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Integration by Parts Contradiction

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neo_
#1
Oct22-05, 06:33 AM
P: 3
Ok guys, this is my first post. Please go easy...

This question is from Morris Kline's Calculus: An Intuitive and Physical Approach and unfortunately there aren't solutions for all questions (really annoying).

I'm not even sure if this counts as a contradiction but anyway:

Let us evaluate int.(1/x)dx by parts. If we let u=1/x and dv=1dx, we obtain int.(dx/x)=1 + int.(dx/x). Then 1=0. What is wrong?

I would really appreciate a simple explanation from any of you experienced brains out there! Thanks.
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VietDao29
#2
Oct22-05, 06:44 AM
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Here's what I think. Let:
[tex]g(x) := \int f(x)dx \quad \mbox{and} \quad h(x):= \int f(x)dx[/tex]. Then, you do not have g(x) - h(x) = 0, you will have g(x) - h(x) = C, where C is some constant.
So here's the same, you can say that:
[tex]\int \frac{dx}{x} - \int \frac{dx}{x} = C[/tex], where C is some constant.
So it's not a contradiction...
Viet Dao,
neo_
#3
Oct22-05, 06:56 AM
P: 3
I get you Viet Dao... I don't think I would not have thought that way at all on my own... not tonight anyway. Thanks.

Here's a thought I just had:
Could int.(dx/x) = C + int.(dx/x) , where C is some constant other than 1, be eventuated from Int. by Parts?

VietDao29
#4
Oct22-05, 07:05 AM
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Integration by Parts Contradiction

You can continue integrating by parts, something like:
[tex]\int \frac{dx}{x} = 1 + \int \frac{dx}{x} = 1 + \left( 1 + \int \frac{dx}{x} \right) = 1 + 1 + .. + 1 + \left( 1 + \int \frac{dx}{x} \right)[/tex].
So you'll have:
[tex]\Leftrightarrow \int \frac{dx}{x} - \int \frac{dx}{x} = 1 + 1 + 1 + ... + 1[/tex].
Viet Dao,
neo_
#5
Oct22-05, 07:15 AM
P: 3
Gotcha, excellent explanation.


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