Theoretical or intuitive understanding of the integral $ \int_\ \frac{dx}{x} = \ln|x|$?

Question

I see that it works, but I'm not sure I get why. Both the relationship between $\frac1x$ and the $\ln x$ function itself, which seems bizarre, and the absolute-value part (I get how you need it for negative x-values to work, but theoretically I don't see where it comes from.)

Have you looked at http://math.stackexchange.com/questions/498339/ ? — Chappers, Mar 18 '15 at 17:46

score 4 · Answer 1 · answered Mar 18 '15 at 17:57

The absolute-value part is just a convenient kludge to get a formula that works for both $x < 0$ and $x > 0$. When you get to complex variables you abandon it. More generally, $\ln( c x)$ is an antiderivative of $1/x$ for any nonzero constant $c$. You can take different $c$ in the two intervals $x > 0$ and $x < 0$ (since $1/x$ is undefined at $x=0$, there's no necessary connection between what happens in these two intervals), and if you want to avoid logarithms of negative numbers you want $c$ to have the same sign as $x$. There's no particular advantage in taking $c$ to be $+1$ for $x > 0$ and $-1$ for $x < 0$.

Several good points in this answer. – mickep Mar 18 '15 at 18:22 — mickep, Mar 18 '15 at 18:22

score 3 · Accepted Answer · answered Mar 18 '15 at 20:18

The way that I initially got my head around this issue was to flip the statement $$\int\frac{dx}{x}=\ln \rvert x \lvert + C$$ into $$\frac{d}{dx}\ln \rvert x \lvert = \frac{1}{x}$$ For the case where $x\gt0$ consider the following $$e^y = x$$ $$\frac{d}{dx}e^y = \frac{dx}{dx}$$ $$e^y\cdot \frac{dy}{dx} = 1$$ $$\frac{dy}{dx} = \frac{1}{e^y}$$ $$\frac{dy}{dx} = \frac{1}{x}$$ $$\frac{d}{dx} \ln x = \frac{1}{x}$$ Where $x<0$ we have $$e^y = -x$$ $$\frac{d}{dx}e^y = \frac{-dx}{dx}$$ $$e^y\cdot \frac{dy}{dx} = -1$$ $$\frac{dy}{dx} = \frac{1}{-e^y}$$ $$\frac{dy}{dx} = \frac{1}{x}$$ $$\frac{d}{dx} \ln (-x) = \frac{1}{x}$$ This allows us to proclaim that $$\frac{d}{dx} \ln \rvert x \lvert = \frac{1}{x}$$ which, of course, implies that $$\int \frac{dx}{x}=\ln\rvert x \lvert + C$$

score 0 · Answer 3 · answered Mar 18 '15 at 18:20

This answer addresses how we can extend the integral for positive numbers to the negative numbers.

I assume we have proved that for positive $a$ $$ \log a=\int_1^a\frac1t\,\mathrm dt $$

Let $x$ be a negative variable and $\xi$ be some negative constant. And lets now look at the integral: $$ \int_\xi^x\frac1t\,\mathrm dt $$ If you perform the substitution $t=-u$ you get that $dt=-du$ and therefore $$ \int_\xi^x\frac1t\,\mathrm dt=\int_{-\xi}^{-x}\frac1t\,\mathrm dt $$ Since $x$ and $\xi$ are negative, we can use the result with positive values and get. $$ \int_{-\xi}^{-x}\frac1t\,\mathrm dt=\log(-x)-\log(-\xi) $$ Since $-\xi>0$ we have $\log(-\xi)$ be some constant, and since $x$ is negative we have $\log(-x)=\log|x|$. Therefore $$ \int_\xi^x\frac1t\,\mathrm dt=\log|x|+C $$ Since for positive $x$ we have $\log x=\log|x|$, we have that for any $a\ne0$, and $b$ with same sign as $a$ $$ \int_{b}^a\frac1t\,\mathrm dt=\log|a|+C $$

It's important to note that the upper and lower limit of integration must have same sign, as you cant integrate over the singularity at $0$

$\ldots$ and we should recall that $C$ is not a constant but rather a piecewise constant; it is constant on $(-\infty,0)$ and on $(0,\infty)$ separately. ${}\qquad{}$ — Michael Hardy, Mar 18 '15 at 20:29

Theoretical or intuitive understanding of the integral $ \int_\ \frac{dx}{x} = \ln|x|$?

3 Answers3