$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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$\ds{\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x = \color{#f00}{-}\gamma}$.
\begin{align}
&\int_{0}^{\infty}{\gamma + \ln\pars{x} \over \expo{x}}\,
{1 - \cos\pars{x} \over x}\,\dd x =
\int_{0}^{\infty}\bracks{\gamma + \ln\pars{x}}\expo{-x}
\int_{0}^{1}\sin\pars{xt}\,\dd t\,\dd x
\\[5mm] = &\
\Im\int_{0}^{1}\int_{0}^{\infty}\bracks{\gamma + \ln\pars{x}}
\expo{-\pars{1 - \ic t}x}\,\dd x\,\dd t\qquad\qquad\pars{~\ln\mbox{-Principal Branch}~}
\\[5mm] = &\
\gamma\,\Im\int_{0}^{1}{\dd t \over 1 - \ic t} +
\Im\int_{0}^{1}{1 \over 1 - \ic t}\int_{0}^{\pars{1 - \ic t}\infty}
\ln\pars{x \over 1 - \ic t}\expo{-x}\,\dd x\,\dd t
\\[5mm] = &\
\gamma\,\Im\int_{0}^{1}{\dd t \over 1 - \ic t} +
\Im\int_{0}^{1}{1 \over 1 - \ic t}\bracks{%
\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x - \ln\pars{1 - \ic t}}\,\dd t
\\[5mm] = &\
-\,\Im\int_{0}^{1}{\ln\pars{1 - \ic t} \over 1 - \ic t}\,\dd t =
-\,\Im\bracks{\ic\,{1 \over 2}\ln^{2}\pars{1 - \ic t}}_{\ 0}^{\ 1} =
-\,{1 \over 2}\,\Re\ln^{2}\pars{1 - \ic}
\\[5mm] = &\
-\,{1 \over 2}\,\Re\bracks{{1 \over 2}\,\ln\pars{2} - {\pi \over 4}\,\ic}^{2} =
-\,{1 \over 8}\,\ln^{2}\pars{2} + {1 \over 32}\,\pi^{2} =
{1 \over 32}\,\pi^{2} - {1 \over 32}\,\ln^{2}\pars{4}
\\[5mm] = &\
\bbx{\ds{{1\over 2}\,{\pi - \ln\pars{4} \over 4}\,{\pi + \ln\pars{4} \over 4}}}
\end{align}