$\int dx/x =\cdots$ (pedantic nitpicking?)

Question

It seems that "everybody knows" that $\displaystyle\int\frac{dx}{x}=\log|x|+C$ (or if one really must, then $\ln$ instead of they synonymous $\log$). Does any textbook or reference work say that $$ \int\frac{dx}{x} = \log |x| + \left.\begin{cases} A & \text{if }x>0, \\ B & \text{if }x<0. \end{cases}\right\} \text{where $A$ and $B$ are constant ?} $$

And now a subtler question: When and how often would this matter? Maybe never? Or maybe only on examinations (Maybe there's some subtle difference between that and "never".)?

No. What about $\displaystyle\int\sec^2 x,dx$? There's a tacit assumption that we're working on a connected domain. Otherwise the constants of integration can always be different on different components. To be more explicit, $f'=0\implies f=\text{constant}$ only holds on intervals, and I usually do remind my students of that. — Ted Shifrin, Jun 06 '14 at 17:19
This has been discussed here on Math Stack Exchange before and had some really good responses but I can't seem to find it. — Cameron Williams, Jun 06 '14 at 17:20
@Cameron Williams: It's been discussed in How to deal with negative x when integrating reciprocal to log and in Confused about taking absolute value after integrating reciprocal continued — Dave L. Renfro, Jun 10 '14 at 21:03
continuation and in Is putting absolute values around the argument of a log obtained through integration incorrect?. — Dave L. Renfro, Jun 10 '14 at 21:10
@DaveL.Renfro Excellent work! I searched but had no idea what in the world to use as search terms so I came up with nothing. — Cameron Williams, Jun 10 '14 at 22:32
@Cameron Williams: FYI, I searched with google using words such as "stackexchange", "logarithm", "integral", "absolute value", "constant". Over the past 15+ years this has been extensively discussed in the College Board discussion group "AP-calculus", whose 2012 and earlier posts are archived at Math Forum, but I wasn't able to find any good posts in the short time that I spent looking. (The Math Forum search engine only allows a "relevance" search with multiple words (no phrases), so all a search with "integral" and "absolue value" does is prioritize posts with "value" in it several times.) — Dave L. Renfro, Jun 11 '14 at 13:50

N. S. · Answer 1 · 2014-06-06T18:06:10.063

7

It is typically understood that $\int f(x) dx$ denotes the set of antiderivatives of $f$ on an interval where $f$ has an antiderivative.

The formula $\int \frac{dx}{x} = \ln |x|+ C$ is true over any interval $I \subset \mathbb R$ where $f$ is defined.

I am not an expert, but this could come in play when solving differential equations.

edited Jun 06 '14 at 18:06

answered Jun 06 '14 at 17:22

N. S.

132,525

Can you think of any concrete instance in which this would matter in solving differential equations? Might it have a practical bearing on $\displaystyle\frac{dP}{dt} = \frac{1}{P(1-P)}$? Clearly it would affect the way you write the general solution, but if you want a solution for some purpose other than an exercise in finding the solution, when and how much would it matter? ${}\qquad{}$ – Michael Hardy Jun 06 '14 at 17:35
@MichaelHardy Im an thinking about the following: the function $$y = \left.\begin{cases} K_1 e^{-\frac{1}{x^2}} & \text{if }x>0, \ K_2 e^{-\frac{1}{x^2}} & \text{if }x<0. \end{cases}\right.$$ is differentiable, and it should be the solution of some separable differential equation... Or some similar function. – N. S. Jun 06 '14 at 17:56
A simpler problem would be $xy'-3y = 0$, whose solutions are of the form $y = \begin{cases}c_1 x^3 & x \ge 0 \ c_2x^3 & x < 0\end{cases}$. When you integrate $x^{-3}y'-3x^{-4}y = 0$, the left side is not differentiable at $x = 0$, which causes the solution to behave differently on each side of $0$. Of course, the solution is only twice differentiable, not infinitely differentiable at $x = 0$. – JimmyK4542 Jun 10 '14 at 21:10

score 2 · Answer 2 · answered Jun 06 '14 at 17:30

2

Actually this is a good point, there can be two different constants for the positive and negative log. You can get into trouble in some situations if you dont take this into account. It really is glossed over in all books. But these situations are quite rare. All of this stuff about constants of integration is not pedantic, it really affects the mathematics.

answered Jun 06 '14 at 17:30

Rene Schipperus

39,526

That "these situations" actually exist is something that would bear a constructive proof that would contribute to answering the question. Can you cite at least one? – Michael Hardy Jun 06 '14 at 17:31
Actually, it happened to me earlier this year, now I am kicking myself that I didnt write it down. Dont delete this question sometime someone will come across this and give the answer this question deserves. But the other posters are right it only happens over a disconnected interval and you shouldent do this anyway so that is maybe why you dont see it. – Rene Schipperus Jun 06 '14 at 17:35

score 0 · Answer 3 · answered Jun 10 '14 at 20:52

I would say that even the constant of integration is not important and is more of a convention than anything else. Here is a quote from Spivak's Calculus (p. 361):

Most people write $$\int\! x^3 dx = \frac{x^4}{4} + C$$ to emphasize that the primitives of $f(x) = x^4$ are precisely the functions of the form $F(x) = x^4/4 + C$ for some number $C$. Although it is possible to obtain contradictions if this point is disregarded, in practice such difficulties do not arise, and concern for this constant is merely an annoyance.

Spivak continues to write the rest of the chapter without the constant of integration.

From a pedagogical perspective I think there is an advantage to requiring the constant of integration (and possibly even multiple constants of integration) but I don't think that it is very important as long as the user understands what they are doing.

If you interpret $\int f \,dx$ as asking for every function so that the derivative is equal to $f$ then you must use the multiple constants approach. If you interpret it as asking for some function then you can leave out any constant of integration. If you interpret it as every function on some interval where the antiderivative exists then the single constant approach is sufficient.

As long as everyone understands that every antiderivative over an interval differs from any other antiderivative by a constant then the constant of integration is not essential. Similarly for antiderivatives over disjoint unions and the multiple constants of integration approach.

Of course the constant of integration is important. Nobody likes writing the constant of integration, but it's nonetheless necessary. Shall we drop the $dx$s at the integral as well? — beep-boop, Jul 02 '14 at 22:26

$\int dx/x =\cdots$ (pedantic nitpicking?)

3 Answers3