One usually gets several definitions of the logarithm along his studies.
- You might be first introduced to the exponential and then told that the logarithm is its inverse.
- You might be given $$\log x = \int\limits_1^x {\frac{{du}}{u}} $$
- Like Landau does. Let $k = 2^n$, then: $$\log x =\mathop {\lim }\limits_{k \to \infty } k\left( {\root k \of x - 1} \right)$$
- And last, if you ever read, Euler famously wrote: $$ - \log x = \frac{{1 - {x^0}}}{0}$$
Landau's definition (although I find it the most usefull to work with) really baffled me untill just now. Since $$\int\limits_1^x {\frac{{du}}{{{u^{\alpha + 1}}}}} = \frac{{{x^{ - \alpha }} - 1}}{{ - \alpha }}$$
Then being $\frac{1}{k} = -\alpha$ one hopes to have:
$$\mathop {\lim }\limits_{\alpha \to 0} \int\limits_1^x {\frac{{du}}{{{u^{\alpha + 1}}}}} = \int\limits_1^x {\frac{{du}}{u}} = \log x = \mathop {\lim }\limits_{k \to \infty } k\left( {\root k \of x - 1} \right)$$
How can one justify taking the limit before integration? Continuity suffices?