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I was looking for a direct proof that for $x > 0$ we have $$\int \frac{1}{x} dx = \ln(x)$$ that did not rely on the fundamental theorem of calculus but could not find one. Is there such a proof? If so I would be glad if someone could point me to it or reproduce it as an answer.

Thanks!

EDIT: I define $\ln(x)$ to be the inverse function of $e^x$ which I define via the usual summation.

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5 Answers

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Define $F(x) = \int\limits_1^x \frac{1}{t} dt$. The domain of $F$ shall be $(0, \infty)$.

We wish to prove that $F = \ln$.

We begin by noting some obvious facts.

Fact 1: $F$ is continuous and strictly increasing.

Proof: very straightforward.

Fact 2: $\int\limits_a^{ab} \frac{1}{t} dt = F(b)$ for all $a, b > 0$.

Proof: It can be proved by analysing Riemann sums that whenever $a > 0$ and $g$ is continuous on $[c, b]$, we have $\int\limits_{ac}^{ab} g(x/a) dx = a \int\limits_c^b g(x) dx$. This fact shows that $a \int\limits_a^{ab} \frac{1}{t} dt = \int\limits_a^{ab} \frac{a}{t} dt = \int\limits_{a \cdot 1}^{ab} \frac{1}{t/a} dt = a \int\limits_1^b \frac{1}{t} dt = aF(b)$, and thus $\int\limits_a^{ab} \frac{1}{t} dt = F(b)$.

Fact 3: $F(ab) = F(a) + F(b)$ for all $a, b > 0$.

Proof: $F(ab) = \int\limits_1^{ab} \frac{1}{t} dt = \int\limits_1^{a} \frac{1}{t} dt + \int\limits_a^{ab} \frac{1}{t} dt = F(a) + F(b)$.

Fact 4: $F(a^n) = n F(a)$ for all $n \in \mathbb{N}$, $a > 0$.

Proof: Induction on $n$.

Fact 5: $\lim\limits_{x \to \infty} F(x) = \infty$.

Proof: Note that $F$ is increasing, $F(2) > F(1) = 0$, and that $F(2^n) = n F(2)$. So $F$ can be made arbitrarily large by taking large powers of 2.

Fact 6: $F(a^n) = n F(a)$ for all $n \in \mathbb{Z}$, $a > 0$.

Proof: If $n \geq 0$, we have already proved this. If $n < 0$, then we have $0 = F(1) = F(a^n \cdot a^{-n}) = F(a^n) + -n \cdot F(a)$, and thus $F(a^n) = n F(a)$.

Fact 7: $F(a^q) = q F(a)$ for all $q \in \mathbb{Q}$, $a > 0$.

Proof: write $q = \frac{n}{m}$ where $n, m \in \mathbb{Z}$, $m \neq 0$. Then we see that $n F(a) = F(a^n) = F(a^{q \cdot m}) = F((a^q)^m) = m F(a^q)$. Thus, $n F(a) = m F(a^q)$. Thus, $F(a^q) = \frac{n}{m} F(a) = q F(a)$.

Fact 8: $F(a^x) = x F(a)$ for all $x \in \mathbb{R}$, $a > 0$.

Proof: One definition of $a^x$ can be taken as follows: Let $\{x_n\}_{n \in \mathbb{N}}$ be a Cauchy sequence of rationals such that $\lim\limits_{n \to \infty} x_n = x$. Then $a^x = \lim\limits_{n \to \infty} a^{x_n}$.

Under this definition, we see that $F(a^x) = F(\lim\limits_{n \to \infty} a^{x_n})$. By continuity, $F(\lim\limits_{n \to \infty} a^{x_n}) = \lim\limits_{n \to \infty} F(a^{x_n}) = \lim\limits_{n \to \infty} x_n F(a) = (\lim\limits_{n \to \infty} x_n) F(a) = x F(a)$.

Fact 9: There is a unique $e$ such that $F(e) = 1$. This is the definition of the constant known as $e$.

Proof: $F$ is strictly increasing, so there can't be more than 1 such $e$. We see that $F(1) = 0$ and that $\lim\limits_{x \to \infty} F(x) = \infty$. Thus, we can find some $x > 1$ such that $F(x) > 1$. Then $F(1) < 1 < F(x)$, so by the intermediate value theorem, there exists some $e$ between $1$ and $x$ such that $F(e) = 1$.

Fact 10: $F(e^x) = x$ for all $x$. Furthermore, $e^{F(x)} = x$ for all $x > 0$.

Proof: Clearly, we have $F(e^x) = x F(e) = x \cdot 1 = x$. From this fact, we see that $F(e^{F(x)}) = F(x)$. Since $F$ is increasing, it is injective, and therefore $e^{F(x)} = x$.

This is exactly what it means for $F(x) = \log_e(x)$. Therefore, $F$ is the natural log function.

In principle, you could define $e$ using some other definition. This would require proving that this other definition of $e$ agrees with our definition of $e$. I will run through one example of how this might be done. Another common definition of $e$ is the value of $\lim\limits_{n \to \infty} (1 + 1/n)^n$.

Fact 11: $\lim\limits_{n \to \infty} (1 + 1/n)^n = e$ (where $e$ represents our definition of $e$).

Proof: We first show that $\lim\limits_{n \to \infty} F((1 + 1/n)^n) = 1$. Note that $F((1 + 1/n)^n) = n F(1 + 1/n)$. So we're taking $\lim\limits_{n \to \infty} n F(1 + 1/n)$. Clearly, we see that $F(1 + 1/n) = \int\limits_1^{1 + 1/n} \frac{1}{t} dt < \int\limits_1^{1 + 1/n} 1 dt = 1/n$ for all $n$, and thus $n F(1 + 1/n) < 1$ for all $n$. Now consider some $\epsilon > 0$. Take $N$ large enough that $\frac{1}{1 + 1/N} > 1 - \epsilon$ - that is, take $N > (\frac{1}{1 - \epsilon} - 1)^{-1}$. Then for all $n \geq N$, we have that $f(t) > 1 - \epsilon$ for all $t \in [1, 1 + 1/n]$. Then we see that $\int\limits_1^{1 + 1/n} \frac{1}{t} dt > \int\limits_1^{1 + 1/n} (1 - \epsilon) dt = (1 - \epsilon) \frac{1}{n}$, and hence $n F(1 + 1/n) > 1 - \epsilon$. This shows that $\lim\limits_{n \to \infty} n F(1 + 1/n) = 1$.

Now since the function $x \mapsto e^x$ is continuous, we have $\lim\limits_{n \to \infty} (1 + 1/n)^n = \lim\limits_{n \to \infty} e^{F((1 + 1/n)^n)} = e^{\lim\limits_{n \to \infty} F((1 + 1/n)^n)} = e^1 = e$.

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You can use implicit differentiation to show that the derivative of $\ln x$ is $\frac1x$ which is what you want to show: If $y=\ln x$, then $x=e^y$. Differentiate with respect to $x$ to get $1=e^yy'$. So $y'=\frac{1}{e^y}=\frac{1}{x}$.

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$\int \frac 1x\ dx = \ln x+C$

It should be: $\int_1^a \frac 1x\ dx = \ln a$

let $F(a) =\int_1^a \frac 1x\ dx$

Then $F(ab) = \int_1^{ab} \frac 1x\ dx = \int_1^{a} \frac 1x\ dx + \int_a^{ab} \frac 1x\ dx$

We can then perform a u-substitution on the second integral. $u = ax, du = a\ dx$

$F(ab) = \int_1^{a} \frac 1x\ dx + \int_1^{b} \frac 1x\ dx = F(a) + F(b)$

$F(a)$ follows the rules of logarithms, thus $F(a)$ is a logarithm. The question remains, what is the base?

There is some value $e$ such that $F(e) = \int_1^{e} \frac 1x\ dx = 1.$ We can then use numerical methods to pin down the value of $e.$

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As notice all depends on the definition you are referring to, for example defining $\ln x$ as the inverse function of $e^x$ by $t=e^u \implies dt=e^u du$ we have

$$\int_1^x \frac{1}{t} dt = \int_0^{\ln x} \frac{1}{e^u} e^u du=\ln x$$

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Let’s use the Riemann Sum which is not a Fundamental Theorem of Calculus:

$$\int_a^b f(x) dx=\lim_{n\to\infty} \frac {b-a}{n}\sum_{k=1}^n f\left(a+k \frac {b-a}{n}\right) \implies \int_a^b \frac{dx}x=\lim_{n\to\infty} \frac {b-a}{n}\sum_{k=1}^n \frac1{a+k \frac {b-a}{n}} $$

Let’s now find a partial sum:

$$ \int_a^b \frac{dx}x=\lim_{n\to\infty} \frac {b-a}{n}\sum_{k=1}^n \frac1{a+k \frac {b-a}{n}} =\lim_{n\to\infty}\frac{b-a}{n} \frac{n \text H_{-\frac{an}{a-b}}-n\text H_{-\frac{bn}{a-b}}}{a-b}$$

where appear the Harmonic Numbers. Note that the definition of the Harmonic Numbers are:

$$\text H_z=z\sum_{k=0}^z \frac{1}{k}$$

Now let’s cancel and take the limit:

$$\int_a^b \frac{dx}x=\lim_{n\to\infty}\frac{b-a}{n} \frac{n \text H_{-\frac{an}{a-b}}-n\text H_{-\frac{bn}{a-b}}}{a-b}= \lim_{n\to\infty} \text H_{\frac{bn}{b-a}}-\text H_{\frac{an}{b-a}}=\ln(b)-\ln(a)=\ln\left(\frac ba\right)$$

Here is an example. Finally, use the other Fundamental Theorem of Calculus to find that:

$$\int \frac{dx}x = \lim_{n\to\infty} \text H_{xn}-\text H_n=\ln(x)$$

Please correct me and give me feedback!

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