Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As I understand, $\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\ln(x)\end{eqnarray} $ is generally specified as $\begin{eqnarray} \frac{1}{x} \end{eqnarray}$. Would it be more appropriate to state it as $\begin{eqnarray} \frac{1}{x}, x>0\end{eqnarray}$ since $\ln(x)$ is undefined for $x\leq0$? If not, why not? In addition, what does this imply about the indefinite integral of or the definite integral with negative limits of integration of $\begin{eqnarray} \frac{1}{x} \end{eqnarray}$?

Thank you!

share|cite|improve this question
If $0 \lt a \le b$ then by symmetry $\int_{-b}^{-a} \frac{1}{x} \mathrm{d}x = -\int_{a}^{b} \frac{1}{x} \mathrm{d}x = \log_e(a)-\log_e(b)$ – Henry Dec 30 '11 at 22:42
up vote 9 down vote accepted

Whenever we write something like $\ln(x)$, we are implicitly asserting that $x$ is restricted to those $x$s for which the expression makes sense.

When we write $$\frac{d}{dx}\ln(x) = \frac{1}{x}$$ we are then implicitly and automatically saying that $x$ is positive for the equation to make sense. As for integrals, that's why $$\int\frac{1}{x}\,dx = \ln|x|+C,$$ with the absolute values: note that $$\frac{d}{dx}\ln(-x) = \frac{1}{-x}(-x)' = \frac{1}{x}$$ as well by the Chain Rule, so that $$\frac{d}{dx}\ln|x| = \frac{1}{x}.$$

For a definite integral to make sense, you usually need the function defined on the entire interval of integration (at least, for the usual Riemann integrals), so $\int_a^b\frac{1}{x}\,dx$ cannot have $a\lt 0 \lt b$ and make sense. Either $0\lt a\lt b$ or $a\lt b\lt 0$.

You can try to do an integral $\int_a^b \frac{1}{x}\,dx$ with $a\lt 0\lt b$ as an improper integral, but you will find that the integral does not converge; neither do $\int_0^b\frac{1}{x}\,dx$ nor $\int_a^0\frac{1}{x}\,dx$.

share|cite|improve this answer

Another answer. Some strange people use complex numbers, not just real numbers. For them, log(x) makes sense other than for $x>0$. And for them its derivative is $1/x$. More importantly: for them, $\int (1/x)\,dx = \log(|x|)+C$ is WRONG!

share|cite|improve this answer
Yes: mathematicians are strange people. – Michael Hardy Dec 30 '11 at 23:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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