\(% 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}}\)

2016.3.1 Question 1

Notice that \[ I_n = \int _{-\infty }^{\infty } \frac {\Diff x}{(x^2 + 2ax + b)^n} = \int _{-\infty }^{\infty } \frac {\Diff x}{((x+a)^2 + (b - a^2))^n}. \]

1. Let \(x + a = \sqrt {b - a^2} \tan u\). When \(x \to -\infty \), \(u \to -\frac {\pi }{2}\), and when \(x \to \infty \), \(u \to \frac {\pi }{2}\). We have also \begin {align*} \Diff x = \Diff (x+a) & = \Diff \sqrt {b - a^2} \tan u \\ & = \sqrt {b - a^2} \Diff \tan u \\ & = \sqrt {b - a^2} \sec ^2 u \Diff u. \end {align*}

   Therefore, we have \begin {align*} I_1 & = \int _{-\infty }^{\infty } \frac {\Diff x}{(x+a)^2 + (b - a^2)} \\ & = \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\sqrt {b - a^2} \sec ^2 u \Diff u}{\left (\sqrt {b - a^2} \tan u\right )^2 + (b - a^2)} \\ & = \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\sqrt {b - a^2} \sec ^2 u \Diff u}{(b - a^2)(\tan ^2 u + 1)} \\ & = \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\sec ^2 u \Diff u}{\sqrt {b - a^2}\sec ^2 u} \\ & = \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sqrt {b - a^2}} \\ & = \frac {\pi }{\sqrt {b - a^2}}, \end {align*}

   as desired.

2. Using the same substitution, we have \begin {align*} I_n & = \int _{-\infty }^{\infty } \frac {\Diff x}{[(x+a)^2 + (b - a^2)]^n} \\ & = \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\sqrt {b - a^2} \sec ^2 u \Diff u}{\left [(b - a^2) \sec ^2 u\right ]^n} \\ & = \frac {1}{\sqrt {b - a^2}} \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\left [(b - a^2) \sec ^2 u\right ]^{n - 1}}. \end {align*}

   Therefore, \[ 2n(b - a^2) I_{n + 1} = (2n - 1) I_n, \] is equivalent to \[ 2n \sqrt {b - a^2} \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\left [(b - a^2) \sec ^2 u\right ]^{n}} = (2n-1) \frac {1}{\sqrt {b - a^2}} \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\left [(b - a^2) \sec ^2 u\right ]^{n - 1}} \] is equivalent to \[ 2n (b - a^2) \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\left [(b - a^2) \sec ^2 u\right ]^{n}} = (2n-1) \int _{-\frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\left [(b - a^2) \sec ^2 u\right ]^{n - 1}} \] is equivalent to \[ 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n} u} = (2n - 1) \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n - 2} u}. \]

   Notice that \begin {align*} \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n - 2} u} & = \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\sec ^2 u \Diff u}{\sec ^{2n} u} \\ & = \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff \tan u}{\sec ^{2n} u} \\ & = \lim _{\substack {a \to \frac {\pi }{2} \\b \to -\frac {\pi }{2}}}\left [\frac {\tan u}{\sec ^{2n}u}\right ]^{a}_{b} - \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \tan u \Diff \sec ^{-2n}u\\ & = \lim _{\substack {a \to \frac {\pi }{2} \\b \to -\frac {\pi }{2}}}\left [\sin u \cos ^{2n - 1} u\right ]^{a}_{b} - \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} -2n \sec u \tan u \sec ^{-2n - 1}u \tan u \Diff u\\ & = 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\tan ^2 u \Diff u}{\sec ^{2n} u} \\ & = 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {(\sec ^2 u - 1) \Diff u}{\sec ^{2n} u} \\ & = 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n - 2} u} - 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n} u}. \\ \end {align*}

   This means \[ (2n - 1) \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n - 2} u} = 2n \int _{- \frac {\pi }{2}}^{\frac {\pi }{2}} \frac {\Diff u}{\sec ^{2n} u}, \] which is exactly what was desired.

3. Proof by induction:

   • Base Case. When \(n = 1\), \[ \LHS = I_1 = \frac {\pi }{\sqrt {b - a^2}}, \] \[ \RHS = \frac {\pi }{2^{2 \cdot 1 - 2} (b - a^2)^{1 - \frac {1}{2}}} \binom {2 \cdot 1 - 2}{1 - 1} = \frac {\pi }{\sqrt {b - a^2}} \binom {0}{0} = \frac {\pi }{\sqrt {b - a^2}}. \]
   • Induction Hypothesis. Assume for some \(n = k \in \NN \), we have \[ I_n = \frac {\pi }{2^{2n - 2} (b - a^2)^{n - \frac {1}{2}}} \binom {2n - 2}{n - 1}. \]
   • Induction Step. When \(n = k + 1\), \begin {align*} I_n & = I_{k + 1} \\ & = \frac {2k + 1}{2(k + 1)(b - a^2)} I_k \\ & = \frac {2k + 1}{2(k + 1)(b - a^2)} \cdot \frac {\pi }{2^{2k - 2} (b - a^2)^{k - \frac {1}{2}}} \binom {2k - 2}{k - 1} \\ & = \frac {\pi }{2^{2k} (b - a^2)^{k + \frac {1}{2}}} \frac {(2k-2)!}{(k-1)!(k-1)!} \frac {(2k + 1)(2k + 2)}{(k + 1)^2} \\ & = \frac {\pi }{2^{2k} (b - a^2)^{k + \frac {1}{2}}} \frac {2k!}{k!k!} \\ & = \frac {\pi }{2^{2k} (b - a^2)^{k + \frac {1}{2}}} \binom {2k}{k} \\ & = \frac {\pi }{2^{2n - 2} (b - a^2)^{n - \frac {1}{2}}} \binom {2n - 2}{n - 1}. \end {align*}

   Therefore, by the principle of mathematical induction, for \(n \in \NN \), \[ I_n = \frac {\pi }{2^{2n - 2} (b - a^2)^{n - \frac {1}{2}}} \binom {2n - 2}{n - 1}, \] as desired.