\[\gamma=-\int_{0}^{\infty}e^{-t}\ln t\,\mathrm{d}t=\int_{0}^{\infty}\left(\frac%
{1}{1+t}-e^{-t}\right)\frac{\,\mathrm{d}t}{t}=\int_{0}^{1}(1-e^{-t})\frac{\,%
\mathrm{d}t}{t}-\int_{1}^{\infty}e^{-t}\frac{\,\mathrm{d}t}{t}=\int_{0}^{%
\infty}\left(\frac{e^{-t}}{1-e^{-t}}-\frac{e^{-t}}{t}\right)\,\mathrm{d}t.\]