\(% 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 integration by parts, we notice that \begin {align*} (n + 1) I_n & = (n + 1) \int _{0}^{1} x^n \arctan x \Diff x \\ & = \int _{0}^{1} \arctan x \Diff x^{n + 1} \\ & = \left [\arctan x \cdot x^{n + 1}\right ]_{0}^{1} - \int _{0}^{1} x^{n + 1} \Diff \arctan x \\ & = \arctan 1 \cdot 1^{n + 1} - \arctan 0 \cdot 0^{n + 1} - \int _{0}^{1} \frac {x^{n + 1}}{1 + x^2} \Diff x \\ & = \frac {\pi }{4} - \int _{0}^{1} \frac {x^{n + 1}}{1 + x^2} \Diff x. \end {align*}
Set \(n = 0\), and we have \begin {align*} I_0 & = (0 + 1) I_0 \\ & = \frac {\pi }{4} - \int _{0}^{1} \frac {x}{1 + x^2} \Diff x \\ & = \frac {\pi }{4} - \frac {1}{2} \cdot \left [\ln (1 + x^2)\right ]_{0}^{1} \\ & = \frac {\pi }{4} - \frac {1}{2} \cdot \left [\ln 2 - \ln 1\right ] \\ & = \frac {\pi }{4} - \frac {\ln 2}{2}. \end {align*}
Using the result in the previous part, \begin {align*} (n + 3) I_{n + 2} + (n + 1) I_n & = \left (\frac {\pi }{4} - \int _{0}^{1} \frac {x^{n + 3}}{1 + x^2} \Diff x\right ) + \left (\frac {\pi }{4} - \int _{0}^{1} \frac {x^{n + 1}}{1 + x^2} \Diff x\right ) \\ & = \frac {\pi }{2} - \int _{0}^{1} \frac {x^{n + 1} + x^{n + 3}}{1 + x^2} \Diff x \\ & = \frac {\pi }{2} - \int _{0}^{1} \frac {x^{n + 1} \left (1 + x^2\right )}{1 + x^2} \Diff x \\ & = \frac {\pi }{2} - \int _{0}^{1} x^{n + 1} \Diff x \\ & = \frac {\pi }{2} - \frac {1}{n + 2} \left [x^{n + 2}\right ]_{0}^{1} \\ & = \frac {\pi }{2} - \frac {1}{n + 2}. \end {align*}
Letting \(n = 0\), and we have \[ 3 I_2 + I_0 = \frac {\pi }{2} - \frac {1}{2}. \]
Letting \(n = 2\), and we have \[ 5 I_4 + 3 I_2 = \frac {\pi }{2} - \frac {1}{4}. \]
Subtracting the first one from the second one, and hence \[ 5 I_4 - I_0 = \frac {1}{4}. \]
Hence, \[ I_4 = \frac {1}{5} \cdot \left [\frac {1}{4} + \left (\frac {\pi }{4} - \frac {\ln 2}{2}\right )\right ] = \frac {1}{20} + \frac {\pi }{20} - \frac {\ln 2}{10}. \]
Let \(n = 1\), and the statement says \begin {align*} (4n + 1) I_{4n} & = 5 I_4 \\ & = A - \frac {1}{2} \sum _{r = 1}^{2 \cdot 1}(-1)^r \frac {1}{r} \\ & = A - \frac {1}{2} \left (- \frac {1}{1} + \frac {1}{2}\right ) \\ & = A + \frac {1}{4}. \end {align*}
Comparing to the previous expression, we claim that \[ A = \frac {\pi }{4} - \frac {\ln 2}{2}. \]
This shows the base case for \(n = 1\). For the induction step, we first introduce a lemma. Since \[ (n + 5) I_{n + 4} + (n + 3) I_{n + 2} = \frac {\pi }{2} - \frac {1}{n + 4}, (n + 3) I_{n + 2} + (n + 1) I_{n} = \frac {\pi }{2} - \frac {1}{n + 2}, \] subtracting the second one from the first one will give us \[ (n + 5) I_{n + 4} - (n + 1) I_{n} = \frac {1}{n + 2} - \frac {1}{n + 4}. \]
Setting \(n = 4m\), we have \begin {align*} (4(m + 1) + 1) I_{4 (m + 1)} & = (4m + 1) I_{4m} + \frac {1}{4m + 2} - \frac {1}{4m + 4} \\ & = (4m + 1) I_{4m} - \frac {1}{2} \cdot \left (-\frac {1}{2m + 1} + \frac {1}{2m + 2}\right ) \\ & = (4m + 1) I_{4m} - \frac {1}{2} \cdot \left [(-1)^{2m + 1} \frac {1}{2m + 1} + (-1)^{2m + 2} \frac {1}{2m + 2}\right ]. \end {align*}
Now we show the inductive step. Assume the statement is true for some \(n = k \geq 1\), i.e. \[ (4k + 1) I_{4k} = A - \frac {1}{2} \sum _{r = 1}^{2n} (-1)^r \frac {1}{r}. \]
Using the identity above, we have \begin {align*} (4(k + 1) + 1) I_{4 (k + 1)} & = (4k + 1) I_{4k} - \frac {1}{2} \cdot \left [(-1)^{2k + 1} \frac {1}{2k + 1} + (-1)^{2k + 2} \frac {1}{2k + 2}\right ] \\ & = A - \frac {1}{2} \sum _{r = 1}^{2k} (-1)^r \frac {1}{r} - \frac {1}{2} \cdot \left [(-1)^{2k + 1} \frac {1}{2k + 1} + (-1)^{2k + 2} \frac {1}{2k + 2}\right ] \\ & = A - \frac {1}{2} \sum _{r = 1}^{2(k + 1)} (-1)^r \frac {1}{r}. \end {align*}
Hence, the original statement is true for \(n = 1\) (as shown when determining the value of \(A\)), and given the original statement holds for some \(n = k \geq 1\), it holds for \(n = k + 1\). By the principle of mathematical induction, this statement holds for all \(n \geq 1\), where \[ A = \frac {\pi }{4} - \frac {\ln 2}{2}. \]