\(% 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}}\)

2018.2.5 Question 5

1. For \(\abs *{x} < 1\), we have \[ \frac {1}{1 + x} = 1 - x + x^2 - x^3 + \cdots = \sum _{t = 0}^{\infty } (-x)^t. \]

   Since \(\ln (1 + x)\) differentiates to \(\frac {1}{1 + x}\), by integration, we have \begin {align*} \ln (1 + x) & = \int \frac {1}{1 + x} \Diff x \\ & = \int \sum _{t = 0}^{\infty } (-x)^t \Diff x \\ & = \sum _{t = 0}^{\infty } (-1)^t \int x^t \Diff x \\ & = C + \sum _{t = 0}^{\infty } (-1)^t \frac {x^{t + 1}}{t + 1} \\ & = C - \sum _{t = 1}^{\infty } \frac {(-x)^t}{t}. \end {align*}

   Let \(x = 0\), and we see \(\ln (1 + x) = \ln 1 = 0\), and the sum on the right-hand side evaluates to \(0\), and hence \(C = 0\). This gives the Maclaurin expansion for \(\ln (1 + x)\) \[ \ln (1 + x) = - \sum _{t = 1}^{\infty } \frac {(-x)^t}{t}. \]

2. We have \[ e^{-ax} = \sum _{t = 0}^{\infty } \frac {(-ax)^t}{t!}, \] and hence \begin {align*} & \phantom {=}\int _{0}^{\infty } \frac {(1 - e^{-ax})e^{-x}}{x} \Diff x \\ & = \int _{0}^{\infty } \frac {-\sum _{t = 1}^{\infty } \frac {(-ax)^t}{t!} \cdot e^{-x}}{x} \Diff x \\ & = \sum _{t = 1}^{\infty } \int _{0}^{\infty } \frac {- (-ax)^t e^{-x}}{t!x} \Diff x \\ & = \sum _{t = 1}^{\infty } \int _{0}^{\infty } \frac {(-x)^{t - 1} a^t e^{-x}}{t!} \Diff x \\ & = \sum _{t = 1}^{\infty } \frac {(-1)^{t - 1} a^t}{t!} \int _{0}^{\infty } x^{t - 1} e^{-x} \Diff x. \end {align*}

   We aim to find an expression for \[ I_t = \int _{0}^{\infty } x^{t} e^{-x} \Diff x. \]

   Using integration by parts, we have \begin {align*} I_t & = \int _{0}^{\infty } x^{t} e^{-x} \Diff x \\ & = - \int _{0}^{\infty } x^{t} \Diff e^{-x} \\ & = - \left [(x^t e^{-x})_{0}^{\infty } - \int _{0}^{\infty } e^{-x} \Diff x^{t}\right ] \\ & = t \int _{0}^{\infty } e^{-x} x^{t - 1} \Diff x \\ & = t I_{t - 1}, \end {align*}

   and further noticing that \[ I_0 = \int _{0}^{\infty } e^{-x} \Diff x = \left [-e^{-x}\right ]_{0}^{\infty } = 1, \] we can see \[ I_t = t!, \] and hence \begin {align*} & \phantom {=}\int _{0}^{\infty } \frac {(1 - e^{-ax})e^{-x}}{x} \Diff x \\ & = \sum _{t = 1}^{\infty } \frac {(-1)^{t - 1} a^t}{t!} \int _{0}^{\infty } x^{t - 1} e^{-x} \Diff x \\ & = \sum _{t = 1}^{\infty } \frac {(-1)^{t - 1} a^t}{t!} (t - 1)! \\ & = \sum _{t = 1}^{\infty } \frac {(-1)^{t - 1} a^t}{t} \\ & = - \sum _{t = 1}^{\infty } \frac {(-a)^t}{t} \\ & = \ln (1 + a), \end {align*}

   precisely as desired.

3. Using a substitution \(x = e^{-u}\), when \(x = 1\), \(u = 0\), and when \(x \to 0^{+}\), \(u \to \infty \). Also, \[ \DiffFrac {x}{u} = -e^{-u}, \] and hence \begin {align*} & \phantom {=} \int _{0}^{1} \frac {x^p - x^q}{\ln x} \Diff x \\ & = \int _{\infty }^{0} \frac {e^{-up} - e^{-uq}}{\ln e^{-u}} \cdot (- e^{-u}) \Diff u \\ & = \int _{\infty }^{0} \frac {\left (e^{-up} - e^{-uq}\right ) e^{-u}}{u} \Diff u \\ & = \int _{0}^{\infty } \frac {\left [\left (1 - e^{-up}\right ) + \left (1 - e^{-uq}\right )\right ]e^{-u}}{u} \Diff u \\ & = \int _{0}^{\infty } \frac {\left (1 - e^{-up}\right ) e^{-u}}{u} \Diff u - \int _{0}^{\infty } \frac {\left (1 - e^{-uq}\right ) e^{-u}}{u} \Diff u \\ & = \ln (1 + p) - \ln (1 + q). \end {align*}