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I have noticed that some integrals and derivatives have absolute value. For example: $$\int\frac 1 x \, dx=\ln|x|+c$$ for $x$ is not equal to zero. So, the point is why is there the absolute value of $x$?

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    $\begingroup$ I'd prefer $\ln|x|+c_1+c_2\operatorname{sgn}x$ $\endgroup$ Commented Dec 11, 2016 at 18:53
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    $\begingroup$ @HagenvonEitzen Can you elaborate on that please? $\endgroup$
    – Ovi
    Commented Dec 12, 2016 at 5:16
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    $\begingroup$ @Ovi The function is not defined at $0$, so the antiderivative is defined on two distinct intervals -- the interval for $x > 0$, and the interval for $x < 0$. The constant ($+ c$) can be different on the two intervals, so there should really be two constants. This is the effect of Hagen's suggestion of adding $+ c_2 \operatorname{sgn} x$. It may be conceptually cleaner to instead say that the indefinite integral is $\ln x + c_1$ for $x > 0$, and $\ln (-x) + c_2$ for $x < 0$. $\endgroup$ Commented Dec 12, 2016 at 10:08
  • $\begingroup$ @6005 Oh that's very clever, this way we have a system of two equations with 2 unknowns $$c_1 + c_2 = C_{x>0}$$ $$c_1 - c_2 = C_{x<0}$$ $\endgroup$
    – Ovi
    Commented Dec 12, 2016 at 16:51

5 Answers 5

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Because if $x < 0$, $\log x$ is not defined, so certainly it's derivative it's not equal to $\frac 1x$; on the other hand $\log -x$ works just fine (check it!)

So one should write $\int \frac 1x$ equal to $\log x $ if $x > 0$ and equal to $\log -x$ if $x<0$. We summarize this by putting the absolute value of logarithm, even though in $0$ the function is not defined and not differentiable

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  • $\begingroup$ But we don't use the absolute value for Square Roots? if $x<0, {\sqrt{x} is also not defined??? or am i wrong here??? $\endgroup$ Commented Dec 11, 2016 at 18:44
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    $\begingroup$ @SarmadRafique: If you take the derivative of $\sqrt{|x|}$ you get something which still has absolute values in it, so it's not something that you're likely to run into “by accident” when computing antiderivatives. On the other hand, it's a very common situation to have to take the antiderivative of $1/x$ (or similar functions like $1/(x-a)$), and then you need to know what the answer is, not just for $x>0$ but also for $x<0$. $\endgroup$ Commented Dec 11, 2016 at 19:27
  • $\begingroup$ @SarmadRafique both have solution in complex numbers. But for real numbers an integration for 1/x is quite common. And the derivative (or integral) of sqrt(x) are straight forward (just modifications of the formula similar to x^2) while ln(x) is very special. $\endgroup$
    – Kami Kaze
    Commented Dec 12, 2016 at 14:11
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The antiderivative of $f(x)$ is a function $F(x)$ such that $F'(x)=f(x)$ for all $x$ in the domain of $f$.

Thus $\ln|x|$ is an antiderivative of $\frac{1}{x}$ because

  • if $x>0$, $\frac{d}{dx}(\ln|x|)=\frac{d}{dx}(\ln(x))=\frac{1}{x}$
  • if $x<0$, $\frac{d}{dx}(\ln|x|)=\frac{d}{dx}(\ln(-x))=\frac{1}{-x} \cdot \frac{d}{dx}(-x)=\frac{1}{-x}\cdot (-1)=\frac{1}{x}.$

It would be wrong to say that $\ln(x)$ is an antiderivative for $\frac{1}{x}$, because it is not defined when $x<0$. Hence $\frac{d}{dx}\ln(x)$ is not defined when $x<0$, so it cannot be equal to $\frac{1}{x}$ (which is defined for $x<0$).

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  • $\begingroup$ and why don't we use the absolute in the square root? $\endgroup$ Commented Dec 11, 2016 at 18:55
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    $\begingroup$ @SarmadRafique the square root of what exactly? $\endgroup$ Commented Dec 11, 2016 at 18:56
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    $\begingroup$ Are you asking why $\int\sqrt{x} \ dx = \frac{2}{3}x^{3/2}+C$ and not $\frac{2}{3}|x|^{3/2}+C$? The reason is that $\sqrt{x}$ and $x^{3/2}$ are both only defined for $x \geq 0$. $\endgroup$
    – kccu
    Commented Dec 12, 2016 at 2:38
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Simply because taking the derivative of those functions with absolute values will yield the original integrand.

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Suppose we know that $\dfrac d {dx}\ln x = \dfrac 1 x,$ and that that of course presupposes that $x$ is positive.

Now suppose we want an antiderivative of $1/x$ on the interval $(-\infty,0)$, i.e. all negative values of $x.$

$$ f'(x) = \frac 1 x, \quad x<0. $$ For $a<b<0$ we have $$ \int_a^b \frac 1 x \, dx = f(b) - f(a). $$

Let $u = -x$, so $du = -dx$. As $x$ goes from $a$ (which is negative) to $b$ (which is negative), then $u$ goes from $-a$ (which is positive) to $-b$ (which is positive), and we have $$ \int_a^b \frac 1 x \, dx = \int_{-a}^{-b} \frac 1 {(-u)}\, (-du) = \int_{-a}^{-b} \frac 1 u \, du = \ln(-b) - \ln(-a) = \ln|b| - \ln|a|. $$ Differentiating with respect to $b$ yields $$ \frac d {db} (\ln |b| - \ln |a|) = \frac d {db} \int_a^b \frac 1 x \, dx = \frac 1 b. $$ And this is with $b<0$.

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Another way to see this is by looking at the graph of $f(x) = \frac 1 x$ which has two, not connected parts, as 0 is not part of the domain.

The same way but more obviously, the function (with the same domain) and definition:
$ F(x) = lnx + c $ (when x>0), $ ln(-x) + c $ (when x<0)
has two, non connected parts. But we can rewrite it more compactly as $ F(x) = ln |x| + c $.

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    $\begingroup$ Actually, you can choose different constants for each connected component. This shows show a first order equation $xy'=1$ may have a two dimensional space of solutions because of a singularity. $\endgroup$
    – Pedro
    Commented Dec 12, 2016 at 4:54

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