Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$

A logarithm of base b for x is defined as the number u such that $b^u=x$. Thus, the logarithm with base $e$ gives us a $u$ such that $e^u=b$.

In the presentations that I have come across, the author starts with the fundamental property $f(xy) = f(x)+f(y)$ and goes on to construct the natural logarithm as $\ln(x) = \int_1^x \frac{1}{t} dt$.

It would be suprising if these two definitions ended up the same, as is the case. How do we know that the are? The best that I can think of is that they share property $f(xy) = f(x)+f(y)$, and coincide at certain obvious values (x=0, x=1). This seems weak. Is their a proof?

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Which presentations have you come across where the author chooses to do things in this order and does not give a proof that the way he's doing it works? – Henning Makholm Feb 15 '12 at 2:29
How do you define $e^u$? Without this information we do not know what the first definition of $\ln x$ means. – Jonas Meyer Feb 15 '12 at 2:30
@HenningMakholm Apostol's "One Variable Calculus". If there is a proof, I'm missing it. – Nathan Feb 15 '12 at 2:35
@JonasMeyer I'll see if I can find something about $e^u$. Maybe that will help me understand this. – Nathan Feb 15 '12 at 2:36
It’s right there in Apostol. He defines the function $\ln x$, shows that there is a unique real number $e$ such that $\ln e=1$, shows that $\ln$ has an inverse function that he temporarily calls $E$, proves that $E(r)=e^r$ for rational $r$, and finally defines $e^x$ to be $E(x)$ for irrational $x$. It is then automatic that $e^{\ln x}=x$. In my edition this is completed in Section 3.7. – Brian M. Scott Feb 15 '12 at 2:48

If you want the logarithm to be expressed as a function $f$ then the most important properties have to hold, which are

$$\log(xy) =\log(x)+\log(y)$$

$$\log(x^a) =a\log(x)$$

$$f(1)=0$$

And... suppose $f$ admits a derivative. Then fixing $y$ and differentiating the first equation gives:

$$yf'(xy) =f'(x)$$

Putting $x=1$ gives

$$f'(y) =\dfrac{f'(1)}{y}$$ for each $y\neq0$

From this equation we se the derivative is monotonous en each interval not containing the origin. Morover, $f'$ is continuous in $(c,x)$ with $c>0$ so we can apply the second $\mathcal{FTC}$:

$$f(x) - f(c) = \int_c^x f'(t) dt =f'(1) \int_c^x\frac{1}{t}dt$$

If $x>0$ this equation is valid for each nonnegative $c$, so choosing $c=1$ gives:

$$f(x) = f'(1) \int_1^x \frac{dt}{t}$$

You can readily check from this equation that the previous properties are met. Moreover, you can check that the logarithm will be the unique function that will satisfy the above requierements and $f'(1)=1$, which will give you the desired definition:

$$\log x = \int_1^x \frac{dt}{t}$$

What I gave you is rather a stub from Apostol's Calculus, pages 278-281, 2nd ed.

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I really like this "constructive" approach to the problem because a lot of mathematical problems really boil down to recognizing what properties we need something to have for a certain definition of it to make sense. – Mathemagician1234 Feb 15 '12 at 6:56
@Nathan Thanks for accepting. I'm glad this helped you. – Pedro Tamaroff Apr 23 '12 at 2:04

If x,y>0, Define $x=e^u,y=e^v$,so $f(e^{u+v})=f(e^u)+f(e^v)$,Define $g(u)=f(e^u)$,then $$g(u+v)=g(u)+g(v)$$ this is Cauchy Functional Equation ,g(u)=cu, the proof you can see How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable I think you can finish the following.

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$g(u)$ is continuous, and it is very easy to solve the Cauchy equation in this case... – N. S. Feb 15 '12 at 2:49

The following properties uniquely determine the natural log:

1) $f(1) = 0$.

2) $f$ is continuous and differentiable on $(0, \infty)$ with $f'(x) = \frac{1}{x}$.

3) $f(xy) = f(x) + f(y)$

We will show that the function $f(x) = \int_1^x \frac{1}{t} dt$ obeys properties 1,2, and 3, and is thus the natural log.

1) This is easy, since $f(1) = \int_1^1 \frac{1}{t} dt = 0$.

2) Defining $f(x) = \int_1^x \frac{1}{t} dt$, we note that since $\frac{1}{t}$ is continuous on any interval of the form $[a,b]$, where $0 < a \leq b$, then the Fundamental Theorem of Calculus tells us that $f(x)$ is (continuous and) differentiable with $f'(x) = \frac{1}{x}$ for all $x \in [a,b]$.

3) \begin{align} f(xy) = \int_1^{xy} \frac{1}{t}dt &= \int_1^x \frac{1}{t} dt + \int_x^{xy} \frac{1}{t} dt \\ &= f(x) + \int_{1}^{y} \frac{1}{u} du \\ &= f(x) + f(y) \end{align} where in the last step we perform the substitution $t = ux$ (viewing $x$ as constant).

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Put $l(x) = \int_1^x dt/t.$ Then $l'(x) = 1/x$ if $x > 0$ by the fundamental theorem of calculus. Since $l' > 0$ on $(0,\infty)$, $l$ is 1-1 and therefore has an inverse. Denote its inverse by $m$. We have $l(m(x))= x;$ differentiate to see that $$1 = l'(m(x))m'(x) = m'(x)/m(x).$$
Multiply to get $m'(x) = m(x).$ It is not at all hard to see that $m(0) = 1.$ Therefore, $m(x) = e^x$.

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Thank you. This is a really clear explanation! – Nathan Feb 15 '12 at 5:09

Let $f(x) = e^{\int_1^x \frac{1}{t} dt}$.

Then $f'(x)=\frac{1}{x}e^{\int_1^x \frac{1}{t} dt}=\frac{f(x)}{x}$.

hence

$$x f'(x)-f(x)=0 \Rightarrow \frac{x f'(x)-f(x)}{x^2}=0 \Rightarrow \left(\frac{f(x)}{x}\right)'=0 \,.$$

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I've done this before, but here I go again.

If $f(xy) = f(x)+f(y)$, $f(1)=0$ and

$f(x+h)-f(x) = f\left(x\left(1+\dfrac{h}{x}\right)\right)-f(x)$

$= f(x)+f\left(1+\dfrac{h}{x}\right)-f(x)$

$= f\left(1+\dfrac{h}{x}-f(1)\right)-$ so

$\dfrac{f(x+h)-f(x)}{h} = \dfrac{f\left(1+\dfrac{h}{x}\right)-f(1)}{h} = \dfrac{1}{x} \dfrac{f\left(1+\dfrac{h}{x}\right)-f(1)}{\dfrac{h}{x}}$.

Letting $h \to 0$, $f'(x) = \dfrac{f'(1)}{x}$.

Choose $f'(1) = 1$ to get the natural log.

Similarly, starting with $f(x+y) = f(x)f(y)$ we get $f'(x) = f'(0)f(x)$.

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