 Solution

Substitute $x=e^{-t}$ to getwhere the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.

On $(0,1)$, $-x\log x\ge 0$. Expand $x^{-x}=e^{-x\log x}={\sum }_{j=0}^{\infty }(-x\log x{)}^j/j!$. The partial sums increase, so the monotone convergence theorem permits termwise integration. The integral of the $j$-th term is $1/(j+1{)}^{j+1}$, including $j=0$. Reindexing givesThis is finite, for instance because the summands with $k\ge 2$ are bounded by $2^{-k}$.