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$\ds{\lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}= -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t:\ {\large ?}}$
\begin{align} \sum_{k = 1}^{n}{1 \over k} &= \sum_{k = 1}^{n}\int_{0}^{1}t^{k - 1}\,\dd t = \int_{0}^{1}\sum_{k = 1}^{n}t^{k - 1}\,\dd t =\int_{0}^{1}{1 - t^{n - 1} \over 1 - t}\,\dd t =\int_{\infty}^{1}{1 - t^{1 - n} \over 1 - 1/t}\,\pars{-\,{\dd t \over t^{2}}} \\[3mm]&=\int_{1}^{\infty}{t^{-1} - t^{-n} \over t - 1}\,\dd t =\int_{0}^{\infty}{\pars{1 + t}^{-1} - \pars{1 + t}^{-n} \over t}\,\dd t \\[3mm]&=-\int_{0}^{\infty}\ln\pars{t} \bracks{-\pars{1 + t}^{-2} + n\pars{1 + t}^{-n - 1}}\,\dd t \\[3mm]&=\int_{0}^{\infty}{\ln\pars{t} \over \pars{1 + t^{2}}}\,\dd t -\int_{0}^{\infty}\ln\pars{t \over n}\pars{1 + {t \over n}}^{-n - 1}\,\dd t \end{align}
The first integral vanishes out: Just split
$\ds{\pars{0,\infty}}$ in
$\ds{\pars{0,1}}$ and
$\ds{\pars{1,\infty}}$ and we'll see that the 'pieces' cancels each other: \begin{align} \sum_{k = 1}^{n}{1 \over k} - \ln\pars{n} & = \ln\pars{n}\bracks{\overbrace{\int_{0}^{\infty}\pars{1 + {t \over n}}^{-n - 1}\,\dd t}^{\ds{=\ 1\,,\ \forall\ n\ >\ 0}}\ -\ 1}\ -\ \int_{0}^{\infty}\ln\pars{t}\pars{1 + {t \over n}}^{-n - 1}\,\dd t \end{align}
Note that $\ds{\lim_{n \to \infty}\pars{1 + {t \over n}}^{-n - 1} = \expo{-t}}$ and $\ds{\int_{0}^{\infty}\expo{-t}\,\dd t = 1}$: $\bbox[15px,border:1px dotted navy]{\displaystyle \lim_{n \to \infty}\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}= -\int_{0}^{\infty}\expo{-t}\ln\pars{t}\,\dd t} $