Antiderivative of 1/x: ln(x) or ln(|x|)?

[Alettix]

Homework Statement

Calculate the integral:
$\int_{a}^{b} \frac{1}{x} dx$

Homework Equations

-

The Attempt at a Solution

In high school we learned that:
$\int_{a}^{b} \frac{1}{x} dx = ln(|x|) + C$
because the logarithm of a negative number is undefined.

However, in my current maths course about complex numbers we leared to take the logarithms of negative numbers as well, using: $ln(-x) = ln(x) + \pi i$. With this extension, can we say:
$\int_{a}^{b} \frac{1}{x} dx = ln(x) + C$
or is this still wrong?

Thank you very much!

 

[andrewkirk]

If $a=b$ the integral exists, trivially, and is equal to zero.
If $a<0<b$ the definite integral does not exist because the explosion at 0 makes the function not integrable over the interval $[a,b]$.
If $a<b=0$ or $0=a<b$ is 0, the proper integral does not exist because $\log 0$ and $\log|0|$ are undefined. The improper integral doesn't exist either, since $\lim_{x\to 0}\log x$ and $\lim_{x\to 0}\log |x|$ do not exist.

If $0<a<b$ then the integral exists and both formulas give the same result, since on that range of integration $|x|=x$.

If $a<b<0$, again the integral exists and gives the same result from both formulas, because the constant $\pi i$ cancels out.

PS:

With this extension, can we say:
$\int_{a}^{b} \frac{1}{x} dx = ln(x) + C$
or is this still wrong?

For one-dimensional integration the integration constant $C$ should not be used in a definite integral. It is used in an indefinite integral, but always cancels out when one makes a definite integral (ie inserts lower and upper integration limits).

 

[Mark44]

@Alettix, please post problems involving integrals and other calculus topics in the Calculus & Beyond section, not the Precalc section where you posted this thread. I have moved your thread.

 

[pasmith]

Homework Statement

Calculate the integral:
$\int_{a}^{b} \frac{1}{x} dx$

Homework Equations

-

The Attempt at a Solution

In high school we learned that:
$\int_{a}^{b} \frac{1}{x} dx = ln(|x|) + C$
because the logarithm of a negative number is undefined.

However, in my current maths course about complex numbers we leared to take the logarithms of negative numbers as well, using: $ln(-x) = ln(x) + \pi i$. With this extension, can we say:
$\int_{a}^{b} \frac{1}{x} dx = ln(x) + C$
or is this still wrong?

Thank you very much!

If $a < b < 0$ then $$\int_{a}^{b} \frac1x\,dx = \left[ \ln x \right]_{a}^{b} = \left[ \ln|x| + i\pi \right]_{a}^{b} = \ln|b| - \ln|a|,$$ consistent with the first rule you were taught. Integrals where $a < 0 < b$ are not defined due to the singularity at the origin.

 

[I like Serena]

If we're talking about complex numbers, we have $\ln b - \ln a$.
If we're talking about real numbers, we need to consider the undefined point we have at 0, so that $a$ and $b$ both have to be either positive, or negative.
Otherwise we have to get to more "tricks" to deal with the discontinuity.
Either way, $\ln |x| + C$ is a poor man's anti-derivative since the integration constant can be different on either side of $0$.

 

[Ray Vickson]

Homework Statement

Calculate the integral:
$\int_{a}^{b} \frac{1}{x} dx$

Homework Equations

-

The Attempt at a Solution

In high school we learned that:
$\int_{a}^{b} \frac{1}{x} dx = ln(|x|) + C$
because the logarithm of a negative number is undefined.

However, in my current maths course about complex numbers we leared to take the logarithms of negative numbers as well, using: $ln(-x) = ln(x) + \pi i$. With this extension, can we say:
$\int_{a}^{b} \frac{1}{x} dx = ln(x) + C$
or is this still wrong?

Thank you very much!

I hope nobody in high school told you that formula, because you have $a$ and $b$ on one side but no $a$ or $b$ on the other side of the equation. That cannot be right.

Anyway, if $0 < a < b$. then $|x| = x$ for any $x$ between $a$ and $b$, so it makes no difference whether you use $\ln |x|$ or $\ln x$. If $a < b < 0$ the integral $\int_a^b dx/x$ computes the (negative) area from $x = a$ to $x = b$ below the $x$-axis and above the curve $y = 1/x$ (which lies below the axis). If you put $t = -x$ the integral becomes $\int_{-a}^{-b} (-dt)/(-t)),$ which equals $\int_{|a|}^{|b|} dt/t = \ln |b| - \ln |a| < 0$ because $|b| < |a|$. That is as it should be.