# Thread: absolute values for integrals equal to log?

1. ## absolute values for integrals equal to log?

In the HSC, would we ever be marked down if we forget to include the absolute values?

2. ## Re: absolute values for integrals equal to log?

Probably not for leaving them out per se, but there are some instances where you end up unable to complete the question if you forget the absolute value.

For example: let $\ddot x = 5 - 10v$, and $t=0,\, v=1$ (this was a question from my trials).

$\frac{d}{dv} t = \frac {1}{5-10v} \implies t = -\frac{1}{10}\log | 5-10v | + C$

If you forgot the absolute value you would have trouble subbing in the initial values.

So just remember to include them.

3. ## Re: absolute values for integrals equal to log?

cool, thanks!

4. ## Re: absolute values for integrals equal to log?

Unlikely. An integral leading to a log function doesn't actually require absolute values. The HSC only teaches it as a hack to get around that issue of having a negative argument inside the log. You can still solve the problem with negative arguments, just that the HSC doesn't teach it, so they get around the issue by adding in absolute values.

5. ## Re: absolute values for integrals equal to log?

Originally Posted by blyatman
Unlikely. An integral leading to a log function doesn't actually require absolute values. The HSC only teaches it as a hack to get around that issue of having a negative argument inside the log. You can still solve the problem with negative arguments, just that the HSC doesn't teach it, so they get around the issue by adding in absolute values.
Hack??
1/x is defined for all real x except x=0, ln(x) is defined only for x>0. You need the abs value. Try the integral I of 1/x from x=-3 to x=-4 and see how it goes without the abs value. As an interesting question is I negative or positive?

6. ## Re: absolute values for integrals equal to log?

Lol this guy

He's talking about actually being able to evaluate logs of negative numbers, but only in the complex field

$log(z) = ln|z| + iArg(z)$

But in HSC 2U and 3U you would only integrate with real numbers, iArg(z) would always cancel out, thus you're left with just the absolute value (which is the modulus).

7. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
Hack??
1/x is defined for all real x except x=0, ln(x) is defined only for x>0. You need the abs value. Try the integral I of 1/x from x=-3 to x=-4 and see how it goes without the abs value. As an interesting question is I negative or positive?
As integral95 said, you extend the log to the complex domain to solve the problem. The complex argument will always cancel out, leading to a real solution. So it essentially just a hack to fix the issue of extending the limits to the negative domain without using complex numbers. But the problem can be solved without absolute values, so there is technically no need for them.

So back to the original point, I don't think HSC markers would take marks off because you didn't include it, since they aren't necessary in the first place. That being said, there's no harm in putting them in, so I guess it's better safe than sorry (not to mention most students won't be able to solve the problem for x<0 unless they know how to extend logs to the complex domain).

8. ## Re: absolute values for integrals equal to log?

Yes, we can extend the log function to C but what about the concept of the integral? Restricting contour integrals back down to the real line is fraught with technical issues. Let's see the evaluation of the simple integral I above using your method? It is certainly trivial using ln|x|.

9. ## Re: absolute values for integrals equal to log?

$\noindent Incidentally, you should end up with the right answer (also to fan96's question) if you blindly apply log laws like \ln a - \ln{b} = \ln\left(a/b\right), without worrying about absolute values or negative arguments in the log.$

10. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
Yes, we can extend the log function to C but what about the concept of the integral? Restricting contour integrals back down to the real line is fraught with technical issues. Let's see the evaluation of the simple integral I above using your method? It is certainly trivial using ln|x|.
If we used the above definition, we'd just get I = ln(-3) - ln(-4) = ln(3) + i*pi - (ln(4) + i*pi) = ln(3) - ln(4), which is the correct answer.

I'm not disagreeing that its trivial (or wrong) using ln|x|, just that schools justify it by saying "we want to incorporate x<0 so add absolute values", which is essentially a hack since their explanation is more or less along the lines of "it works and it fixes the issue". For negative x,y values, ln(x) - ln(y) = ln|x| + i*pi - (ln|y|+i*pi) = ln|x|-ln|y|, so essentially, the integral of 1/x is equal to ln(x), which simplifies down to ln|x|. I'm just saying that including the absolute values isn't critical since you can solve the integral without it, but including it is also OK since the expression reduces to ln|x| in the end.

11. ## Re: absolute values for integrals equal to log?

Originally Posted by blyatman
If we used the above definition, we'd just get I = ln(-3) - ln(-4) = ln(3) + i*pi - (ln(4) + i*pi) = ln(3) - ln(4), which is the correct answer.

I'm not disagreeing that its trivial (or wrong) using ln|x|, just that schools justify it by saying "we want to incorporate x<0 so add absolute values", which is essentially a hack since their explanation is more or less along the lines of "it works and it fixes the issue". For negative x,y values, ln(x) - ln(y) = ln|x| + i*pi - (ln|y|+i*pi) = ln|x|-ln|y|, so essentially, the integral of 1/x is equal to ln(x), which simplifies down to ln|x|. I'm just saying that including the absolute values isn't critical since you can solve the integral without it, but including it is also OK since the expression reduces to ln|x| in the end.
It's not a hack. It is just a fact!

d/dx ln|x|=1/x for all non-zero x in exactly the same way as d/dx 1/x=-1/x^2 for all non zero x.

You need the abs value or else it will only be true for positive x. There is no funny business here, just reality.

If the absolute value concerns you let Fred(x)=ln|x|. Then the integral of 1/x is Fred(x)+C

It doesn't need to be "justified"....it is just true.

12. ## Re: absolute values for integrals equal to log?

Originally Posted by blyatman
schools justify it by saying "we want to incorporate x<0 so add absolute values", which is essentially a hack since their explanation is more or less along the lines of "it works and it fixes the issue".
What would the "proper" way of integrating $1/x$ be, without leaving $\mathbb{R}$?

13. ## Re: absolute values for integrals equal to log?

Originally Posted by fan96
What would the "proper" way of integrating $1/x$ be, without leaving $\mathbb{R}$?

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

It doesn't work because it "fixes the issue" .......it works because it's true.

14. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout

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

It doesn't work because it "fixes the issue" .......it works because it's true.

A nice thing to think about while we are here:

The derivative of ln(x) approaches zero as x approaches infinity.
Therefore the ln(x) function has a tangent with arbitrarily small gradient as x goes to infinity.
Therefore ln(x) is horizontal in the limit.
Therefore ln(x) has a horizontal asymptote?

15. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
It's not a hack. It is just a fact!

d/dx ln|x|=1/x for all non-zero x in exactly the same way as d/dx 1/x=-1/x^2 for all non zero x.

You need the abs value. There is no funny business here, just reality.

It doesn't need to be "justified"....it is just true.
I always justify things to my students. By understanding where it comes from, I think it helps them appreciate and understand the material better.

I just demonstrated how to calculate it without the absolute value lol, so in this example, you don't "need it", but it just does help speed things along if you have it. The only time I saw absolute values being used for the integral of 1/x was in the HSC, I don't think I ever encountered it in any of the math courses I did at university, so I'm still extremely skeptical about it lol. ALTHOUGH, I'm still open to being convinced otherwise, but I'd need to see a problem which necessitates the absolute value (i.e. cannot be solved without it).

16. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout

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

It doesn't work because it "fixes the issue" .......it works because it's true.
Yes I'm not saying it's not true! I'm just saying that the way they teach it at school is by adding absolute values and showing it works, therefore it's true. I don't particularly agree with that type of teaching lol.

17. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
A nice thing to think about while we are here:

The derivative of ln(x) approaches zero as x approaches infinity.
Therefore the ln(x) function has a tangent with arbitrarily small gradient as x goes to infinity.
Therefore ln(x) is horizontal in the limit.
Therefore ln(x) has a horizontal asymptote?
Yeh I think that's an interesting one haha. It's kind of like the harmonic series: 1 + 1/2 + 1/3 + ... , which looks like it converges since 1/n -> 0.

18. ## Re: absolute values for integrals equal to log?

Originally Posted by fan96
What would the "proper" way of integrating $1/x$ be, without leaving $\mathbb{R}$?
My understanding of the issue is that the integral of 1/x = log(x) = log|x| upon simplification after passing through the complex field (see discussion where the imaginary terms cancel out). So the result log|x| is correct, and the final result is in R (even though the proof to show this went outside of R).

19. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
A nice thing to think about here:

The derivative of ln(x) approaches zero as x approaches infinity.
Therefore the ln(x) function has a tangent with arbitrarily small gradient as x goes to infinity.
Therefore ln(x) is horizontal in the limit.
Therefore ln(x) has a horizontal asymptote?
This is something I noticed recently while working on a problem.

Oddly enough, if a series converges then each successive term approaches zero but the converse is not necessarily true.

We could take the definition

$\int_1^x \frac 1 u \, du = \log x$

So we could learn more about the behaviour of $\log x$ for increasing $x$ by examining the series

$\sum_{n=1}^\infty \frac{1}{n} = 1+\frac12 +\frac13 + ...$

But this series actually diverges even though each term is getting smaller, so hence $\log x$ does not actually have a horizontal asymptote.

20. ## Re: absolute values for integrals equal to log?

Originally Posted by blyatman
My understanding of the issue is that the integral of 1/x = log(x) = log|x| upon simplification after passing through the complex field (see discussion where the imaginary terms cancel out). So the result log|x| is correct, and the final result is in R (even though the proof to show this went outside of R).
No! not after simplification passing through the complex field. It has NOTHING to do with the complex field.

The derivative of ln|x| is 1/x therefore the integral of 1/x is ln|x|+C.

There is nothing here...it is a simple fact.

ln|x| is a real function...it is a real function whose derivative is 1/x...therefore the integral of 1/x is ln|x|+C

No complex field, no contour integration no nothing. The derivative of x^3 is 3x^2. Therefore the integral of 3x^2 is x^3. That is all that is going on. You really are doing your students a disservice by ploughing through complex numbers when it has nothing to do with complex numbers.

Let y=ln|x|. Then y=ln(x) for x>0 and ln(-x) for x<0. Thus y'=1/x for x>0. For x<0 and via the chain rule
y'=1/(-x)(-1)=1/x. That is y'=1/x ALWAYS for non zero x. So the derivative of ln|x| is 1/x and thus the integral of 1/x is ln|x|

It... .has ..nothing.... to ....do... with .....complex..... numbers!.

21. ## Re: absolute values for integrals equal to log?

Originally Posted by fan96
This is something I noticed recently while working on a problem.

Oddly enough, if a series converges then each successive term approaches zero but the converse is not necessarily true.

We could take the definition

$\int_1^x \frac 1 u \, du = \log x$

So we could learn more about the behaviour of $\log x$ for increasing $x$ by examining the series

$\sum_{n=1}^\infty \frac{1}{n} = 1+\frac12 +\frac13 + ...$

But this series actually diverges even though each term is getting smaller, so hence $\log x$ does not actually have a horizontal asymptote.
Yes but it is still true that the tangent is arbitrarily close to the horizontal for sufficiently large x. There is no need to consider sums.

22. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
Yes but it is still true that the tangent is arbitrarily close to the horizontal for sufficiently large x. There is no need to consider sums.
I found that it was a more intuitive way to try to understand the paradox.

But yeah... sometimes I tend to overthink things.

23. ## Re: absolute values for integrals equal to log?

Originally Posted by peter ringout
No! not after simplification passing through the complex field. It has NOTHING to do with the complex field.

The derivative of ln|x| is 1/x therefore the integral of 1/x is ln|x|+C.

There is nothing here...it is a simple fact.

ln|x| is a real function...it is a real function whose derivative is 1/x...therefore the integral of 1/x is ln|x|+C

No complex field, no contour integration no nothing. The derivative of x^3 is 3x^2. Therefore the integral of 3x^2 is x^3. That is all that is going on. You really are doing your students a disservice by ploughing through complex numbers when it has nothing to do with complex numbers.

Let y=ln|x|. Then y=ln(x) for x>0 and ln(-x) for x<0. Thus y'=1/x for x>0. For x<0 and via the chain rule
y'=1/(-x)(-1)=1/x. That is y'=1/x ALWAYS for non zero x. So the derivative of ln|x| is 1/x and thus the integral of 1/x is ln|x|

It... .has ..nothing.... to ....do... with .....complex..... numbers!.
Yes, but if I differentiate ln(x), I also get 1/x, so one can argue that ln(x) is also the integral of 1/x. It will also have the same properties you described: ln(x) is also a real function, it can also solve for negative x values as I demonstrated earlier. It fits all the properties you describe, the only difference is that the absolute value version can be derived using complex numbers from the more general case of ln(x).

The derivative of ln|x| is also 1/x, but so is the derivative of ln(x). They're both correct since they satisfy the same properties for reals: For positive reals, they're equal, and for negative reals, ln(x) simplifies to ln|x|. So they are essentially one and the same. Both expressions will yield the same result when you apply it to any particular problem.

Lol I feel like this argument is going in circles. I'm not disputing your answer that ln|x| is the integral of 1/x. I'm merely stating that ln(x) is also an an integral, since it shares exactly the same properties as you mentioned, and will also simplify to ln|x| for x<0. Again, I'm not disputing that ln|x| is wrong, I'm merely stating that both are correct.

To definitely state that ln|x| is the one true answer would require stating why the ln(x) solution is wrong. Both satisfy all the same properties you mentioned and can be used to solve the same problems. A sound mathematical justification needs to be provided if you reject one case and not the other. The "it's true because it works" (and vice versa) argument can also be applied to the ln(x) case!

This has actually been pretty enlightening, I never actually thought too deeply about it until this discussion.

24. ## Re: absolute values for integrals equal to log?

I'll ask a few of my friends/colleagues who are mathematics post-doc research fellows and math lecturers. I'll ask for justification on why one (or both) are correct.

It's an interesting topic, so I'm happy to change my mind as long as I get a reason lol.

Will keep you posted.

25. ## Re: absolute values for integrals equal to log?

In the HSC, just use absolute values, you don't want to get caught up on this. But, since you all seem genuinely interested in the maths...

As one of the previous replies alluded to, there is a way to extend the real logarithm to a complex logarithm, by ln(z) = ln(|z|) + iArg(z). The point is that C is the natural place to think about integrating 1/x. On the real line, both functions have a break at 0, but in the complex numbers, 1/z and ln(z) have what's called an isolated singularity at 0, and we can do calculus on the functions in a more consistent way.

In particular, in the complex numbers, it really does only make sense to say d/dz (ln(z)) = 1/z, and you evaluate integrals of 1/z by plugging numbers into ln(z) in the normal way. So for instance, you can evaluate the integral from -4 to -3 of 1/z by using the complex log - but you can also integrate from 3i to 4i.

However, this no longer works for ln|z|, because ln|z| is not in fact complex differentiable. d/dz(ln|z|) cannot be evaluated for complex numbers z. ln|x| just happens to be okay for real integrals in (\infty, 0) because it differs from the complex ln(x) by a constant, so it doesn't matter. This is the sense in which it is a 'hack' - it still works 100% for any application you'd meet (like any other hack), but it's hiding the deeper story underneath.

If you're interested, https://en.wikipedia.org/wiki/Complex_logarithm has more (but is not very readable if you haven't done complex analysis before). Hopefully this helps!

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