Puzzle about integration by parts (H2 maths)
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I came across this puzzle while considering the integral ∫1/(x lnx) dx.
This integral has the form f '(x)/f(x), where f(x) = ln x and f '(x) = 1/x. So the answer is just ln(ln x) + c. It's quite straightforward if we do it this way.
But we can also try integration by parts. But unfortunately we end up with a nonsensical answer.
We are integrating ∫ 1/x·1/ln x dx
Let u = 1/ln x and dv/dx = 1/x. Then du/dx = −1/(x(ln x)2) and v = lnx.
The formula is ∫ u dv/dx dx = uv − ∫ v du/dx dx
⇒ ∫ 1/x·1/ln x dx = 1 + ∫ 1/x·1/ln x dx
⇒ 0 = 1.
What does this mean and why doesn't integration by parts work in this case? There is some discussion of this on the web but I thought it is more rewarding if we figure out ourselves. Any ideas?
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Hi,
From Wolfram, IBP may fail "because it leads back to the original integral" and one example has been quoted by Apostol 1967, p. 219. Please see:
http://mathworld.wolfram.com/IntegrationbyParts.html
In another source, the reason is not so straightforward, please see page 259 of the book:
http://books.google.com.sg/books?id=-yQXwhE6iWMC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false
Here's another where an example is given that illustrates the non-existence of integrals even though the functions are differentiable:
http://faculty.uml.edu/jpropp/parts.pdf
Thanks!
P.S. We can also find examples where integration by substitution does not work because it leads back to the original integral. In fact, one student raised this with me today during a lesson and I'd thought of this query on the forum :)
Cheers,
Wen Shih -
Thanks for those references. Still looking at them. Some are easy to understand but one or two are a bit too cheem for me. But I look a bit more later ... no time these few days!
The issue of "leading back to the original integral" however is not a problem in this case because the equation can still proceed. Leading back to the original integral is a problem only if you get back to square one, but, in this case, that doesn't happen.
There's another famous example of this with ∫ ex sinx dx, isn't there? If you integrate this by parts, you will eventually find yourself back with the original integral. But the equation can still proceed and you can still find the correct eventual answer.
So I think the issue must be something else. I guess this is beyond A-levels, but just for interest and understanding, I am curious ... I had never seen an example before where integration by parts leads to nonsense!
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Dear Mad Hat,
Let's revisit your example where we wish to find ∫ 1/x·1/ln x dx
Let u = 1/ln x and dv/dx = 1/x. Then du/dx = −1/(x(ln x)2) and v = lnx.
The formula is ∫ u dv/dx dx = uv − ∫ v du/dx dx
⇒ ∫ 1/x·1/ln x dx = 1 + ∫ 1/x·1/ln x dx
⇒ 0 = 1.
We have to be careful that when we integrate 1/x, it should give ln |x|, so that
v = ln |x|,
uv = (ln |x|) / (ln x)
rather than 1.
Put simply, IBP may not work because
1. it leads back to the same integral (here we exclude the e^x . sin x type of cases),
2. the function(s) may not be integrable,
3. the function(s) may not have nice graphical behaviours.
Thanks!
Cheers,
Wen Shih -
Hi Madhat, are you using bbcode to write the above formula or just copy and paste it?
And you might like to visit this interesting local maths site which cover topic from Kindergarten 1 to Secondary 4 :
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Originally posted by M the name:
Hi Madhat, are you using bbcode to write the above formula or just copy and paste it?
And you might like to visit this interesting local maths site which cover topic from Kindergarten 1 to Secondary 4 : http://sg.ixl.com
No, I just use normal html code. It's quite hard to format. I wish I could use latex or something, then very easy. Ya, thanks for showing that site. It looks really well-designed and user-friendly. But it looks like an ang-moh site trying to capture the sg market. (Like gumtree.) I would love to come up with something like that for H2 maths, but I don't seem to have the time!
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Originally posted by wee_ws:
Dear Mad Hat,
Let's revisit your example where we wish to find ∫ 1/x·1/ln x dx
Let u = 1/ln x and dv/dx = 1/x. Then du/dx = −1/(x(ln x)2) and v = lnx.
The formula is ∫ u dv/dx dx = uv − ∫ v du/dx dx
⇒ ∫ 1/x·1/ln x dx = 1 + ∫ 1/x·1/ln x dx
⇒ 0 = 1.
We have to be careful that when we integrate 1/x, it should give ln |x|, so that
v = ln |x|,
uv = (ln |x|) / (ln x)
rather than 1.
Put simply, IBP may not work because
1. it leads back to the same integral (here we exclude the e^x . sin x type of cases),
2. the function(s) may not be integrable,
3. the function(s) may not have nice graphical behaviours.
Thanks!
Cheers,
Wen ShihYa WS, thanks for that insight. I was worried a bit about this modulus issue but didn't think very hard about it. Looks now like it might be a problem! Also this issue about whether the function is differentiable/integrable everywhere is another problem! It's quite deep for me man ... my math education is not that deep, as you know. But I try!
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If you are interested in more maths puzzle, you can visit science 2.0 : http://www.science20.com/mathematics
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Dear Mad Hat,
Your deep sense of curiosity is good enough :) Have a great day ahead!
Cheers,
Wen Shih -
Mad Hat,
Maybe you can try NUS Pure Math Analysis and Algebra modules on your own or take some uni modules from maybe Coursera. Very fun, but cannot really make money from it.
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Hey guys thanks for those suggestions. I'm not that serious in maths actually. Frankly, the subject scares me because it seems infinitely deep
I just get whatever kick out of it, as and when! By the way, sometimes when I try to login to this forum, I get this scary-looking warning. Is it serious? Do you get it once in a while also?
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Maths Puzzle : Connect The Towns : http://www.boreme.com/posting.php?id=30207
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Hi M,
Thanks for sharing this interesting content :)
Here's a paper to share:
http://www.parabola.unsw.edu.au/vol20_no2/vol20_no2_1.pdf
Enjoy!
Cheers,
Wen Shih