For the integral:
$$-\gamma = \int_0^\infty \!\! e^{-u} \ln u\,\mathrm du = \int_0^1 \!\! e^{-u} \ln u\,\mathrm du + \int_0^\infty \!\! e^{-u}\ln u \,\mathrm du $$ $$ =[(-e^{-u} +1)\ln u]_0^1 - \int_0^1 \!\! \frac {(-e^{-u}+1)}{u}\,\mathrm du + [-e^{-u}\ln u]_1^\infty \!\! - \int_1^\infty \!\! \frac{-e^{-u}}{u}\,\mathrm du$$
When I integrate by parts on the two integrals I get: $$\int_0^1\!\! e^{-u} \ln u \,\mathrm du = [-e^{-u} \ln u]_0^1 + \int_0^1 \!\!\frac{ e^{-u}}{u} \,\mathrm du +[-e^{-u}\ln u]_1^\infty \!\! -\int_1^\infty \!\! \frac{e^{-u}}{u}\,\mathrm du $$ I don't see where on the first of the two integrals $$ [(-e^{-u} +1)\ln u]_0^1 - \int_0^1 \!\! \frac{(-e^{-u}+1)}{u} \,\mathrm du $$ comes from. It is the $$ -e^{-u} +1 $$ that keeps messing me up.