231. The connection between the logarithmic and inverse trigonometrical functions

We found in Ch.VI that the integral of a rational or algebraical function \(\phi(x, \alpha, \beta, \dots)\), where \(\alpha\), \(\beta\), … are constants, often assumes different forms according to the values of \(\alpha\), \(\beta\), …; sometimes it can be expressed by means of logarithms, and sometimes by means of inverse trigonometrical functions. Thus, for example, \[\begin{equation*} \int \frac{dx}{x^{2} + \alpha} = \frac{1}{\sqrt{\alpha}} \arctan \frac{x}{\sqrt{\alpha}} \tag{1} \end{equation*}\] if \(\alpha > 0\), but \[\begin{equation*} \int \frac{dx}{x^{2} + \alpha} = \frac{1}{2\sqrt{-\alpha}} \log \left|\frac{x – \sqrt{-\alpha}}{x + \sqrt{-\alpha}}\right| \tag{2} \end{equation*}\] if \(\alpha < 0\). These facts suggest the existence of some functional connection between the logarithmic and the inverse circular functions. That there is such a connection may also be inferred from the facts that we have expressed the circular functions of \(\zeta\) in terms of \(\exp i\zeta\), and that the logarithm is the inverse of the exponential function.

Let us consider more particularly the equation \[\int \frac{dx}{x^{2} – \alpha^{2}} = \frac{1}{2\alpha} \log \left(\frac{x – \alpha}{x + \alpha}\right),\] which holds when \(\alpha\) is real and \((x – \alpha)/(x + \alpha)\) is positive. If we could write \(i\alpha\) instead of \(\alpha\) in this equation, we should be led to the formula \[\begin{equation*} \arctan \left(\frac{x}{\alpha}\right) = \frac{1}{2i} \log\left(\frac{x – i\alpha}{x + i\alpha}\right) + C, \tag{3} \end{equation*}\] where \(C\) is a constant, and the question is suggested whether, now that we have defined the logarithm of a complex number, this equation will not be found to be actually true.

Now (§ 221) \[\log(x \pm i\alpha) = \tfrac{1}{2} \log(x^{2} + \alpha^{2}) \pm i(\phi + 2k\pi),\] where \(k\) is an integer and \(\phi\) is the numerically least angle such that \(\cos\phi = x/\sqrt{x^{2} + \alpha^{2}}\) and \(\sin\phi = \alpha/\sqrt{x^{2} + \alpha^{2}}\). Thus \[\frac{1}{2i} \log\left(\frac{x – i\alpha}{x + i\alpha}\right) = -\phi – l\pi,\] where \(l\) is an integer, and this does in fact differ by a constant from any value of \(\arctan(x/\alpha)\).

The standard formula connecting the logarithmic and inverse circular functions is \[\begin{equation*} \arctan x = \frac{1}{2i} \log\left(\frac{1 + ix}{1 – ix}\right), \tag{4} \end{equation*}\] where \(x\) is real. It is most easily verified by putting \(x = \tan y\), when the right-hand side reduces to \[\frac{1}{2i} \log\left(\frac{\cos y + i\sin y}{\cos y – i\sin y}\right) = \frac{1}{2i} \log(\exp 2iy) = y + k\pi,\] where \(k\) is any integer, so that the equation is ‘completely’ true (Ex. XCIII. 3). The reader should also verify the formulae \[\begin{equation*} \arccos x = -i \log\{x \pm i\sqrt{1 – x^{2}}\},\quad \arcsin x = -i \log\{ix \pm \sqrt{1 – x^{2}}\}, \tag{5} \end{equation*}\] where \(-1 \leq x \leq 1\): each of these formulae also is ‘completely’ true.

Example. Solving the equation \[\cos u = x = \tfrac{1}{2}\{y + (1/y)\},\] where \(y = \exp(iu)\), with respect to \(y\), we obtain \(y = x \pm i\sqrt{1 – x^{2}}\). Thus: \[u = -i \log y = -i \log\{x \pm i\sqrt{1 – x^{2}}\},\] which is equivalent to the first of the equations (5). Obtain the remaining equations (4) and (5) by similar reasoning.