\(% 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}}\)
Consider the substitution \(u = \frac {1}{v}\).
When \(u \to 0^{+}\), \(v \to \infty \).
When \(u = x\), \(v = \frac {1}{x}\).
We also have \[ \Diff u = - \frac {1}{v^2} \Diff v. \] Therefore, \begin {align*} T(x) & = \int _0^x \frac {\Diff u}{1 + u^2} \\ & = \int _{\infty }^{\frac {1}{x}} -\frac {1}{v^2} \cdot \frac {1}{1 + \frac {1}{v^2}} \Diff v \\ & = \int _{\frac {1}{x}}^{\infty } \frac {\Diff v}{1 + v^2} \\ & = \int _{0}^{\infty } \frac {\Diff v}{1 + v^2} - \int _{0}^{\frac {1}{x}} \frac {\Diff v}{1 + v^2} \\ & = T_{\infty } - T(x^{-1}), \end {align*}
as desired.
When \(u \neq a^{-1}\), we have \begin {align*} \frac {\Diff v}{\Diff u} & = \frac {\Diff }{\Diff u} \frac {u+a}{1-au} \\ & = \frac {1 \cdot (1-au) + a \cdot (u+a)}{(1-au)^2} \\ & = \frac {1 - au + au + a^2}{(1-au)^2} \\ & = \frac {1 + a^2}{(1-au)^2}. \end {align*}
Also, notice that \begin {align*} \frac {1 + v^2}{1 + u^2} & = \frac {1 + \left (\frac {u + a}{1 - au}\right )^2}{1 + u^2} \\ & = \frac {(1-au)^2 + (u+a)^2}{(1+u^2) (1-au)^2} \\ & = \frac {1 - 2au + a^2u^2 + u^2 + 2au + a^2}{(1+u^2) (1-au)^2} \\ & = \frac {(1 + a^2) (1 + u^2)}{(1 - au)^2 (1 + u^2)} \\ & = \frac {1 + a^2}{(1 - au)^2}. \end {align*}
Therefore, \(\frac {\Diff v}{\Diff u} = \frac {1 + v^2}{1 + u^2}\) as desired.
Consider the substitution \(v = \frac {u + a}{1 - au}\). When \(u = 0\), \(v = a\). When \(u = x, v = \frac {x + a}{1 - ax}\). Therefore, \begin {align*} T(x) & = \int _0^x \frac {\Diff u}{1 + u^2} \\ & = \int _a^{\frac {x+a}{1-ax}} \frac {1+u^2}{1+v^2} \cdot \frac {\Diff v}{1 + u^2} \\ & = \int _a^{\frac {x+a}{1-ax}} \frac {\Diff v}{1 + v^2} \\ & = \int _0^{\frac {x+a}{1-ax}} \frac {\Diff v}{1 + v^2} - \int _0^a \frac {\Diff v}{1 + v^2} \\ & = T\left (\frac {x+a}{1-ax}\right ) - T(a), \end {align*}
as desired.
If we substitute \(T(x) = T_{\infty } - T(x^{-1})\) and \(T(a) = T_{\infty } - T(a^{-1})\), we can see that \begin {align*} T(x) & = T\left (\frac {x+a}{1-ax}\right ) - T(a) \\ T_{\infty } - T(x^{-1}) & = T\left (\frac {x+a}{1-ax}\right ) - \left [T_{\infty } - T(a^{-1})\right ] \\ T(x^{-1}) & = 2T_{\infty } - T\left (\frac {x+a}{1-ax}\right ) - T(a^{-1}), \end {align*}
as desired.
Now, let \(y = x^{-1}\) and \(b = a^{-1}\). Then \begin {align*} \frac {x + a}{1 - ax} & = \frac {y^{-1} + b^{-1}}{1 - b^{-1} y^{-1}} \\ & = \frac {b + y}{by - 1}. \end {align*}
This gives us \[ T(y) = 2T_{\infty } - T\left (\frac {b + y}{by - 1}\right ) - T(b), \] as desired.
Let \(y = b = \sqrt {3}\). We can easily verify that \(b > 0\) and \(y > \frac {1}{b}\). Therefore, \[ T(\sqrt {3}) = 2 T_{\infty } - T\left (\frac {\sqrt 3 + \sqrt 3}{3 - 1}\right ) - T(\sqrt {3}), \] which simplified, gives us \(T(\sqrt {3}) = \frac {2}{3} T_{\infty }\) as desired.
In \(T(x) = T\left (\frac {x + a}{1 - ax}\right ) - T(a)\), let \(x = a = \sqrt {2} - 1\), we can verify that \(a > 0\) and \(x < \frac {1}{a}\), therefore we have \begin {align*} T(\sqrt {2} - 1) & = T\left (\frac {(\sqrt {2} - 1) + (\sqrt {2} - 1)}{1 - (\sqrt {2} - 1) \cdot (\sqrt {2} - 1)}\right ) - T(\sqrt {2} - 1), \\ T(\sqrt {2} - 1) & = T\left (\frac {2\sqrt {2} - 2}{1 - \left (2 + 1 - 2\sqrt {2}\right )}\right ) - T(\sqrt {2} - 1), \\ T(\sqrt {2} - 1) & = T\left (\frac {2\sqrt {2} - 2}{2\sqrt {2} - 2}\right ) - T(\sqrt {2} - 1), \\ 2T(\sqrt {2} - 1) & = T(1). \end {align*}
In \(T(x) = T_{\infty } - T(x^{-1})\), let \(x = 1\). We have \begin {align*} T(1) & = T_{\infty } - T(1), \\ 2T(1) & = T_{\infty }. \end {align*}
Therefore, \(T(\sqrt {2} - 1) = \frac {1}{4} T_{\infty }\), as desired.