Problem source

The functions $\s(x)$ ($0\le x<1$) and $t(x)$ ($x\ge0$), 
and the real number $p$, are defined 
by
\[
\s(x) = \int_0^x \frac 1  {\sqrt{1-u^2}}\, \d u\;, \ \ \ \ 
t(x) = \int_0^x \frac 1  {1+u^2}\, \d u\;, \ \ \ \ 
p= 2 \int_0^\infty \frac 1 {1+u^2}\, \d u \;.
\]  
For this question, do not evaluate any of the above integrals explicitly in terms of inverse trigonometric functions or the number $\pi$.
\begin{questionparts}
\item  Use the substitution $u=v^{-1}$ to show that $\displaystyle t(x) =\int_{1/x}^\infty\frac 1 {1+v^2}\, \d v \, $. Hence evaluate $t(1/x) + t(x)$ in terms of $p$ and deduce that $2t(1)= \frac12 p\,$.
\item Let $y=\dfrac{u}{\sqrt{1+u^2}}$. Express $u$ in terms  of $y$,
and show that $\displaystyle \frac{\d u}{\d y} = \frac 1 {\sqrt{(1-y^2)^3}}$. By making a substitution  in the integral for $t(x)$, show that
\[
t(x) = \s\left(\frac{x}{\sqrt{1+x^2}}\right)\!.
\]
Deduce that $\s\big(\frac1{\sqrt2}\big) =\frac1 4 p\,$.
\item Let $z= \dfrac{u+ \frac1{\sqrt3}}{1-\frac 1{\sqrt3}u}\,$. Show that $\displaystyle t(\tfrac1{\sqrt3}) = \int_{\frac1{\sqrt3}}^{\sqrt3} \frac1 {1+z^2} \,\d z\;, $ and hence  that $3t(\frac1{\sqrt3}) = \frac12 p\,$.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\
u = v^{-1}, \d u = -v^{-2} \d v&&&= \int_{v = \infty}^{v = 1/x} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\
&&&= \int_{1/x}^\infty \frac{1}{1+v^2} \d v \\
\\
\Rightarrow && t(x) + t(1/x) &= \int_0^x \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\
&&&= \int_{1/x}^{\infty} \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\
&&&= \int_0^{\infty} \frac{1}{1+u^2} \d u \\ 
&&&= \frac12 p \\
\\
\Rightarrow && t(1) +t(1/1) = 2t(1) &= \frac12 p
\end{align*}

\item $\,$
\begin{align*}
&& y &= \frac{u}{\sqrt{1+u^2}} \\
\Rightarrow && y^2 &= \frac{u^2}{1+u^2} \\
&&&= 1-\frac{1}{1+u^2} \\
\Rightarrow && 1+u^2 &= \frac{1}{1-y^2} \\
\Rightarrow && u &= \frac{y}{\sqrt{1-y^2}} \\
\\
&& \frac{\d u}{\d y} &= \frac{\sqrt{1-y^2} + y^2(1-y^2)^{-1/2}}{1-y^2} \\
&&&= \frac{1}{(1-y^2)^{3/2}} \\
\\
&& t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\
&&&= \int_0^{y = x/\sqrt{1+x^2}} \frac{1}{1 + \frac{y^2}{1-y^2}} \frac{1}{(1-y^2)^{3/2}} \d y \\
&&&= \int_0^{x/\sqrt{1+x^2}} \frac{1-y^2}{(1-y^2)^{3/2}} \d y \\
&&&= \int_0^{x/\sqrt{1+x^2}} \frac{1}{(1-y^2)^{1/2}} \d y \\
&&&= s\left ( \frac{x}{\sqrt{1+x^2}} \right) \\
\\
\Rightarrow && s\left ( \frac{1}{\sqrt{2}} \right) &= t(1) = \frac14p
\end{align*}

\item $\,$  
\begin{align*}
 && z &= \frac{u + \frac1{\sqrt{3}}}{1- \frac{1}{\sqrt{3}} u}\\
\Rightarrow && z - \frac{z}{\sqrt{3}}u &= u + \frac{1}{\sqrt{3}} \\
\Rightarrow && u &= \frac{z-\frac{1}{\sqrt{3}}}{1 + \frac{z}{\sqrt{3}}} \\
\\
\Rightarrow && \frac{\d u}{\d z} &= \frac{\sqrt{3}(\sqrt{3}+z ) -(\sqrt{3}z-1)}{\left (\sqrt{3}+z \right)^2} \\
&&&= \frac{4}{(\sqrt{3}+z)^2} \\
\\
\Rightarrow && t \left ( \frac{1}{\sqrt{3}} \right) &= \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \d u \\
&&&= \int_{z=1/\sqrt{3}}^{z=\sqrt{3}} \frac{1}{1 + \left ( \frac{\sqrt{3}z-1}{\sqrt{3}+z}\right)^2} \frac{4}{(\sqrt{3}+z)^2} \d z\\
&&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{(\sqrt{3}+z)^2+(\sqrt{3}z-1)^2} \d z \\
&&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{4+4z^2} \d z \\
&&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \d z
\end{align*}

Notice that $t(1/\sqrt{3})+t(\sqrt{3}) = \frac12p$ and also notice that $t(1/\sqrt{3}) + t(1/\sqrt{3}) =t(\sqrt{3})$ so $3t(1/\sqrt{3}) = \frac12p$
\end{questionparts}