1. ## Strange Integrals

What is the most general antiderivative of $\frac{1}{x}$? I'll be you are expecting that:

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

BUT

Note that $\frac{d}{dx}ln ( -|x| ) = \frac{1}{x}$.

Now, I'll be the first to admit that $ln ( -|x| )$ has no real values. But it does manage to solve the differential equation $y^{\prime} = \frac{1}{x}$.

So what, if anything, does this all mean? And what does it mean to have a function with no (real) domain, but that has a real valued derivative? There's something I'm just not picturing right here.

-Dan

2. Oh Tospq, you are so affected by Physics, that you look for a meaning behind everything in the Formalistic Playground of Maths!

3. Originally Posted by Rebesques
Oh Tospq, you are so affected by Physics, that you look for a meaning behind everything in the Formalistic Playground of Maths!
But of course! I wouldn't have it any other way.

-Dan

4. Originally Posted by topsquark
What is the most general antiderivative of $\frac{1}{x}$? I'll be you are expecting that:

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

BUT

Note that $\frac{d}{dx}ln ( -|x| ) = \frac{1}{x}$.

Now, I'll be the first to admit that $ln ( -|x| )$ has no real values. But it does manage to solve the differential equation $y^{\prime} = \frac{1}{x}$.

So what, if anything, does this all mean? And what does it mean to have a function with no (real) domain, but that has a real valued derivative? There's something I'm just not picturing right here.

-Dan
I am not sure what you are doing.

$\ln |x|$ is the most general anti-derivative. You need to prove that $f: \mathbb{R} - \{ 0 \}\mapsto \mathbb{R}$ has this an an anti-derivative which it does.

What you wrote it not a function.

5. Originally Posted by ThePerfectHacker
I am not sure what you are doing.

$\ln |x|$ is the most general anti-derivative. You need to prove that $f: \mathbb{R} - \{ 0 \}\mapsto \mathbb{R}$ has this an an anti-derivative which it does.

What you wrote it not a function.
I am aware of all this. Especially the last part. What I am interested in is why is the $ln(-|x^2|)$ solution even there at all?

-Dan

6. Originally Posted by topsquark
I am aware of all this. Especially the last part. What I am interested in is why is the $ln(-|x^2|)$ solution even there at all?

-Dan
How is it a solution?

7. Originally Posted by ThePerfectHacker
How is it a solution?
Whoops! Made an oopsie. I meant, of course, that $y = ln(-|x|)$ is a solution.

Then
$y^{\prime} = \frac{1}{-|x|} \cdot -sign(x)$

When x is negative this means that $y^{\prime} = \frac{1}{-(-x)} \cdot -(-1) = \frac{1}{x}$

When x is positive this means that $y^{\prime} = \frac{1}{-x} \cdot -(1) = \frac{1}{x}$

(Hah! I did it right after all! )

-Dan

8. Originally Posted by topsquark
Whoops! Made an oopsie. I meant, of course, that $y = ln(-|x|)$ is a solution.

Then
$y^{\prime} = \frac{1}{-|x|} \cdot -sign(x)$

When x is negative this means that $y^{\prime} = \frac{1}{-(-x)} \cdot -(-1) = \frac{1}{x}$

When x is positive this means that $y^{\prime} = \frac{1}{-x} \cdot -(1) = \frac{1}{x}$

(Hah! I did it right after all! )

-Dan
The chain rule for $f(g(x))$ works only for functions so that $g$ is differenciable and $f$ is differenciable on the range of $g$. Now the range of $g$ over here is $(-\infty,0)$ certainly $f$ is not differenciable on this interval.

Another explanation is that you just cannot use the chain rule here. The "function" $\ln (-|x|)$ is simply now a function. It is like taking the derivative of $\sqrt{-(1+x^2)}$.

9. Originally Posted by ThePerfectHacker
The chain rule for $f(g(x))$ works only for functions so that $g$ is differenciable and $f$ is differenciable on the range of $g$. Now the range of $g$ over here is $(-\infty,0)$ certainly $f$ is not differenciable on this interval.

Another explanation is that you just cannot use the chain rule here. The "function" $\ln (-|x|)$ is simply now a function. It is like taking the derivative of $\sqrt{-(1+x^2)}$.
Hmmm....

Never thought about it that way. (sigh) I suppose that is the more sensible approach. Though not nearly as intriguing...

All right, problem solved to my satisfaction.

-Dan