5
$\begingroup$

In Europe, in the 20th century, $\int\frac{1}{x}dx$ equalled $\ln{x}+C$. (I have references from Poland for 1930-1947 and the UK for the 1960s and 1970s).

Now, if one mentions $\int\frac{1}{x}dx=\ln{x}+C$ in Mathematics Stack Exchange, one is lynched. The doctrine is now that $\int\frac{1}{x}dx=\ln|x|+C$, and any suggestion to the contrary is a crime.

I don't want to discuss the pros and cons of this alteration, but I am interested to know: who made this change, and when? Was it intentional, or just an influential textbook? Or is this less of an "old millennium/new millennium" and more of an "Old World/New World" kind of thing?

There is an argument for posting this to Maths Stack Exchange rather than here, and I will if asked; but it is a historical question and so this seems a reasonable forum for it.

$\endgroup$
  • 11
    $\begingroup$ The doctrine is $\ln|x| + C$? I thought it is $\ln|x| + C_1$ for $x > 0$ and $\ln|x| + C_2$ for $x < 0$, where $C_1$ and $C_2$ are constants that are not necessarily equal. $\endgroup$ – KCd Aug 24 '16 at 4:16
  • 7
    $\begingroup$ This question is of the form "when did X happen?," where X never happened. What's being discussed here is not a "convention." Definitions could be described as conventions, but the value of $\int dx/x$ is not a definition, it's a consequence of a definition. The inflammatory language about lynching and crimes is not helpful in starting a reasoned discussion. $\endgroup$ – Ben Crowell Aug 24 '16 at 5:48
  • $\begingroup$ @Kcd: shouldn't that be $\ln x + C_1$ for $x > 0$ and $\ln (-x) + C_2$ for $x < 0$? $\endgroup$ – Jonathan Cast Aug 24 '16 at 15:06
  • 2
    $\begingroup$ This is not a "convention": the first formula is only true when $x>0$ (if we stay in the real domain. $\endgroup$ – Alexandre Eremenko Aug 24 '16 at 17:29
  • 1
    $\begingroup$ About the "convention". If $\ln$ is taken as an analytic branch the first formula is correct. If one restricts $\ln$ to positive integers only, then one needs the absolute value to get a formula that is also valid for negatives. But what one gets then is not a restriction of any single branch of $\ln$ to the union of positives and negatives. The domain, the meaning of $\ln$, and the particular antiderivative in front of $C$ do not follow from definitions, and are conventional. Textbooks switching from $\ln x$ to $\ln|x|$ did happen, it was gradual rather than an event though. $\endgroup$ – Conifold Aug 24 '16 at 21:01
2
$\begingroup$

Mathematicians say $\int \frac{1}{x}\;dx = \log x + C$. It works even for the complex case.

Calculus instructors say $\int \frac{1}{x}\;dx = \ln|x|+C$ for some reason, but it is WRONG in the complex case. (Perhaps that "ln" in there gives us a hint that they are writing for engineers and physicists rather than mathematicians.)

So my guess is that the change came around 1950 or so, when calculus textbook writers were no longer research mathematicians first, and textbook writers later.

$\endgroup$
  • 4
    $\begingroup$ The topic of this site is the history of science and mathematics. The question is not a question about the history of mathematics, and this answer is not an answer about the history of mathematics. So my guess is that the change came around 1950 A guess is not an answer. $\endgroup$ – Ben Crowell Aug 24 '16 at 5:54
-2
$\begingroup$

I think that´s because now we use a special integral to integrate over the complex plane. It´s called Contour Integral. I think then when we´d want to integrate over complex plane we can write a Contour integral and then it´s clear that the answer must be ln x and not the absolute value. But with this integral there is a problem. When you evaluate that integral by contour integration (over complex plane) the answer is then 2pii.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.