MHB What is Stoll's definition of the natural logarithm function?

[Math Amateur]

I am reading Manfred Stoll's book: Introduction to Real Analysis.

I need help with Stoll's definition of the natural logarithm function (page 234 -235)

The relevant section of Stoll reads as follows:

View attachment 3975
In this section we read:

" ... ... To prove (a), consider the function $$L(ax), x \gt 0$$. By the Chain Rule (Theorem 5.1.6)

$$\frac{d}{dx} L(ax) = \frac{1}{ax} a = \frac{1}{x} = L'(x) $$

... ... "I am somewhat puzzled by the above text ... I hope someone can clarify it for me ...

I will try to make my problem clear ... as follows ... ...

In the above text that I have quoted, the Chain Rule and the Fundamental Theorem of the Calculus (FTC) are used ...

... BUT ... the FTC asserts that if:

$$F(x) = \int_a^x f$$(t) dt

then

$$F'(c) = f(c)$$ where $$c$$ belongs to an interval $$[a,b] $$... BUT ... in the text on the natural logarithm function, Stoll seems (confusingly in my opinion!) that

$$F'(ax) = f(ax)$$ ... ... !?

... BUT ... $$x$$ is the upper limit of the integral ...

How then are we to interpret this use of FTC? ... ... indeed, if $$a \gt 1$$, then $$ax \gt x$$ where $$x$$ is the upper limit of the integral $$\int_a^x f(t) dt$$ ... ...

I thus find the above puzzling and confusing ...

Can someone please clarify the above issue for me ... ...

Peter***EDIT***

Since my post above refers to Stoll's statement of the Fundamental Theorem of the Calculus, I am providing Stoll's statement of the relevant version of the Fundamental Theorem of the Calculus as follows:View attachment 3976
View attachment 3977

 

[Dethrone]

Hi Peter,

If I am understanding your question correctly, you are confused as to why we are able to apply FTC when $ax$ is not in the interval $(a,x)$ in $\displaystyle \int_1^x f(t) \,dt$? You can view the $x$ in the FTC and the $x$ in $ax$ are different $x$'s. In the case of the $ax$, we are to interpret that $ax\in(0,\infty)$. I think, more clearly, we can say $L(s)=\int_{1}^{s} f(t) \,dt$, and so let $s=ax \in(0,\infty)$, then $L'(ax)=\frac{1}{x}$.

So indeed, it follows from the definition where $c=ax$ and $c$ belongs in the interval $[a,b]$. Again, this $a$ is different from the above $a$.

 

[Prove It]

While entirely correct, this definition of the natural logarithm seems extremely convoluted. It is much more concise to define the natural exponential function $\displaystyle \begin{align*} y = \mathrm{e}^x \end{align*}$ as being the exponential function which is equal to its own derivative. Since this function is one-to-one, its inverse is a function, which is defined as the natural logarithm.

 

[Prometheus1]

While entirely correct, this definition of the natural logarithm seems extremely convoluted. It is much more concise to define the natural exponential function $\displaystyle \begin{align*} y = \mathrm{e}^x \end{align*}$ as being the exponential function which is equal to its own derivative. Since this function is one-to-one, its inverse is a function, which is defined as the natural logarithm.

It's the book (what analysis text uses x as both bound and free variable? I mean, really?).

The definition he gave is actually nice. This is the proof he should have presented:

Consider $ \displaystyle \int_1^{ab}\frac{1}{t}\;{dt} = \int_1^a\frac{1}{t}\;{dt}+\int_a^{ab}\frac{1}{t}\;{dt}. $ Let $t = au$ for the last integral

We have $ \displaystyle \int_a^{ab}\frac{1}{t}\;{dt} = \int_1^{b} \frac{1}{u}\;{du}$, therefore $\displaystyle \int_1^{ab}\frac{1}{t}\;{dt}=\int_1^a\frac{1}{t}\;{dt}+\int_1^b\frac{1}{t}\;{dt}.$

 

[chisigma]

The definition of 'natural logarithm' through a definite integral has, in my opinion, criticalities when the variable is a complex number ...

$\displaystyle \ln z = \int_{1}^{z} \frac{d s}{s}\ (1)$

Let suppose we want to calculate $\displaystyle \ln z$ with z=-1 using (1)... referring to the figure we see that starting from z = 1 we arrive at z = -1 along two different paths, the top or bottom half circle ... in the first case we have...

$\displaystyle \ln -1 = i\ \int_{0}^{\pi} e^{- i\ \theta}\ e^{i\ \theta}\ d \theta = i\ \int_{0}^{\pi} d \theta = \pi\ i\ (2)$

... and in the second case we have...

$\displaystyle \ln -1 = i\ \int_{0}^{- \pi} e^{- i\ \theta}\ e^{i\ \theta}\ d \theta = i\ \int_{0}^{- \pi} d \theta = - \pi\ i\ (3)$

Kind regards

$\chi$ $\sigma$