231. The connection between the logarithmic and inverse trigonometrical functions

We found in Ch.VI that the integral of a rational or algebraical function $\varphi (x,\alpha ,\beta ,\dots )$, 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, $$\begin{array}{}\text{(1)}& \int \frac{dx}{x^2+\alpha }=\frac{1}{\sqrt{\alpha }}\arctan \frac{x}{\sqrt{\alpha }}\end{array}$$ if $\alpha >0$, but $$\begin{array}{}\text{(2)}& \int \frac{dx}{x^2+\alpha }=\frac{1}{2\sqrt{-\alpha }}\log \left|\frac{x–\sqrt{-\alpha }}{x+\sqrt{-\alpha }}\right|\end{array}$$ 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 $$\begin{array}{}\text{(3)}& \arctan \left(\frac{x}{\alpha }\right)=\frac{1}{2i}\log \left(\frac{x–i\alpha }{x+i\alpha }\right)+C,\end{array}$$ 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) $$\log (x\pm i\alpha )=\frac{1}{2}\log (x^2+{\alpha }^2)\pm i(\varphi +2k\pi ),$$ where $k$ is an integer and $\varphi$ is the numerically least angle such that $\cos \varphi =x/\sqrt{x^2+{\alpha }^2}$ and $\sin \varphi =\alpha /\sqrt{x^2+{\alpha }^2}$. Thus $$\frac{1}{2i}\log \left(\frac{x–i\alpha }{x+i\alpha }\right)=-\varphi –l\pi ,$$ 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 $$\begin{array}{}\text{(4)}& \arctan x=\frac{1}{2i}\log \left(\frac{1+ix}{1–ix}\right),\end{array}$$ 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 $$\begin{array}{}\text{(5)}& \arccos x=-i\log \{x\pm i\sqrt{1–x^2}\},\arcsin x=-i\log \{ix\pm \sqrt{1–x^2}\},\end{array}$$ where $-1\le x\le 1$: each of these formulae also is ‘completely’ true.

Example. Solving the equation $$\cos u=x=\frac{1}{2}\{y+(1/y)\},$$ 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.