Definite integral on fractional part

Question

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 $1-\gamma_0$.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{1}\braces{1 \over x}\ln\pars{x}\,\dd x = \gamma_{0} + \gamma_{1} - 1:\ {\LARGE ?}}$ 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.