\(% 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}}\)
We have \begin {align*} I_n - I_{n + 1} & = \int _0^{\infty } \frac {1}{(1 + u^2)^n} \Diff {x} - \int _0^{\infty } \frac {1}{(1 + u^2)^{n + 1}} \Diff x \\ & = \int _0^{\infty } \frac {(1 + u^2) - 1}{(1 + u^2)^{n + 1}} \Diff {x} \\ & = \int _0^{\infty } \frac {u^2}{(1 + u^2)^{n + 1}} \Diff {x}. \end {align*}
Notice that \[ \DiffFrac {(1 + u^2)^{-n}}{x} = - \frac {2un}{(1 + u^2)^{n + 1}}, \] and therefore, \[ \frac {u\Diff {x}}{(1 + u^2)^{n + 1}} = - \frac {\Diff {(1 + u^2)^{-n}}}{2n}. \]
Using integration by parts, we have \begin {align*} I_n - I_{n + 1} & = \int _0^{\infty } \frac {u^2}{(1 + u^2)^{n + 1}} \Diff {x} \\ & = \int _0^{\infty } \left [ - \frac {u\Diff {(1 + u^2)^{-n}}}{2n}\right ] \\ & = \frac {1}{2n}\left [\int _0^{\infty } \frac {\Diff {u}}{(1 + u^2)^n} - \left [u \cdot (1 + u^2)^{-n}\right ]^{\infty }_0\right ] \\ & = \frac {1}{2n} \left [I_n - (0 - 0)\right ] \\ & = \frac {1}{2n} I_n, \end {align*}
as desired.
Hence, we have \[ I_{n + 1} = \left (1 - \frac {1}{2n}\right ) I_n = \frac {2n - 1}{2n} I_n. \]
Notice that \[ I_1 = \int _0^{\infty } \frac {\Diff u}{1 + u^2} = \left [\arctan u\right ]^{\infty }_0 = \frac {\pi }{2}, \] and therefore, we have \begin {align*} I_{n + 1} & = \frac {2n - 1}{2n} I_n \\ & = \frac {2n - 1}{2n} \cdot \frac {2n - 3}{2n - 2} \cdot I_{n - 1} \\ & = \frac {2n - 1}{2n} \cdot \frac {2n - 3}{2n - 2} \cdot \frac {2n - 5}{2n - 4} \cdot I_{n - 2} \\ & \phantom {=} \vdots \\ & = \frac {2n - 1}{2n} \cdot \frac {2n - 3}{2n - 2} \cdot \frac {2n - 5}{2n - 4} \cdots \frac {2 \cdot 1 - 1}{2 \cdot 1} \cdot I_1 \\ & = \frac {(2n - 1)(2n - 3)(2n - 5) \cdots 3 \cdot 1}{(2n) \cdot (2n - 2) \cdot (2n - 4) \cdots 4 \cdot 2} \cdot \frac {\pi }{2}. \end {align*}
Let \begin {align*} A & = (2n - 1)(2n - 3)(2n - 5) \cdots 3 \cdot 1, \\ B & = (2n) \cdot (2n - 2) \cdot (2n - 4) \cdots 4 \cdot 2. \end {align*}
Notice that \[ AB = (2n)!, B = 2^n \cdot n!, \] and therefore \[ A = \frac {(2n)!}{2^n \cdot n!}, \] and hence \begin {align*} I_{n + 1} & = \frac {(2n - 1)(2n - 3)(2n - 5) \cdots 3 \cdot 1}{(2n) \cdot (2n - 2) \cdot (2n - 4) \cdots 4 \cdot 2} \cdot \frac {\pi }{2} \\ & = \frac {(2n)! / (2^n \cdot n!)}{2^n \cdot n!} \cdot \frac {\pi }{2} \\ & = \frac {(2n)! \pi }{2^{2n + 1} (n!)^2}, \end {align*}
as desired.
If we do the substitution \(u = \frac {1}{x}\), we will have \(u \to 0^{+}\) as \(x \to \infty \), and \(u \to \infty \) as \(x \to 0^{+}\). We have \(\Diff u = - \frac {1}{x^2} \Diff x\). Therefore, \begin {align*} J & = \int _{0}^{\infty } f((x - x^{-1})^2) \Diff x \\ & = \int _{\infty }^{0} f((u^{-1} - u)^2) \left (- x^2 \Diff u\right ) \\ & = \int _{0}^{\infty } u^{-2} f((u - u^{-1})^2) \Diff u, \end {align*}
which is exactly the first equal sign as desired (since \(u\) is just an arbitrary variable).
For the second equal sign, notice that \begin {align*} 2J & = J + J \\ & = \int _{0}^{\infty } f((x - x^{-1})^2) \Diff x + \int _{0}^{\infty } x^{-2} f((x - x^{-1})^2) \Diff x \\ & = \int _{0}^{\infty } \left (1 + x^{-2}\right ) f((x - x^{-1})^2) \Diff x, \end {align*}
and therefore \[ J = \frac {1}{2} \int _{0}^{\infty } \left (1 + x^{-2}\right ) f((x - x^{-1})^2) \Diff x. \]
For the final equal sign, consider the substitution \(u = x - x^{-1}\). Note \(\Diff u = \left (1 + x^{-2}\right ) \Diff x\), and when \(x \to 0^{+}\), \(u \to -\infty \), when \(x \to \infty \), \(u \to \infty \). Therefore, \begin {align*} J & = \frac {1}{2} \int _{0}^{\infty } \left (1 + x^{-2}\right ) f((x - x^{-1})^2) \Diff x \\ & = \frac {1}{2} \int _{-\infty }^{\infty } f(u^2) \Diff u. \end {align*}
Since \(f(u^2) = f((-u)^2)\) for all \(u \in \RR \), we therefore have \[ \int _{-\infty }^{0} f(u^2) \Diff u = \int _{0}^{\infty } f(u^2) \Diff u, \] and hence \[ J = \int _{0}^{\infty } f(u^2) \Diff u, \] as desired.
Notice that the integrand satisfies \begin {align*} \frac {x^{2n - 2}}{(x^4 - x^2 + 1)^n} & = \frac {1}{x^2} \cdot \frac {(x^2)^{n}}{(x^4 - x^2 + 1)^n} \\ & = \frac {1}{x^2} \cdot \frac {1}{(x^2 - 1 + x^{-2})^n} \\ & = \frac {1}{x^2} \cdot \frac {1}{[(x - x^{-1})^2 + 1]^n}. \end {align*}
Therefore, consider the function \(f_n(x) = \frac {1}{(x + 1)^n}\), we have \begin {align*} \int _{0}^{\infty } \frac {x^{2n - 2}}{(x^4 - x^2 + 1)^n} \Diff x & = \int _{0}^{\infty } \frac {1}{x^2} \cdot \frac {1}{[(x - x^{-1})^2 + 1]^n} \cdot \Diff x \\ & = \int _0^{\infty } x^{-2} f_n((x - x^{-1})^2) \Diff x \\ & = \int _0^{\infty } f_n(u^2) \Diff u \\ & = \int _0^{\infty } \frac {\Diff u}{(u^2 + 1)^n} \\ & = \frac {(2n - 2)! \pi }{2^{2n - 1} ((n - 1)!)^2}. \end {align*}