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:

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

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

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

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    $\begingroup$ 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. $\endgroup$
    – abnry
    Jul 12, 2013 at 17:51
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    $\begingroup$ $e$ was invented such that that happens. The fact that the area of a hyperbola is a logarithm was long known. $\endgroup$
    – OR.
    Jul 12, 2013 at 17:59
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    $\begingroup$ @JoeHobbit I'm curious to know how you imagine the power series for $\ln x$ would have been known before its first derivative. $\endgroup$
    – Erick Wong
    Jul 12, 2013 at 18:01
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    $\begingroup$ 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. $\endgroup$ Jul 12, 2013 at 18:12
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    $\begingroup$ 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 $\log$. $\endgroup$ Jul 12, 2013 at 18:27

3 Answers 3


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: $$\int_{1}^{bd}\frac{\text{d}t}{t}=\int_{1}^{b}\frac{\text{d}t}{t}+\int_{1}^{d}\frac{\text{d}t}{t}$$


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

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

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


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.

  • $\begingroup$ This is a nice answer. $\endgroup$ Jul 12, 2013 at 20:56
  • $\begingroup$ Well Done, Joe Hobbit! Very well done indeed! $\endgroup$ Jul 13, 2013 at 5:03

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


"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 $\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n$.

  • $\begingroup$ What were you meaning in your comment by " splitting up the integral and then a substitution"? $\endgroup$
    – User3910
    Jul 12, 2013 at 18:16
  • $\begingroup$ That wasn't me. But what they meant was to prove $\int_{1}^{xy}\frac{dt}{t}=\int_{1}^{x}\frac{dt}{t}+\int_{1}^{y}\frac{dt}{t}$. $\endgroup$
    – OR.
    Jul 12, 2013 at 18:18
  • $\begingroup$ @RBG how dos that help? $\endgroup$
    – User3910
    Jul 12, 2013 at 18:20
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    $\begingroup$ It doesn't. Do you need help proving it? Substitute $t:=u/x$ in the last integral. $\endgroup$
    – OR.
    Jul 12, 2013 at 18:23
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    $\begingroup$ 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. $\endgroup$
    – OR.
    Jul 12, 2013 at 19:26

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.

  • $\begingroup$ How does $(\frac{dy}{dx}ln(x))*e^{ln(x)} = 1$ inply that $\frac{dy}{dx}ln(x) = \frac1x$ ? $\endgroup$
    – User3910
    Jul 12, 2013 at 18:05
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    $\begingroup$ @JoeHobbit Since $e^{\ln(x)} = x$? $\endgroup$
    – Arthur
    Jul 12, 2013 at 18:07
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    $\begingroup$ This doesn't answer the question at all. $\endgroup$ Jul 12, 2013 at 18:14
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    $\begingroup$ Why the upvotes? $\endgroup$
    – Did
    Jul 12, 2013 at 19:31

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