I know that $$ \int_0^\infty \frac{\log x}{\exp x} = -\gamma $$ where $ \gamma $ is the Euler-Mascheroni constant, but I have no idea how to prove this.
The series definition of $ \gamma $ leads me to believe that I should break the integrand into a series and interchange the summation and integration, but I can't think of a good series. The Maclaurin series of $ \ln x $ isn't applicable as the domain of $ x $ is not correct and I can't seem to manipulate the integrand so that such a Maclaurin series will work.
Another thing I thought of was using $ x \mapsto \log u $ to get $ \int\limits_{-\infty}^\infty \frac{\log \log u}{u^2} \ du $ and use some sort of contour integration, but I can't see how that would work out either.