Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

Hello again

I've been doing a bit of practice of integration by substitution and it was all going very well until I got to a question involving the natural logarithm and I got a bit stuck because it's been rather a long time...Anyway, i'm complaining again Here's the question:

Well, this one wasn't that easy, but my working so far (which I think is on the right track, but maybe here's where my problem is after all) goes like this:

From which I get:

But the fact that I can't get the correct answer from here and that wolfram alpha tells me that:

Rather makes me suspect that I'm wrong about this I'm sure it must have something to do with the fact that the natural logarithm isn't defined for *z* ≤ 0, but i just can't seem to work out why this is the correct answer and my textbook disperses it's calculus over the course of the book just a bit so it's not that easy to find this information, at least not without starting from page one and working through to the very end

*Last edited by Au101 (2013-06-12 14:00:07)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

Hi;

Did you differentiate your answer and Wolframs?

I got

for both. Not worrying about the 1 / 2.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

I hadn't, but having done so, my answer:

Gives me:

And wolfram's gives:

Which, surely, is equivalent to:

*Last edited by Au101 (2013-06-12 14:51:59)*

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

Sorry bobbym, yes, i agree, i get the same for both.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

An antiderivative is a class of functions, there can be more than one. In definite integration it all gets absorbed into the constant of integration. This is how I understand it.

Use your answer and working. Mathematica uses different algorithms than we use by hand. There will sometimes be differences.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

Okay, thanks a lot bobbym, i'll give it a try in the morning, perhaps i just made a mistake when i put:

Back into the equation, which is why I couldn't get the answer which the book has:

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

Hi;

When I differentiate that book answer it is correct. I check integrations by differentiating, use Wolfram if you need to. You could also use it to check every step of your solution.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

Hmmm...that's what i was trying to do, well, I really need to get off to bed, but for what it's worth, our old friend Wolfram gives, as its answer to

This:

Which I can get from my answer, so I guess it's just a question of laws of logs, I don't suppose anyone has any ideas?

*Last edited by Au101 (2013-06-12 15:28:10)*

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

They are all antiderivatives of that integrand. This can be proven by differentiation. I would do it the way you did.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,116

hi Au101

you wrote:

the natural logarithm isn't defined for z ≤ 0

Strictly, it is the expression within the log that isn't defined (eg. 1-z) And log has been extended into complex numbers to allow for the log of a negative. This is just as well in view of what I do below.

Your expression and Wolfram's are the same except for a constant:

This may seem strange but (i) you are right to think it's all down to the laws of logs and (ii) logs still obey those laws even for values that are undefined in real numbers.

For example

even though you might think those negative logs shouldn't 'exist'.

And all this means that your first answer is correct.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

Thank you very much to both bobs, tehe, I only wonder if anybody knows how:

Could have been obtained. If only out of interest

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

I think those two answers are algebraically equal.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**Au101****Member**- Registered: 2010-12-01
- Posts: 270

i think you're right, thank you very much bobbym

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,601

Hi;

I filled in post #12 with the proof.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

**zetafunc.****Guest**

Au101 wrote:

As an aside, you can speed up the partial fractions bit by noting that

in other words, whenever the products in the denominator differ by one, you can do this. In general;

with x ≠ -a,-b and a ≠ b.