In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions.
The function \(\T\) is defined for \(x>0\) by
\[
\T(x) = \int_0^x \! \frac 1 {1+u^2} \, \d u\,,
\]
and $\displaystyle T_\infty =
\int_0^\infty \!\! \frac 1 {1+u^2} \, \d u\,$ (which has a finite value).
By making an appropriate substitution in the integral for \(\T(x)\),
show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
Let \(v= \dfrac{u+a}{1-au}\), where \(a\) is a constant. Verify that, for
\(u\ne a^{-1}\),
\[
\frac{\d v}{\d u} = \frac{1+v^2}{1+u^2}
\,.
\]
Hence show that, for \(a>0\) and \(x< \dfrac1a\,\),
\[
\T(x) = \T\left(\frac{x+a}{1-ax}\right) -\T(a) \,.
\]
Deduce that
\[
\T(x^{-1})
= 2\T_\infty -\T\left(\frac{x+a}{1-ax}\right)
-\T(a^{-1})
\]
and hence that, for
\(b>0\) and \(y>\dfrac1b\,\),
\[
\T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,.
\]
Use the above results to show that
\(\T(\sqrt3)= \tfrac23 \T_\infty \,\)
and
\(\T(\sqrt2 -1)= \frac14 \T_\infty\,\).
Solution:
\(\,\) \begin{align*}
&& T(x) &= \int_0^x \! \frac 1 {1+u^2} \, \d u \\
&&&= \int_0^{\infty} \frac{1}{1+u^2} \d u - \int_x^\infty \frac{1}{1+u^2} \d u \\
&&&= T_\infty - \int_x^\infty \frac{1}{1+u^2} \d u \\
u = 1/v, \d u = -1/v^2 \d v: &&&= T_\infty - \int_{v=x^{-1}}^{v=0} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\
&&&= T_\infty - \int_{0}^{x^{-1}} \frac{1}{1+v^2} \d v \\
&&&= T_\infty - T(x^{-1})
\end{align*}