I want to find the following integral.
$$ \frac{1}{\Gamma(\alpha) \beta^\alpha} \int^{\infty}_{5} (x-5)^{a-1} e^{tx} e^{- \frac{x-5}{\beta}} dx$$
Based on some research I have done, I believe that it could be a good idea to use the gamma function, but I am unsure how to exactly use it to solve the integral in this case. How would I go about this?
$$I=\int_5^\infty(x-5)^{\alpha-1}e^{tx}e^{-(x-5)/\beta}dx$$ $u=x-5$, $du=dx$ $$I=e^{5t}\int_0^\infty u^{\alpha-1}e^{tu}e^{-u/\beta}du=e^{5t}\int_0^\infty u^{\alpha-1}e^{-(\beta^{-1}-t)u}du=e^{5t}\int_0^\infty u^{\alpha-1}e^{-\gamma u}du$$ for ease I am going to get $\gamma=\beta^{-1}-t$ then we will need to make the substitution $v=\gamma u$ $dv=\gamma du$ and so: $$I=e^{5t}\gamma^{2-\alpha}\int_0^\infty v^{\alpha-1}e^{-v}dv=e^{5t}\gamma^{2-\alpha}\Gamma(\alpha)=e^{5t}\left(\frac1\beta-t\right)^{2-\alpha}\Gamma(\alpha)$$ now you should be able to do the rest pretty easily