33

Who discovered the non-obvious $\int \frac 1x dx=\ln(x)+c$ ? Were power series involved? The series look similar on opposite sides of 1:

$$ \frac 1x =\sum_{n=0}^\infty (-1+x)^n \text{ for } |x-1|<1 $$

$$ \ln(x) = \ln(-1+x)-\sum_{k=1}^\infty \frac{(-1)^k}{k(-1+x)^k} \text{ for } |x-1|>1 $$

$$ \ln(x) = -\sum_{k=1}^{\infty}\frac{(-1)^k(-1+x)^k}k \text{ for } |x-1|<1 $$

Michael Hardy
  • 1
User3910
  • 2,390
  • 2
    I don't know who realized it, but I suspect it was discovered by looking at $f(x)=\int_1^x \frac{1}{t} dt$ and proving $f(xy)=f(x)+f(y)$ through splitting up the integral and then a substitution. Then the idea for a log would follow naturally. – abnry Jul 12 '13 at 17:51
  • 3
    $e$ was invented such that that happens. The fact that the area of a hyperbola is a logarithm was long known. – OR. Jul 12 '13 at 17:59
  • 3
    @JoeHobbit I'm curious to know how you imagine the power series for $\ln x$ would have been known before its first derivative. – Erick Wong Jul 12 '13 at 18:01
  • 2
    @ErickWong I'm ignorant and imaginative. – User3910 Jul 12 '13 at 18:02
  • 1
    @JoeHobbit: What is your definition of $\ln x$? One of the most common definitions of $\ln x$ is that it satisfies this equation (integrating from $1$ to $x$), in which case there's nothing to realize. – TMM Jul 12 '13 at 18:03
  • 1
    Wikipedia seems to suggest that this was the first definition of the natural logarithm, so perhaps the question is the one suggested by @RGB: who figured out that the area under a hyperbola is logarithmic? – dfeuer Jul 12 '13 at 18:05
  • 2
    @nayrb Note that the substitution you describe is just a linear scaling, so the identity can be deduced by simple geometric principles rather than any calculus. – Erick Wong Jul 12 '13 at 18:08
  • It is likely that the property $f(xy)=f(x)+f(y)$ was long known. I haven't, but it would be nice to see if Fermat's Treatise on quadratures has it. – OR. Jul 12 '13 at 18:08
  • 2
    Apparently the first guy to get the quadrature of the hyperbola was a guy called Gregoire de Saint-Vincent. – OR. Jul 12 '13 at 18:10
  • 8
    The true story is extremely complicated, would take a long article to do without distortion. One could probably begin with the Mercator projection, an object of mathematical study since the late $16$-th century, with things neatly tied up finally by Wren, before calculus. And yes, there were power series before differentiation. Napier, again before calculus, knew the result, anyone who calculated log tables using interpolation knew the result. – André Nicolas Jul 12 '13 at 18:12
  • @AndréNicolas Who is this Wren fellow you speak of? Google has too many wrens (even Mathematicians). Also how are mercator projections related? – User3910 Jul 12 '13 at 18:19
  • 3
    Wren is the famous architect Wren. He did mathematics too! In the Mercator projection, $\log(\sec x)$ occurs naturally. In a sense $\log(\sec)$ was discovered before $\log$. – André Nicolas Jul 12 '13 at 18:27
  • Related – User3910 Jul 31 '13 at 22:44
  • Well... it's only mostly true. The most precise statement is that $x \mapsto \frac1x$ has $x \mapsto \log |x| + C(x)$—where $C$ is a locally constant function—as an antiderivative on $\mathbb R \setminus {0}$. – kahen Oct 25 '13 at 05:49

3 Answers3

20

By about 1640, the solution to the "area problem" for curves with equation $Y^n = aX^m$ was known by Fermat for all integer cases except when $n = 1, m = -1$.

I.e., the only unsolved area problem was for $Y = \frac 1X$ - the standard equation for the graph of a hyperbola.

In 1647, Gregoire de St. Vincent showed the following special property for hyperbolas:

enter image description here

If: $\frac ab = \frac cd$ then $\int_a^b\frac 1x dx=\int_c^d\frac 1x dx$

In 1649 Alfonso Antonio de Sarasa recognized this feature in Gregoire's work and connected it to the properties of logarithms. In particular he recognized the following additive property of these measurements which had previously be a key feature in the study of common logarithms (base 10): the area determined by a product of two numbers , bd, is equal to the sum of the areas determined by b and d separately.

If $\frac ab = \frac cd$ and $a=1$ and $c=1$ then $\frac 1b \not= \frac 1d$, but $\frac 1b + \frac 1d = \frac {1}{bd}$

This is the same as saying what RGB did: $$\int_{1}^{bd}\frac{\text{d}t}{t}=\int_{1}^{b}\frac{\text{d}t}{t}+\int_{1}^{d}\frac{\text{d}t}{t}$$

$$f(bd)=f(b)+f(d)$$

$$ A^b A^d=A^{b+d}$$

Thus a logarithmic relationship was implied and A was found to be e.

$$ \int_1^b \frac 1x dx = \ln(b)$$ $$ \int_1^d \frac 1x dx = \ln(d)$$ $$ \int_1^{bd} \frac 1x dx = \ln(b)+\ln(d)$$

Source

The Hyperbolic Angle "$u$" is defined in relation to the function $\frac 1x$:

enter image description here

The magnitude of the hyperbolic angle "$u$" is the area of the corresponding hyperbolic sector (red above, yellow below) which is ln x since it is defined as the integral of the projection onto the axis from a=1 or y=x to b: $\int_1^b \frac 1x dx=\ln(b)$ enter image description here

While power series were not involved in the discovery of the natural logarithm, they can be used to show the relationship between $\frac 1x$ and e:

If a function is its own derivative then: $ y=f(x), \frac{dy}{dx}=y$

Such a function is a series: $ y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$

$$\frac{dy}{dx}=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+\frac{4x^3}{4!}\dots=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \dots$$

If we differentiate $y^p$: $ \frac{d(y^p)}{dx}=py^{p-1}\frac{dy}{dx}$

Since $ \frac{dy}{dx}=y$ it follows: $ \frac{d(y^p)}{dx}=py^{p-1}y=py^p$

Given: $y_1=1+z+\frac{(z)^2}{2!}+ \frac{(z)^3}{3!}+\dots$

$$\frac{dy_1}{dz}= 1+z+\frac{(z)^2}{2!}+ \frac{(z)^3}{3!}+\dots=y_1$$

Let $z=ax$,$\frac{dy_1}{d(ax)}=y_1$ $\frac{dy_1}{d(x)}=ay_1$

$$y_1=y^p= 1+px+\frac{(px)^2}{2!}+ \frac{(px)^3}{3!}+\dots=y_1$$

If we let $p=\frac 1x$ then, $y^{\frac 1x}=1+1+\frac{(1)^2}{2!}+ \frac{(1)^3}{3!}+\dots=e$

And Hence replacing $e^1$ with $e^x$ yeilds:

$e^x=1+x+\frac{(x)^2}{2!}+ \frac{(x)^3}{3!}+\dots$

This all lead to the development of the Slide Rule as well as hyperbolic trig functions.

Community
  • 1
User3910
  • 2,390
17

It is a combination of ideas. First is the property that the quadrature of the hyperbola satisfies $f(xy)=f(x)+f(y)$. This seems to have been discovered by Gregoire de Saint-Vincent in his Opus geometricum quadrature circuli et sectionum coni (1647).

In it

 [He made the] quadrature of a hyperbola to its asymptotes, 
 and showed that as the area increased in arithmetic series 
 the abscissas increased in geometric series.

This is nothing else than the equation

$$\int_{1}^{xy}\frac{\text{d}t}{t}=\int_{1}^{x}\frac{\text{d}t}{t}+\int_{1}^{y}\frac{\text{d}t}{t}$$

"Arithmetic series" means the addition on the right hand side, "geometric series" means the multiplication $xy$ on the left.

Then there is the idea of logarithms to ease computations and the idea of using the most natural basis. The natural logarithms were first called hyperbolic logarithms. Logarithms used to be tabulated in the times of by-hand computations. Then, the name natural logarithm seems to come from Mercator in his Logarithmotechnia (1668).

Apparently it was Euler the one who put together $\ln$ and $e$ in $\ln=\log_e$; a constant hinted in a publication of Napier, and computed first by Jakob Bernoulli as $\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n$.

OR.
  • 5,941
  • What were you meaning in your comment by " splitting up the integral and then a substitution"? – User3910 Jul 12 '13 at 18:16
  • That wasn't me. But what they meant was to prove $\int_{1}^{xy}\frac{dt}{t}=\int_{1}^{x}\frac{dt}{t}+\int_{1}^{y}\frac{dt}{t}$. – OR. Jul 12 '13 at 18:18
  • @RBG how dos that help? – User3910 Jul 12 '13 at 18:20
  • 1
    It doesn't. Do you need help proving it? Substitute $t:=u/x$ in the last integral. – OR. Jul 12 '13 at 18:23
  • It suggests an homomorphism between $\langle\mathbb{R}\setminus{0},\cdot\rangle$ and $\langle\mathbb{R}, +\rangle$. It's common to see this with exponentiation ($e^{x+y}=e^xe^y$) so such a discovery could lead one to hypothesize that the two are related. – Alexander Gruber Jul 12 '13 at 18:25
  • I don't quite comprehend how the idea for a log naturally follows from splitting said integral. – User3910 Jul 12 '13 at 18:41
  • It is what Gruber was telling you above, but ignore the unnecessary terminology/notation. $f(xy)=f(x)+f(y)$ is the property that distinguishes logarithms. They are supposed to be inverse of exponentials, and exponentials satisfy $a^{x+y}=a^{x}a^{y}$ for every $x,y$. Since this is for every $x$, $y$, replace $x$ by $f(x)$, $y$ by $f(y)$ and apply $f$ on both sides. You get $f(a^{f(x)f(y)})=f(a^{f(x)}a^{f(y)})$. Use that $f(a^x)=a^{f(x)}=x$, and you get $f(xy)=f(x)+f(y)$. – OR. Jul 12 '13 at 19:12
  • Logarithms had appeared well before Saint-Vincent's book: Napier, 1614 and Briggs, 1617. I don't think they used the term hyperbolic logarithms. Do you have a reference for this terminology? – egreg Jul 12 '13 at 19:22
  • 1
    Saint-Vincent's is the origin of the property of the quadrature of the hyperbola, being him the first to compute them. This is what connects the integral and logarithms. – OR. Jul 12 '13 at 19:26
1

This is actually easier to prove than it appears, assuming we understand that $\frac{d}{dx}e^x = e^x$.

To prove it, we will prove that $\frac {d} {dx} \ln(x) = \frac 1 x $.

We start with

$$e^{\ln(x)} = x$$

Now we take the derivative of both sides. $$\frac{d}{dx}e^{\ln(x)}=1$$

Applying the chain rule $$\left(\frac{d}{dx}\ln(x)\right) \cdot e^{\ln(x)} = 1$$

Divide both sides by $e^{\ln(x)}$

$$\frac{d}{dx}\ln(x) = \frac1{e^{\ln(x)}}$$

Finally substituting back $e^{\ln(x)} = x$

$$\frac{d}{dx}\ln(x) = \frac1x$$

from here we know that $\int \frac 1x dx = \ln(x) + c$.

I'm not entirely sure who first figured this out, but it is a very important equation and not too difficult, so I'm assuming it was soon after calculus was first invented.

Christopher Mahoney
  • 1
Zar
  • 539