The function $$\log x$$ satisfies the functional equation $\begin{equation*} f(xy) = f(x) + f(y). \tag{1} \end{equation*}$ For, making the substitution $$t = yu$$, we see that \begin{aligned} \log xy &= \int_{1}^{xy} \frac{dt}{t} = \int_{1/y}^{x} \frac{du}{u} = \int_{1}^{x} \frac{du}{u} – \int_{1}^{1/y} \frac{du}{u}\\ &= \log x – \log(1/y) = \log x + \log y,\end{aligned} which proves the theorem.

Example LXXXIII
1. It can be shown that there is no solution of the equation (1) which possesses a differential coefficient and is fundamentally distinct from $$\log x$$. For when we differentiate the functional equation, first with respect to $$x$$ and then with respect to $$y$$, we obtain the two equations $yf'(xy) = f'(x),\quad xf'(xy) = f'(y);$ and so, eliminating $$f'(xy)$$, $$xf'(x) = yf'(y)$$. But if this is true for every pair of values of $$x$$ and $$y$$, then we must have $$xf'(x) = C$$, or $$f'(x) = C/x$$, where $$C$$ is a constant. Hence $f(x) = \int \frac{C}{x}\, dx + C’ = C\log x + C’,$ and it is easy to see that $$C’ = 0$$. Thus there is no solution fundamentally distinct from $$\log x$$, except the trivial solution $$f(x) = 0$$, obtained by taking $$C = 0$$.

2. Show in the same way that there is no solution of the equation $f(x) + f(y) = f\left(\frac{x + y}{1 – xy}\right)$ which possesses a differential coefficient and is fundamentally distinct from $$\arctan x$$.