Show that two functions are identical over the real positive axis

Question

A Mathematica computation yielded the pair of functions \begin{equation} g(u)= \frac{4 \pi \left(\sqrt{u^2-1}+u^2 \csc ^{-1}(u)\right)}{3 u^2} \end{equation} and \begin{equation} h(u)= \frac{2}{3} \pi \left(2 \sqrt{1-u^2} u+2 i \log \left(u+i \sqrt{1-u^2}\right)+\pi. \right) \end{equation} It appears that $h(u)=g(1/u)$ (and, so of course $h(1/u)=g(u)$ for $u \in [0,\infty]$). This does not seem immediately obvious. If I issue the \begin{equation} \rm{FullSimplify[g[1/u]-h[u],Assumptions->u >0]} \end{equation} command, Mathematica yields for this difference \begin{equation} -\frac{4}{3} \pi \left(\cos ^{-1}(u)+i \log \left(u+i \sqrt{1-u^2}\right)\right). \end{equation} So, is this last expression equal to zero over the positive real axis? (Obviously, the $-\frac{4}{3} \pi$ coefficient is irrelevant to this question.)

Answer 1

Consider $$y=\cos ^{-1}(u)+i \log \left(u+i \sqrt{1-u^2}\right)$$ and differentiate it with respect to $u$.

You will get $$y'=-\frac{1}{\sqrt{1-u^2}}+\frac{i \left(1-\frac{i u}{\sqrt{1-u^2}}\right)}{u+i \sqrt{1-u^2}}$$ Simplify this expression and get $y'=0$. So , $y$ is a constant.

Now, use $u=0$ in $y$, simplify again and you will see that the constant is $0$.