I had a question from a student which I'm unable to answer.

We were practicing the rule $\int \frac{f'(x)}{f(x)} \, dx=\ln(f(x))$.

A student noticed that if applied naively it gives the following two results $$\int \frac{2}{2x} \, dx = \ln(2x) +c$$ and $$\int \frac{2}{2x} \, dx = \int \frac{1}{x} \, dx = \ln(x) +c$$ which appear contradictory.

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    $\begingroup$ By the way, the rule is $\int \frac{f'(x)}{f(x)}\, dx = \ln | f(x) | + C$. $\endgroup$ – trancelocation Oct 17 '18 at 15:59
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    $\begingroup$ @trancelocation: the rule $\int\frac{f'(x)}{f(x)}\, dx=\ln(f(x))+C$ is not incorrect. The absolute value is not necessary to make the rule correct. In fact, in the context where $x$ might be a complex variable $z$, the rule without the absolute value is preferred. $\endgroup$ – user52817 Oct 17 '18 at 23:35
  • $\begingroup$ @user52817: Fact is, that $(\ln (-x))' = \frac{1}{x}$ and the additive constant $C$ in $\ln x + C$ can only produce a positive factor in front of $x$ as $C = \ln c$ for $c>0$ and, hence, $\ln x + C = \ln x + \ln c = \ln (cx)$. Please, give a clear mathematical reasoning why going through the complex justifies the omitting of the absolute value in the rule. If you think of using a rule like $\log (ab) = \log a + \log b$ with complex numbers $a$ and $b$, then I have bad news for you: This rule is not valid anymore for complex numbers as we know it from the reals. $\endgroup$ – trancelocation Oct 18 '18 at 1:26
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    $\begingroup$ @trancelocation: the point is that $G(z)=\ln(|z|)$ is not analytic as a function of a complex variable $z$. In other words, $G'(z)$ does not exist, so $G(z)$ cannot be an antiderivative for $\ln(z)$. The antiderivative for $\frac{1}{z}$ is $\ln(z)$. $\endgroup$ – user52817 Oct 18 '18 at 20:22
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    $\begingroup$ I removed a bunch of comments above, some as I felt they were inappropriate some as they became obsolete or went too far from the subject at hand. Sorry it took so long. $\endgroup$ – quid Oct 27 '18 at 16:36

As guest mentioned in a comment, the two expressions are equivalent. Suppose $x > 0$ so the logarithm is defined without any problems.

For logarithms, $\log(2x) = \log (2) + \log (x)$. Consider graphing $x \mapsto \log (x)$ and $x \mapsto \log (2x)$, and maybe their difference, to illustrate. A good window would be $x=1\ldots 10$, $y=0\ldots 5$, as recommended by user52817.

Hence, the first integral equals $\log (x) + \log (2) + C_1$, where $C_1$ is the constant of integration.

The second integral equals $\log (x) + C_2$, where $C_2$ is the constant of integration.

To see that these are the same thing, choose $C_1 = C_2 - \log(2)$.

In particular, as $C_1$ goes through all the real numbers, and as $C_2$ does so, you get precisely the same collections of functions from both integrals.

This would be a nice moment to discuss the meaning of the constant of integration, and of the indefinite integral in general. You might want to see this questions and answers for motivation: Should we avoid indefinite integrals?

This might also be useful: Explaining the symbols in definite and indefinite integrals

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  • $\begingroup$ Extra credit if you mix in "mantissa" and "characteristic". ;-) $\endgroup$ – guest Oct 17 '18 at 16:22
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    $\begingroup$ Many thanks. I will gladly use the opportunities offered. In my defense for not seeing this straight away I'm citing fatigue due to being at the end of a long term :) $\endgroup$ – Paul Dean Oct 19 '18 at 8:47
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    $\begingroup$ @PaulDean Never be ashamed of asking "stupid" questions. Not asking the question does not teach you anything, while asking does, potentially. $\endgroup$ – Tommi Oct 19 '18 at 8:53

Show your students this integral as well (maybe later): Find $\int \sec^2x\tan x\,dx$.

Method 1: $u = \sec x, \,du = \sec x \tan x \,dx$

$\int \sec^2x\tan x\,dx = \int u\,du = \frac{u^2}{2} + C = \frac{\sec^2 x}{2} + C$.

Method 2: $u = \tan x, \,du = \sec^2 x \,dx$

$\int \sec^2x\tan x\,dx = \int u\,du = \frac{u^2}{2} + C = \frac{\tan^2 x}{2} + C$.

How can this be?

Well, those two answers are the same up to a constant. $\sec^2 x = \tan^2 x + 1$.

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  • $\begingroup$ @Jasper Yes, I interpreted the question as "How can I turn this issue into a concept that I can learn and pass on?" $\endgroup$ – Chris Cunningham Oct 31 '18 at 12:19

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