# The definition of the logarithm.

One usually gets several definitions of the logarithm along his studies.

1. You might be first introduced to the exponential and then told that the logarithm is its inverse.
2. You might be given
$$\log x = \int\limits_1^x {\frac{{du}}{u}}$$
3. Like Landau does. Let $k = 2^n$, then:
$$\log x =\mathop {\lim }\limits_{n \to \infty } k\left( {\root k \of x – 1} \right)$$
4. 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?

#### Solutions Collecting From Web of "The definition of the logarithm."

You can use the fact that it’s a uniform limit for $u \in [1,x]$, or use Dominated Convergence or Monotone Convergence.