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 \(\arctan x\).


$\leftarrow$ 196–197. The logarithmic function Main Page 199–201. The behaviour of $\log x$ as $x$ tends to infinity or to zero $\rightarrow$
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