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Question in title really, something I always found strange when I was learning calculus.

I can see that $\int \frac{1}{x} dx$ can't be $\frac{x^0}{0}$ since this is not defined, and then the definite integral $\int_1^t \frac{1}{x} dx$ comes down to $$\lim_{\delta \rightarrow 0} \frac{t^{\delta}-1}{\delta} = \log (t).$$

But this understanding just comes from l'Hopital's rule, and also it still just seems really bizaare that the log function should fit into the set of power functions like this. Can anyone de-mystify this at all?

noname
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  • Something has to fit in there. Are you sure that you're specifically mystified that $\log x$ happens to be the function that fits? Or are you just trying to wrap your head around the idea that there is something that fits that isn't of the form $x \mapsto x^n$? –  Jun 26 '14 at 10:24
  • In $\frac{\mathrm d\ln x}{\mathrm dx}=\frac1x$ substitute $x=e^t$, so $\frac{\mathrm dt}{\mathrm de^t}=\frac1{e^t}$, or $\frac{\mathrm de^t}{\mathrm dt}={e^t}$. So the appearance of $\ln$ is explained by the relation between reciprocals and inverse functions – Hagen von Eitzen Jun 26 '14 at 10:27
  • First one, why log? I get that there is a phase transition between the integral of $x^{-1 - \delta}$ and $x^{-1 + \delta}$. It isn't intuitively clear why log fills that gap though. – noname Jun 26 '14 at 10:31
  • Note the limit you wrote doesn't exist: (for positive $x$) it's a limit of the form $1/0$, so L'Hôpital's rule doesn't apply. –  Jun 26 '14 at 10:36
  • Oh yeah, I missed the lower integral, edited... – noname Jun 26 '14 at 10:40
  • Hans, nice link, I think the second answer there is pretty close to what I was looking for, was maybe hoping for something a bit more geometrical, but you can mark closed if you think there isn't going to be any better discussion... – noname Jun 26 '14 at 10:45

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It is beautiful that this works out so nicely.

Here's a thought. The key property of $\log$ is that it's the inverse of $e^x$, and the key property of $e^x$ is that $\frac{d}{dx} e^x = e^x$. Because the derivative of $e^x$ is so simple, we might expect some correspondingly nice result for the derivative of its inverse function.

Differentiating both sides of \begin{equation} \log(e^x) = x \qquad (\spadesuit) \end{equation} yields \begin{align} & \log'(e^x) e^x = 1 \\ \implies & \log'(y) = \frac{1}{y} \quad \text{for all } y > 0. \end{align} So, the fact that the derivative of $\log(y)$ is $\frac{1}{y}$ is almost just a restatement of $(\spadesuit)$.

littleO
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Note the limit you wrote doesn't exist: it's of the form $1/0$.

However, let's instead consider the definite integral

$$ \int_1^x t^{a-1} \, dt = \frac{x^a - 1}{a} \qquad \qquad a \neq 0 $$

Taking the limit of this as $a \to 0$ we can do, and we can rationalize this being the logarithm as follows:

$$\begin{align} f(xy) = \lim_{a \to 0} \frac{(xy)^a - 1}{a} &= \lim_{a \to 0} \frac{x^a (y^a - 1) + (x^a - 1)}{a} \\&= \lim_{a \to 0} x^a \frac{y^a - 1}{a} + \lim_{a \to 0} \frac{x^a - 1}{a} \\&= 1 \cdot f(y) + f(x) \\&= f(x) + f(y) \end{align}$$

so it ought to satisfy the right algebraic identity. In fact, there's even a theorem that says any sufficiently nice function (I think continuous is enough) that satisfies $f(xy) = f(x) + f(y)$ must be a logarithm.

  • I edited in response to your comment as you were writing, so I made a correction, that's a great answer though! – noname Jun 26 '14 at 10:46