$$\int\frac{1}{x}dx=\ln| x |+C$$

Why the absolute value? Why is the following not valid:

$$\int\frac{1}{x}dx=\ln x+C$$

  • 1
    $\begingroup$ $\int_{-2}^{-1}\frac{1}{x}dx$ has a value, however $\ln(-1)$ and $\ln(-2)$ will be more complicated to evaluate... $\endgroup$ – Surb Aug 25 '14 at 8:28
  • 11
    $\begingroup$ Because ln is not defined for negative values of $x$, whereas the function in the integral is. $\endgroup$ – Amitai Yuval Aug 25 '14 at 8:28
  • $\begingroup$ Btw, why is it that the base of the log is e. Couldn't it be anything? $\endgroup$ – Nick Aug 25 '14 at 8:47
  • $\begingroup$ @Nick - I believe it has something to do with the customary proof of $\frac{d}{dx}b^{x}$ for $b > 0$ using the first principles of differentiation. It came up in a previous question I posted a couple days ago: math.stackexchange.com/a/905510/170148 $\endgroup$ – StudentsTea Aug 25 '14 at 9:08
  • 1
    $\begingroup$ @Nick. It's because the Euler number (or Napier's constant) is defined as the unique number such that the area between the hyperbola $y=1/x$, the x-axis, and the vertical lines $x=1$ and $x=e$, is $1$. That is: $\int\limits_1^e x^{-1} \operatorname{d} x = 1$. $\endgroup$ – Graham Kemp Aug 25 '14 at 9:32

$\int_{-2}^{-1}\frac{1}{x}dx$ has a value, however $\ln(-1)$ and $\ln(-2)$ will be more complicated to evaluate since $\ln(x)$ is only defined on $\mathbb{R}$ for positive numbers... Actually since $\frac{1}{x} = -\frac{1}{-x}$ for every $x$, we have $$\ln(|-1|)-\ln(|-2|)=\int_{-2}^{-1}\frac{1}{x}dx=\int_2^1\frac{1}{x}dx = \ln(1)-\ln(2) = \ln\left(\frac{1}{2}\right)<0.$$

  • $\begingroup$ Also, plugging the limits in directly works as well, even if it's not the most technically correct method: $\ln(-1) - \ln(-2) = \ln(\frac{-1}{-2}) = \ln(\frac{1}{2})$ $\endgroup$ – Mateen Ulhaq Apr 13 '15 at 9:42
  • 2
    $\begingroup$ I would mention it still works out the same if you use the complex answer. $\ln(-1) = \ln(1) - i\pi$, so if $a,b > 0$ then $\ln(-b) - \ln(-a) = (\ln(b) - i\pi) - (\ln(a) -i\pi) = \ln(b) - \ln(a)$. So this is really just a technique for avoiding complex numbers. $\endgroup$ – Joseph Garvin Aug 3 '18 at 14:13

For $x$ positive: $ \frac{d}{dx}\ln{x}=\frac{1}{x} $

For $x$ negative: $ \frac{d}{dx}\ln{(-x)}=\frac{-1}{-x}=\frac{1}{x} $

So when you're integrating $\frac{1}{x}$, if $x$ is positive you'll get $\ln{x}+C$, and if $x$ is negative you'll get $\ln{(-x)}+C$. To summarize $\ln{|x|} + C$.

And if you want to know $\int\frac{1}{x}dx$ is not exactly equal to $\ln|x|+C$. The constants could be different for positive or negative $x$.

$$ \int\frac{1}{x}dx = \begin{cases} \ln{x} + C_1 \qquad \text{for $x$ positive} \\ \ln{(-x)} + C_2 \qquad \text{for $x$ negative} \end{cases} $$

  • $\begingroup$ Thank you! I've always thought that, if $\frac{a}{b} = \frac{-a}{-b}$, it seemed a bit superfluous to express that concept as $|\ \frac{a}{b}\ |$. Why don't we simply express it as $\frac{a}{b}$? $\endgroup$ – StudentsTea Aug 25 '14 at 9:11
  • 4
    $\begingroup$ @LMiz, $\frac{a}{b}$ is the same as $\frac{-a}{-b}$ and we don't express it as $|\frac{a}{b}|$; we simply express it as $\frac{a}{b}$. In $\ln{|x|}$, the absolute value sign is there, because $\ln{x}$ is not defined for negative $x$. You're confusion is probably because, $\ln{x}$ and $\ln{(-x)}$ are different functions with similar looking but different derivatives. $\endgroup$ – kptlronyttcna Aug 25 '14 at 9:16

Your range of integration can't include zero, or the integral will be undefined by most of the standard ways of defining integrals. So we have to think of a range of integration which is strictly positive, or strictly negative.

What you wrote is perfectly valid for strictly positive x, so let's think about strictly negative x. We have

$\int_{-a}^{-b}\frac{1}{x}d x$

where $a>0$ and $b>0$, so the range of integration is strictly negative. Do a change of variables, $y=-x$. Then

$\int_{a}^{b}\frac{1}{y}d y$.

(There is a negative from the $y$ in the denominator, and $d x=-d y$, so the two negatives cancel.) We have converted the integral of $1/x$ over a strictly negative range to an integral of $1/y$ over a strictly positive range. The answer is $\ln b-\ln a$. Since the $y$ is just a variable of integration, we can replace it with $x$ if we like, and

$\int_{-a}^{-b}\frac{1}{x}d x=\int_{a}^{b}\frac{1}{x}d x$.

That's the definite integral; the analogous result for the indefinite integral is

$\int^{-x}\frac{1}{x}d x=\int^{x}\frac{1}{x}d x$ (to within a constant of integration).


I will offer a very simple intuitive approach.

If we take: $$ln(x)=\int_{1}^x \frac1u du$$

We find that $u=0$ is a point of discontinuity for the function $1/u$. So, you might notice that, for example, that for $ln(x)$, as $x$ approaches $0$, the function $ln(x)$ increases "beyond all bounds" in the negative direction, or that $ln(1)=0$: $$ln(1)=\int_{1}^1 \frac1u du=0$$

Supplement - If we take $F(x)$ as any primitive function and keeping in mind of the fundamental theorem of calculus where any primitive function differ only by a constant (in this case, the indefinite integral(s) of the natural logarithm, differing by a constant, makes intuitive sense) such that: $$F(x)=ln(x)+c=\int_{1}^x \frac1u du +c$$ $$\frac{d}{dx}F(x)=\frac{d}{dx}(ln(x)+c)=1/x$$

To be specific, the definition of the primitive function here is merely just $\frac{d}{dx}F(x)=f(x)$.

This answer is just to offer some basic intuitive sprinkle on the agreeable assertion "Your range of integration can't include zero..."@Роберт

Further Supplement - For further basic intuition, notice that $1/u$ as defined for the domain $(0,1]$ lacks a uniform modulus of continuity. However, the existence of a uniform modulus of continuity is implicit in the existence of any integral (as suggested by e.g. Bolzano–Weierstrass theorem etc., and in where I would say as according to the traditional definition of an integral as suggested by e.g. $\lim \limits_{\Delta x \to 0}\sum_{i}={f(x_i)}{\Delta x}$ defined for the interval $[a,b]$ such that $a+0\Delta x=a$ and $a+n\Delta x=b$ etc.).


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.