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Integration by Parts Contradiction 
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#1
Oct2205, 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. 


#2
Oct2205, 06:44 AM

HW Helper
P: 1,421

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, 


#3
Oct2205, 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? 


#4
Oct2205, 07:05 AM

HW Helper
P: 1,421

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, 


#5
Oct2205, 07:15 AM

P: 3

Gotcha, excellent explanation.



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