Fig. 1

In my previous blog “A Case of Pre-FTC Definite Integration“, we obtained result

$\int\limits_{a}^{b} {x^p} dx= {{b^{p+1} - a^{p+1}} \over {p+1}}, p\in Z_{0}^{+}\quad\quad\quad(1)$

without the Fundamental Theorem of Calculus.

Let’s now consider the case of $\int\limits_{a}^{b} {x^p} dx$ where $p=-1$.  Namely, $\int\limits_{1}^{x}{1 \over u} du$, the area under the curve from 1 to $x$.

Closed form result (1) is not applicable since $p = -1 \in Z^{-}$.

Attempt of finding the limit of a sum quickly bites the dust too due to the fact that $\sum\limits_{i=1}^{n}{{x-1} \over n}{1 \over {1+{i{{x-1} \over n}}}}=\sum\limits_{i=1}^{n}{{x-1} \over {n + i(x-1)}}$.

However,  let

$\log{x} = \int\limits_{1}^{x}{1 \over u} du\quad\quad\quad(2)$

We see immediately that $\log{1} = \int\limits_{1}^{1}{1 \over u} du$, i.e.,

$\log{1} = 0\quad\quad\quad(3-0)$

Other properties of function $\log{x}$ can be extracted from (2), as shown below:

By definition,

$\log x_{1}x_{2} = \int\limits_{1}^{x_{1}x_{2}} { 1 \over s} ds = \int\limits_{1}^{x_{1}}{1 \over s} ds + \int\limits_{x_{1}}^{x_{1}x_{2}}{1 \over s }ds$,

$\log{x_{1}} = \int\limits_{1}^{x_{1}}{1 \over s} ds$

and

$\int\limits_{x_{1}}^{x_{1}x_{2}}{1 \over s} ds= \lim\limits_{n \to \infty}\sum\limits_{i=1}^{n}{1\over{x_{1}+i{{x_{1}x_{2}-x_{1}} \over n}}}{{x_{1}x_{2}-x_{1}} \over n}$

$= \lim\limits_{n \to \infty}\sum\limits_{i=1}^{n}{1\over{ ({1+i{{x_{2}-1} \over n}}) x_{1} }} {{x_{1}(x_{2}-1)} \over n}$

$= \lim\limits_{n \to \infty}\sum\limits_{i=1}^{n}{1\over{1+i{{x_{2}-1} \over n}}}{{x_{2}-1} \over n}=\int\limits_{1}^{x_{2}}{1 \over s}ds =\log{x_{2}}$.

Therefore,

$\log{x_{1}x_{2}} = \log{x_{1}} + \log{x_{2}}\quad\quad\quad(3-1)$

Let $x_{2}={1 \over x_{1}}$, we have

$\log{x_{1}{1 \over x_{1}}}=\log{1} = 0 = \log{x_{1}} + \log{1 \over x_{1}}$,

i.e.,

$\log{1 \over x_{1}} = - \log{x_{1}}\quad\quad\quad(3-2)$

By (3-2),

$\log {x_{1} \over x_{2}} =\log{x_{1}{1 \over x_{2}}}= \log{x_{1}} + \log{1 \over x_{2}}$

$= \log{x_{1}} - \log{x_{2}}$,

i.e.,

$\log{x_{1} \over x_{2}} = \log{x_{1}} - \log{x_{2}}\quad\quad\quad\quad(3-3)$

Let $p = 0$,

$\log{x^{p}}=\log{x^{0}} = \log{1} = 0 = 0\cdot \log{x} = p\log{x}\quad(3-4)$

When $p = 1$,

$\log{x^{p}}=\log{x} = 1\cdot \log{x}=p\log{x}$.

Assume when $p=k , \log{x^{k}}=k\log{x}$  where  $k \in Z^{+}$,  we have

$\log{x^{k+1}}=\log{x^{k} x} =\log{x^{k}} + \log{x}$

$=k \log{x} + \log{x} = (k+1)\log{x}$

Hence,

$\log{x^{p}} = p \log{x}, \forall p\in Z^{+}\quad\quad\quad\quad(3-5)$

Moreover, $\forall p \in Z^{-}, p = -k$ where $k \in Z^{+}$,

$\log{x^p} =\log{x^{-k}} = \log{1 \over x^{k}} = \log{1}-\log{x^k}$

$=-k\log{x} = p \log{x}$

As result,

$\log{x^p} = p\log{x}, \forall p \in Z^{-}\quad\quad\quad\quad(3-6)$

With (3-4), (3-5) and (3-6), we conclude that

$\log{x^p} = p\log{x}, \forall p \in Z\quad\quad\quad\quad(3-7)$

We will leave this post with the following observation:

$\lim\limits_{ h\to 0}{{\log(x+h)-\log{x}} \over {h}}={1 \over x}\quad\quad\quad(4)$

Fig. 2

This is not difficult to see. $\forall x > 0$, if $h > 0$, then from Fig. 2, we have

area in blue $=\int\limits_{x}^{x+h}{1 \over u} du = \int\limits_{1}^{x+h}{1 \over u} du -\int\limits_{1}^{x}{1 \over u}du$.

The fact that

${1 \over {x +h}} \cdot {h} <$ area in blue $< {1 \over {x}} \cdot {h}$

means

${1 \over x + {h}} { h} < \int\limits_{1}^{x+h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u}du < {1 \over x} {h}$,

or,

${1 \over { x + h}} <{{\int\limits_{1}^{x+h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u}du} \over {h}} < {1 \over {x}}$,

Since $\lim\limits_{h \to 0} {1 \over {x +h}} = {1 \over x}, \lim\limits_{h \to 0} {1 \over {x}} = {1 \over x}$,

$\lim\limits_{h \to 0}{{\int\limits_{1}^{x+h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u}du} \over {h}} = {1 \over x}$

from which (4) is obtained.

When $h < 0$, area in blue is

$\int\limits_{x + h}^{x}{1 \over u} du = -\int\limits_{x}^{x +h}{1 \over u} du = -(\int\limits_{1}^{x + h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u} du)$.

Fig. 3

From Fig. 3 we see that

${-{h} \over x } <-({\int\limits_{1}^{x+h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u}du})< {{-h}\over { x + h}}$.

Hence

${1 \over { x }}<{{\int\limits_{1}^{x+h}{1 \over u}du -\int\limits_{1}^{x}{1 \over u}du} \over {h}} < {1 \over {x +h}}$,

from which (4) is obtained again.

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