Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by Parts Contradiction

  1. Oct 22, 2005 #1
    Ok guys, this is my first post. Please go easy...:redface:

    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. jcsd
  3. Oct 22, 2005 #2

    VietDao29

    User Avatar
    Homework Helper

    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,
     
  4. Oct 22, 2005 #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?
     
  5. Oct 22, 2005 #4

    VietDao29

    User Avatar
    Homework Helper

    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,
     
  6. Oct 22, 2005 #5
    Gotcha, excellent explanation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integration by Parts Contradiction
  1. Integration by parts (Replies: 7)

  2. Integration By Parts (Replies: 2)

Loading...