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.