Need help with substitution for an integral
Solution 1:
If you do$$t=\frac u{1-u}\quad\text{and}\quad\mathrm dt=\frac1{(1-u)^2},$$then$$\int_0^\infty t^{x-1}e^{-t}(\ln t)^n\,\mathrm dt=\int_0^1\left(\frac u{1-u}\right)^{x-1}e^{-u/(1-u)}\left(\ln\left(\frac u{1-u}\right)\right)^n\frac1{(1-u)^2}\,\mathrm du.$$