

Question

Thanks for your attention, I meet an integral involves Gamma function and exponential function as follows:$$\int_a^\infty {{x^\alpha }} {e^{cx}}\Gamma \left( {s,bx} \right)dx$$ where $a > 0,s \leqslant 0,b \geqslant 0$ and $\alpha $ is integer, here is the definition of the Gamma function:$$\Gamma \left( {\alpha ,z} \right) = \int_z^\infty {{t^{\alpha - 1}}{e^{ - t}}dt} $$,I have no idea about how to solve this.

Answer 1

Since there is probably no closed form, a series expension is (subject to convergence conditions) : $$I=\int_a^\infty {{x^\alpha }} {e^{cx}}\Gamma \left( {s,bx} \right)dx$$ $$\Gamma \left( {s,bx} \right)=\Gamma(s)- \sum_{k=0}^{\infty}\frac{(-1)^k b^{s+k} x^{s+k}}{(s+k) k!}$$ Supposing $c<0$ among the conditions for convergence : $$I=\Gamma(s) \int_a^\infty x^\alpha e^{cx} dx- \sum_{k=0}^{\infty} \frac{(-1)^k b^{s+k} }{(s+k) k!} \int_a^\infty x^{s+k+\alpha} e^{cx} dx$$ $$I=\Gamma(s) (-c)^{-\alpha-1} \Gamma\big(\alpha+1,(-c)a\big) - \sum_{k=0}^{\infty} \frac{(-1)^k b^{s+k} }{(s+k) k! (-c)^{\alpha+1+s+k} } \Gamma\big(\alpha+1-s-k,(-c)a\big)$$