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 \(\pi\).
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 \(2t(1)= \frac12 p\,\).
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 \(\s\big(\frac1{\sqrt2}\big) =\frac1 4 p\,\).
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 \(3t(\frac1{\sqrt3}) = \frac12 p\,\).
Solution:
\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*}