Why is $\frac{1}{x} \lt 1$ important in this proof of $\ln\left( \frac{1}{x}\right) = -\ln(x) $?

Question

I am studying the following proof that $\ln( \frac{1}{x}) = -\ln(x) $ using the integral definition of $\ln(x)$, and am wondering why it is important that $\frac{1}{x} \lt 1$, which before the author begins this proof he stresses we assume.

First, we have $$\ln\left(\frac {1}{x}\right) = \int_1^\frac{1}{x}\frac{dt}{t}. $$ Then let $u = \dfrac{1}{t}$, so $du = \dfrac{-dt}{t^2}$ and $\dfrac{du}{u} = \dfrac{-dt}{t}$. Changing endpoints accordingly, the integral becomes $$ -\int_1^x \frac {du}{u} = - \ln(x).$$ I have 3 questions regarding this:

1. Why is $\frac{1}{x} \lt 1$ important here? it seems this proof would work without it. If our intentions are to accord with the domain of $\ln(\frac{1}{x})$, it seems we only need $\frac{1}{x} \ge 0$.

2. What basis do we have to assume this fact? Is the case $x\in(0,1]$ irrelevant?

3. Is it necessary for our proof to only allow for $x$ in the domain of $\ln(x)$? It seems that even if we prove this for all values of $x$, we have shown that it is true for all $x$ in the domain of $\ln(x)$.

Answer 1

There are certain contexts, e.g. differential forms or Lebesgue integration, where you need to be careful about orientation when changing variables. I thought that might be what's going on here, but the proof presented requires a notion of integration over an oriented interval, and orientation-reversing change of variables, whether $\frac{1}{x} < 1$ or not.

So my best guess is that the author made a mistake, or was being unnecessarily cautious. Maybe they meant to write $\frac{1}{x} > 0$ (which is important)?

Answer 2

Though i am new to calculus, but i may have an answer.

I think it isnt required if you just do:

$$\ln\left(\frac1x\right) = \ln(x^{-1}) = -\ln(x)$$

Answer 3

It seems just a WLOG assumption such that

$$\ln\left(\frac {1}{x}\right) = \int_1^\frac{1}{x}\frac{dt}{t}<0$$

and accordingly

$$\int_1^x \frac {du}{u} = \ln(x)>0$$

but all works fine also assuming (wlog) $x<1$.