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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?

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Euler's definition is really the same as Landau's with the limit elided, isn't it? – Rahul Feb 10 '12 at 3:29
Yes it is. But I found it interesting that he put it that way. – Pedro Tamaroff Feb 10 '12 at 3:32
Which Landau? Lev Landau, the physicist? – Ben Crowell Feb 10 '12 at 4:12
@BenCrowell Edmund Landau, the rigorist. =) – Pedro Tamaroff Feb 10 '12 at 4:14
There is also the lemma-definition that log is the unique continuous homomorphism from $(\mathbb{R}^\times, \times)$ to $(\mathbb{R}, +)$ that has unit slope at 1. – Scott Carnahan Feb 10 '12 at 5:02
up vote 3 down vote accepted

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

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Could you expand a little? – Pedro Tamaroff Feb 10 '12 at 3:25
I was going to say dominated convergence or monotone convergence. (But I'm only here 23 hours per day, so Robert Israel beat me to it.) – Michael Hardy Feb 10 '12 at 3:55
@Robert Could you expand a little on your answer? Thank you. – Pedro Tamaroff Feb 21 '12 at 22:01
@MichaelHardy Could you answer and explain a little? Robert doesn't seem to be responding – Pedro Tamaroff Feb 24 '12 at 23:47
$f_\alpha(u) = 1/u^{\alpha+1}$ and $f(u) = 1/u$ are continuous functions on $[1,x]$, and $f_\alpha(u) \to f(u)$ uniformly for $u \in [1,x]$ as $\alpha \to 0$, so $\int_1^x f_\alpha(u)\ du \to \int_1^x f(u)\ du$. In fact, $\left|\int_1^x f_\alpha(u) \ du - \int_1^x f(u)\ du\right| \le |x-1| \sup_{1 \le u \le x} \left|f_\alpha(u) - f(u)\right|$ – Robert Israel Feb 28 '12 at 6:27

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