Question

This is a long-winded way of telling you that
integral of Log[u]^k from 1 to infinity is equal to Gamma[1+k]

Use the transformation based on the substitution
u=E^x
to explain where this result comes from.

Answer 1

\[\int_{1}^{\infty}(\log u)^kdu=\Gamma(1+k)\]by using the sub\[u=e^x\]?

Answer 2

yup?