\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Using the substitution \(t = \frac {1}{x}\), we have \[ \DiffFrac {t}{x} = - \frac {1}{x^2} \implies \Diff {x} = - x^2 \Diff {t} = - \frac {\Diff {t}}{t^2}, \] and when \(x \to 0^{+}\), \(t \to \infty \), and when \(x = 1\), \(t = 1\). Hence, \begin {align*} I & = \int _{0}^{1} \frac {f(x^{-1})}{1 + x} \Diff {x} \\ & = \int _{1}^{\infty } \frac {f(t)}{1 + t^{-1}} \cdot \left (- \frac {\Diff {t}}{t^2}\right ) \\ & = \int _{1}^{\infty } \frac {f(t) \Diff {t}}{t (1 + t)} \\ & = \int _{1}^{2} \frac {f(t) \Diff {t}}{t (1 + t)} + \int _{2}^{3} \frac {f(t) \Diff {t}}{t (1 + t)} + \int _{3}^{4} \frac {f(t) \Diff {t}}{t (1 + t)} + \cdots \\ & = \sum _{n = 1}^{\infty } \int _{n}^{n + 1}\frac {f(t) \Diff {t}}{t (1 + t)}, \end {align*}
as desired.
Since \(f(x) = f(x + 1)\) for all \(x\), we must have that \(f(x) = f(x + n)\) for all \(x\) and integers \(n\). Also, we have \[ \frac {1}{y(1 + y)} = \frac {1}{y} - \frac {1}{1 + y}. \]
Hence, \begin {align*} I & = \sum _{n = 1}^{\infty } \int _{n}^{n + 1} \frac {f(t) \Diff {t}}{t (1 + t)} \\ & = \sum _{n = 1}^{\infty } \int _{0}^{1} \frac {f(n + t) \Diff {t}}{(n + t) (n + t + 1)} \\ & = \sum _{n = 1}^{\infty } \int _{0}^{1} f(t) \cdot \left [\frac {1}{n + t} - \frac {1}{n + t + 1}\right ] \Diff {t} \\ & = \sum _{n = 1}^{\infty } \int _{0}^{1} \frac {f(t) \Diff {t}}{n + t} - \sum _{n = 1}^{\infty } \int _{0}^{1} \frac {f(t) \Diff {t}}{n + t + 1} \\ & = \sum _{n = 1}^{\infty } \int _{0}^{1} \frac {f(t) \Diff {t}}{n + t} - \sum _{n = 2}^{\infty } \int _{0}^{1} \frac {f(t) \Diff {t}}{n + t} \\ & = \int _{0}^{1} \frac {f(t) \Diff {t}}{1 + t}. \end {align*}
For the first integral, simply consider \(f(x) = \{x\}\), and we can immediately see that \(f(x)\) has period of \(1\) from the definition. Hence, \[ \int _{0}^{1} \frac {\{x^{-1}\}}{1 + x} \Diff x = \int _{0}^{1} \frac {f(x^{-1})}{1 + x} \Diff x = \int _{0}^{1} \frac {f(x)}{1 + x} \Diff x = \int _{0}^{1} \frac {\{x\}}{1 + x} \Diff x. \]
Since for \(0 < x < 1\), we have \(\{x\} = x\), and hence \begin {align*} \int _{0}^{1} \frac {\{x^{-1}\}}{1 + x} \Diff x & = \int _{0}^{1} \frac {\{x\}}{1 + x} \Diff x \\ & = \int _{0}^{1} \frac {x}{1 + x} \Diff x \\ & = \int _{0}^{1} \left (1 - \frac {1}{1 + x}\right ) \Diff x \\ & = 1 - \left [\ln (1 + x)\right ]_{0}^{1} \\ & = 1 - (\ln (2) - \ln (1)) \\ & = 1 - \ln 2. \end {align*}
For the second integral, we let \(g(x) = \{2x\}\), and we can see that \(g(x)\) has a period of \(\frac {1}{2}\), and hence it also has a period of \(1\). Hence, \[ \int _{0}^{1} \frac {\{2x^{-1}\}}{1 + x} \Diff x = \int _{0}^{1} \frac {g(x^{-1})}{1 + x} \Diff x = \int _{0}^{1} \frac {g(x)}{1 + x} \Diff x = \int _{0}^{1} \frac {\{2x\}}{1 + x} \Diff x. \]
We split this integral into two parts, \([0, 0.5]\) and \([0.5, 1]\).
\begin {align*} \int _{0}^{1} \frac {\{2x^{-1}\}}{1 + x} \Diff x & = \int _{0}^{1} \frac {\{2x\}}{1 + x} \Diff x \\ & = \int _{0}^{0.5} \frac {\{2x\}}{1 + x} \Diff x + \int _{0.5}^{1} \frac {\{2x\}}{1 + x} \Diff x \\ & = \int _{0}^{0.5} \frac {2x}{1 + x} \Diff x + \int _{0.5}^{1} \frac {2x - 1}{1 + x} \Diff x \\ & = \int _{0}^{0.5} \left [2 - \frac {2}{1 + x}\right ] \Diff x + \int _{0.5}^{1} \left [2 - \frac {3}{1 + x}\right ] \Diff x \\ & = 1 - 2 \left [\ln (1 + x)\right ]_{0}^{0.5} + 1 - 3 \left [\ln (1 + x)\right ]_{0.5}^{1} \\ & = 2 - 2 \ln 1.5 + 2 \ln 1 - 3 \ln 2 + 3 \ln 1.5 \\ & = 2 - 3 \ln 2 + \ln 1.5 \\ & = 2 - 3 \ln 2 + \ln 3 - \ln 2 \\ & = 2 - 4 \ln 2 + \ln 3. \end {align*}