# absolute values for integrals equal to log? (1 Viewer)

#### noformalities

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

#### fan96

##### 617 pages
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 $\bg_white \ddot x = 5 - 10v$, and $\bg_white t=0,\, v=1$ (this was a question from my trials).

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

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

So just remember to include them.

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cool, thanks!

#### blyatman

##### Active Member
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.

#### integral95

##### Well-Known Member
Lol this guy

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

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

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).

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#### blyatman

##### Active Member
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).

#### InteGrand

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

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#### blyatman

##### Active Member
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.

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#### fan96

##### 617 pages
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 $\bg_white 1/x$ be, without leaving $\bg_white \mathbb{R}$?

#### blyatman

##### Active Member
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).

#### blyatman

##### Active Member

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

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.

#### blyatman

##### Active Member
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.

#### blyatman

##### Active Member
What would the "proper" way of integrating $\bg_white 1/x$ be, without leaving $\bg_white \mathbb{R}$?
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).

#### fan96

##### 617 pages
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

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

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

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

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

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#### fan96

##### 617 pages
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.

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#### blyatman

##### Active Member
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.

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#### blyatman

##### Active Member
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.

#### eshingalong

##### New Member
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!

#### blyatman

##### Active Member

First off, always put absolute values in HSC, don't get fancy (Agreed).

Next, log(x) is canonically defined for the positive reals, and is ill-defined otherwise. So log(x) is technically correct only if you tell the reader how to define it (i.e. specify the branch cut - a notion in complex analysis), otherwise you're relying on the reader to make assumptions. However, if both limits are on the negative x-axis, then you'll get the correct solution regardless of the branch cut, but it should still be specified as it tells the reader how you're calculating the complex arguments. That being said, defining log(x) as log|x| + i*pi is rather complicated, and it's often better to just write log|x| for simplicity.

TLDR:
log(x) is correct if you specify the branch cut, and it becomes a bit hazy if you don't specify the branch cut (even if it leads to the correct result). As we're dealing with HSC questions, we don't do complex analysis, so we write log|x| as log|x| requires no extra definitions.

I'm reasonably happy with this explanation as it delves into the crux of the issue. I'll also probably tell my students never to use log(x) even if the limits are positive, since log(x) as an anti-derivative is not mathematically well-defined.

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#### fan96

##### 617 pages
The derivative of ln(x) is not 1/x much as you would like this to be true. If you want to say something along these lines you will need to define a function h with domain x>0 and definition h(x)=1/x for x>0. The the derivative of ln(x) is most certainly h(x).
Could you elaborate on why the domain restriction is necessary?