Past exam of the integration course of the University of Oxford / 2021 / 1 / c / i / Solution

For each nonnegative integer $k$, the change of variables $x=e^{-t}$ givesThe gamma function identity $\Gamma (k+1)=k!$ follows by repeated integration by parts. Since $\log x<0$ on $(0,1)$, the sign of the original integrand is $(-1{)}^k$, soFor $k=0$, the integral is one, consistent with the formula.