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

[noformalities]

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

 

[fan96]

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 , and (this was a question from my trials).



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

So just remember to include them.

 

[noformalities]

cool, thanks!

 

[integral95]

Lol this guy

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



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).

 

[InteGrand]

 

[fan96]

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 be, without leaving ?

 

[fan96]

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



So we could learn more about the behaviour of for increasing by examining the series



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

 

[fan96]

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.

 

[eshingalong]

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!

 

[fan96]

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?

 

[Jonomyster]

No need for complex numbers, no need to extend the log function, no need for any tricks.

 

[stupid_girl]

Technically, in the world of real functions, the anti-derivative of 1/x should be
ln(x)+C1 for x>0
ln(-x)+C2 for x<0
where C1 and C2 are (possibly different) constants.

When the question is only looking at either the positive or negative side, then it can be written as ln|x|+C, where C is a constant.

 

[fan96]

What would be the proper method of dealing with a definite integral of that involves both the positive and negative sides, taking into account the discontinuity at ?

For example, Wolfram Alpha tells me that



does not converge (and I could see why this might be the case), even though one might expect it to be equal to zero.

 

[InteGrand]

What would be the proper method of dealing with a definite integral of that involves both the positive and negative sides, taking into account the discontinuity at ?

For example, Wolfram Alpha tells me that



does not converge (and I could see why this might be the case), even though one might expect it to be equal to zero.

https://en.wikipedia.org/wiki/Cauchy_principal_value.

 

[stupid_girl]

It has nothing to do with either sides. It is a simple fact that the integral of 1/x IS
ln|x|+C.

THIS IS NOT COMPLICATED!!!!


Do we at least agree that the integral of 1/(x^2) IS -1/x??

Technically, the anti-derivative of 1/x^2 should be
-1/x+C1 for x>0
-1/x+C2 for x<0
where C1 and C2 are (possibly different) constants.

It definitely matters which side you are looking at.
Suppose f'(x)=1/x^2 and f(1)=1. You cannot determine f(-1).

 

[stupid_girl]

f'=1/x^2 so f=-1/x+C. f(1)=1 imples C=2 and hence f=-1/x+2. So f(-1)=-1+2=1.

Consider two antiderivatives below.
f(x)=-1/x+2 for x>0
f(x)=-1/x for x<0
f'(x)=1/x^2 for all non-zero real values of x
f(1)=1, f(-1)=1

f(x)=-1/x+2 for x>0
f(x)=-1/x+2 for x<0
f'(x)=1/x^2 for all non-zero real values of x
f(1)=1, f(-1)=3

Both satisfy the requirement. Therefore, f(-1) cannot be uniquely determined.