2017.3.6 Question 6

1. Consider the substitution $u=\frac{1}{v}$.

   When $u\to 0^{+}$, $v\to +\infty$.

   When $u=x$, $v=\frac{1}{x}$.

   We also have

   $$du=-\frac{1}{v^2}dv.$$

   Therefore,

   $$\begin{array}{llll}\hfill T(x)& ={\int \; <!--\; nolimits\; -->}_0^x\frac{du}{1+u^2}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{+\infty }^{\frac{1}{x}}-\frac{1}{v^2}\cdot \frac{1}{1+\frac{1}{v^2}}dv& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_{\frac{1}{x}}^{+\infty }\frac{dv}{1+v^2}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_0^{+\infty }\frac{dv}{1+v^2}-{\int \; <!--\; nolimits\; -->}_0^{\frac{1}{x}}\frac{dv}{1+v^2}& \hfill & \\ \hfill & =T_{\infty }-T(x^{-1}),& \hfill & \end{array}$$

   as desired.

2. When $u\ne a^{-1}$, we have

   $$\begin{array}{llll}\hfill \frac{dv}{du}& =\frac{d}{du}\frac{u+a}{1-\mathit{au}}& \hfill & \\ \hfill & =\frac{1\cdot (1-\mathit{au})+a\cdot (u+a)}{{(1-\mathit{au})}^2}& \hfill & \\ \hfill & =\frac{1-\mathit{au}+\mathit{au}+a^2}{{(1-\mathit{au})}^2}& \hfill & \\ \hfill & =\frac{1+a^2}{{(1-\mathit{au})}^2}.& \hfill & \end{array}$$

   Also, notice that

   $$\begin{array}{llll}\hfill \frac{1+v^2}{1+u^2}& =\frac{1+{\left(\frac{u+a}{1-\mathit{au}}\right)}^2}{1+u^2}& \hfill & \\ \hfill & =\frac{{(1-\mathit{au})}^2+{(u+a)}^2}{(1+u^2){(1-\mathit{au})}^2}& \hfill & \\ \hfill & =\frac{1-2\mathit{au}+a^2{u}^2+u^2+2\mathit{au}+a^2}{(1+u^2){(1-\mathit{au})}^2}& \hfill & \\ \hfill & =\frac{(1+a^2)(1+u^2)}{{(1-\mathit{au})}^2(1+u^2)}& \hfill & \\ \hfill & =\frac{1+a^2}{{(1-\mathit{au})}^2}.& \hfill & \end{array}$$

   Therefore, $\frac{dv}{du}=\frac{1+v^2}{1+u^2}$ as desired.

   Consider the substitution $v=\frac{u+a}{1-\mathit{au}}$. When $u=0$, $v=a$. When $u=x,v=\frac{x+a}{1-\mathit{ax}}$. Therefore,

   $$\begin{array}{llll}\hfill T(x)& ={\int \; <!--\; nolimits\; -->}_0^x\frac{du}{1+u^2}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_a^{\frac{x+a}{1-\mathit{ax}}}\frac{1+u^2}{1+v^2}\cdot \frac{dv}{1+u^2}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_a^{\frac{x+a}{1-\mathit{ax}}}\frac{dv}{1+v^2}& \hfill & \\ \hfill & ={\int \; <!--\; nolimits\; -->}_0^{\frac{x+a}{1-\mathit{ax}}}\frac{dv}{1+v^2}-{\int \; <!--\; nolimits\; -->}_0^a\frac{dv}{1+v^2}& \hfill & \\ \hfill & =T\left(\frac{x+a}{1-\mathit{ax}}\right)-T(a),& \hfill & \end{array}$$

   as desired.

   If we substitute $T(x)=T_{\infty }-T(x^{-1})$ and $T(a)=T_{\infty }-T(a^{-1})$, we can see that

   $$\begin{array}{llll}\hfill T(x)& =T\left(\frac{x+a}{1-\mathit{ax}}\right)-T(a)& \hfill & \\ \hfill T_{\infty }-T(x^{-1})& =T\left(\frac{x+a}{1-\mathit{ax}}\right)-\left[T_{\infty }-T(a^{-1})\right]& \hfill & \\ \hfill T(x^{-1})& =2T_{\infty }-T\left(\frac{x+a}{1-\mathit{ax}}\right)-T(a^{-1}),& \hfill & \end{array}$$

   as desired.

   Now, let $y=x^{-1}$ and $b=a^{-1}$. Then

   $$\begin{array}{llll}\hfill \frac{x+a}{1-\mathit{ax}}& =\frac{y^{-1}+b^{-1}}{1-b^{-1}{y}^{-1}}& \hfill & \\ \hfill & =\frac{b+y}{\mathit{by}-1}.& \hfill & \end{array}$$

   This gives us

   $$T(y)=2T_{\infty }-T\left(\frac{b+y}{\mathit{by}-1}\right)-T(b),$$

   as desired.

3. Let $y=b=\sqrt{3}$. We can easily verify that $b>0$ and $y>\frac{1}{b}$. Therefore,

   $$T(\sqrt{3})=2T_{\infty }-T\left(\frac{\sqrt{3}+\sqrt{3}}{3-1}\right)-T(\sqrt{3}),$$

   which simplified, gives us $T(\sqrt{3})=\frac{2}{3}{T}_{\infty }$ as desired.

   In $T(x)=T\left(\frac{x+a}{1-\mathit{ax}}\right)-T(a)$, let $x=a=\sqrt{2}-1$, we can verify that $a>0$ and $x<\frac{1}{a}$, therefore we have

   $$\begin{array}{llll}\hfill T(\sqrt{2}-1)& =T\left(\frac{(\sqrt{2}-1)+(\sqrt{2}-1)}{1-(\sqrt{2}-1)\cdot (\sqrt{2}-1)}\right)-T(\sqrt{2}-1),& \hfill & \\ \hfill T(\sqrt{2}-1)& =T\left(\frac{2\sqrt{2}-2}{1-\left(2+1-2\sqrt{2}\right)}\right)-T(\sqrt{2}-1),& \hfill & \\ \hfill T(\sqrt{2}-1)& =T\left(\frac{2\sqrt{2}-2}{2\sqrt{2}-2}\right)-T(\sqrt{2}-1),& \hfill & \\ \hfill 2T(\sqrt{2}-1)& =T(1).& \hfill & \end{array}$$

   In $T(x)=T_{\infty }-T(x^{-1})$, let $x=1$. We have

   $$\begin{array}{llll}\hfill T(1)& =T_{\infty }-T(1),& \hfill & \\ \hfill 2T(1)& =T_{\infty }.& \hfill & \end{array}$$

   Therefore, $T(\sqrt{2}-1)=\frac{1}{4}{T}_{\infty }$, as desired.