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

blyatman

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Wow

There really is a bizarre obsession here with the analytic continuation of log function to the complex plane. It is completely misplaced in this context and pedagogically ludicrous as a teaching approach to the integral of 1/x. You CAN mow your lawn with an egg beater, but that doesn't make it a good idea. The problem seems to be in the language so let me be precise.

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). I have proven above using ONLY the chain rule and the definition of the absolute value that the derivative of ln|x| IS 1/x.

Pushing this backward we have two simple easily verified theorems:

The integral of h(x) is ln(x)+C and

yes..............

the integral of 1/x is ln|x|+C

This is not a fudge, or a trick or a hack or an oversight or a waving of the hands or hiding any deeper story at all. It is a simple mathematical fact. Mowing your lawn with a lawn mower does not hide the deeper story that you could also have used an egg beater. Filling your students heads with complex analysis dross on this topic really is a waste of their time and yours.

I don't mean to be nasty here but this sort of narrow-minded obsession with a particular mathematical approach is what we often see when engineers try to do mathematics. You know what they say, "when the only tool you have is a hammer, every problem starts to look like a nail".
I don't think eshingalong is advocating that we teach the integral of 1/x using complex analysis, nor am I. They're just illustrating that the complex number system is a generalisation of the real number system, such that all results from real analysis should be derivable from the complex equivalent by setting z = x. In the real domain, the integral of 1/x = log|x| is fine by me, as is the result of log(x) provided a branch cut is defined. Obviously, we would not bring up the latter result in the context of the HSC, which is why I said (as well as eshingalong) to just use log|x|. I'm just saying that both have exactly the same properties in the real domain and lead to identical results.

Again, I'm NOT advocating for teaching complex numbers in this scenario, we've just been debating the mathematical technicalities for the fun of it (which of course, is clearly NOT for teaching HSC).

I don't believe what we're discussing is a narrow-minded obsession with a particular mathematical approach, but the exact opposite: to me it feels like an open discussion which made me explore a number of ideas, which is what mathematicians should do. In contrast, it's actually the engineers that hate mathematical rigour and who typically accept things at face value without further consideration or justification of the process! That approach seems significantly more close-minded since it's not really forcing any critical thinking or exploring the issue. E.g. engineers use delta functions in Laplace Transforms - mathematicians get heart attacks when they see that, but we just use it because it works even without defining it as rigourously as the mathematicians.

But yeh, the discussion was more about the technicalities more than anything else. In an university mathematics exam, I don't think it'd be wrong to say that the integral of 1/x is ln(x) provided you specify the branch cut (remembering back to my complex analysis class). Is it convoluted and unnecessary? Absolutely! But it's not wrong, as it's perfectly valid as it solves all problems.

Again, I would not teach involve complex numbers when teaching this to my students! I would teach the log|x| result. This discussion was more out of personal interest.
 
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blyatman

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But the integral of 1/x is not ln(x) because as I have shown above the derivative of ln(x) is NOT 1/x. The integral of 1/x is ln|x| for the excruciatingly simple reason that the derivative of ln|x| is 1/x.

Sorry I give up. No more on this one.
Fair enough lol, just agree to disagree then. Good discussion nonetheless.
 

Jonomyster

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

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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

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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

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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

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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

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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.
 

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