198. The functional equation satisfied by (log x)

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

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199–201. The behaviour of $\log x$ as $x$ tends to infinity or to zero $\rightarrow$