Definite integral on fractional part Prove following definite integral:
$$\int_0^1\Bigl\{\frac{1}{x}\Bigr\}\ln(x)\,dx = \gamma_0+\gamma_1-1$$
Found in "Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis" sample page (Problem Books in Mathematics, DOI 10.1007/978-1-4614-6762-5 2, © Springer Science+Business Media New York 2013) but not able to reach for the solution.
Also need if exist, a generalization on it noting that similar integral without logarithm on it retrieve $1-\gamma_0$.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\int_{0}^{1}\braces{1 \over x}\ln\pars{x}\,\dd x =
\gamma_{0} + \gamma_{1} - 1:\ {\LARGE ?}}$ where
  $\ds{\braces{\vphantom{\Large A}\gamma_{n}\ \mid\ n = 0,1,2,\ldots}}$ are 
  Stieltjes Constants.
  
  $\ds{\gamma_{0}}$ is the
  Euler-Mascheroni Constant too.

With $\ds{N \in \mathbb{N}_{\ \geq\ 1}}$:
\begin{align}
\int_{0}^{1}\braces{1 \over x}\ln\pars{x}\,\dd x &
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
-\int_{1}^{\infty}\braces{x}\ln\pars{x}\,{\dd x \over x^{2}}
\\[5mm] & =
-\lim_{N \to \infty}\bracks{\int_{1}^{N}{\ln\pars{x} \over x}\,\dd x -
\int_{1}^{N}\left\lfloor{x}\right\rfloor{\ln\pars{x} \over x^{2}}\,\dd x}
\\[5mm] & =
\lim_{N \to \infty}\bracks{\bbox[10px,#ffd]{\ds{\sum_{k = 1}^{N - 1}k\int_{k}^{k + 1}{\ln\pars{x} \over x^{2}}\,\dd x}} - {1 \over 2}\,\ln^{2}\pars{N}}\label{1}\tag{1}
\end{align}

Note that

\begin{align}
\bbox[10px,#ffd]{\ds{\sum_{k = 1}^{N - 1}k\int_{k}^{k + 1}{\ln\pars{x} \over x^{2}}\,\dd x}} & =
\sum_{k = 1}^{N - 1}\bracks{\ln\pars{k} - \ln\pars{k + 1} + {1 \over 1 + k} +
{\ln\pars{1 + k} \over 1 + k}}
\\[5mm] & =
-\ln\pars{N} + H_{N} - 1 +
\sum_{k = 2}^{N }{\ln\pars{k} \over k}
\label{2}\tag{2}
\end{align}

$\ds{H_{z}}$ is a Harmonic Number. \eqref{1} and \eqref{2} lead to:

\begin{align}
& \bbx{\int_{0}^{1}\braces{1 \over x}\ln\pars{x}\,\dd x} =
\\[5mm] = &\
\underbrace{\lim_{N \to \infty}\bracks{H_{N} - \ln\pars{N}}}
_{\ds{\gamma_{0}}}\ +\
\underbrace{\lim_{N \to \infty}\bracks{\sum_{k = 2}^{N}
{\ln\pars{k} \over k} -
{1 \over 2}\,\ln^{2}\pars{N}}}
_{\ds{\gamma_{1}}} - 1
\\[5mm] = &\ \bbx{\gamma_{0} + \gamma_{1} - 1}
\end{align}

See the $\ds{\gamma_{1}}$ definition in this link.

