From Wikipedia we have the following equation: $$\frac{{d}^{n}}{d{x}^{n}}\Gamma (x)=\int _{0}^{\infty }{t}^{x-1}{e}^{-t}{\left(\ln t\right)}^{n}dt$$
which is clearly an integral going from 0 to infinity. I was wondering if you were able to see how to do a substitution on this integral to change it from one going from 0 to infinity to an integral going from 0 to some finite number. Am doing some numerical work with this integral and it would be much more convenient to compute integrals that do not go to infinity for higher derivatives of $\Gamma(x)$.
To make what I am asking for even more explicit, I am asking if you are able to turn an integral such as $$\int _{0}^{\infty }{e}^{-t}{t}^{x-1}dt$$ into $$\int _{0}^{1}{\left(-\ln t\right)}^{x-1}dt$$
If you do$$t=\frac u{1-u}\quad\text{and}\quad\mathrm dt=\frac1{(1-u)^2},$$then$$\int_0^\infty t^{x-1}e^{-t}(\ln t)^n\,\mathrm dt=\int_0^1\left(\frac u{1-u}\right)^{x-1}e^{-u/(1-u)}\left(\ln\left(\frac u{1-u}\right)\right)^n\frac1{(1-u)^2}\,\mathrm du.$$