# Integrating derivatives over functions problem

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.

• By the way, the rule is $\int \frac{f'(x)}{f(x)}\, dx = \ln | f(x) | + C$. Oct 17 '18 at 15:59
• @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. Oct 17 '18 at 23:35
• @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. Oct 18 '18 at 1:26
• @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 $\ln(z)$. Oct 18 '18 at 20:22
• 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.
– 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

• Extra credit if you mix in "mantissa" and "characteristic". ;-) Oct 17 '18 at 16:22
• 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 :) Oct 19 '18 at 8:47
• @PaulDean Never be ashamed of asking "stupid" questions. Not asking the question does not teach you anything, while asking does, potentially. 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. $$\sec^2 x = \tan^2 x + 1$$.

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