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Improper integral =\

  1. Oct 15, 2004 #1
    I stopped at the last step while calculating this improper integral:

    integral of x^3\ ( x^4 - 3)^1\2 with limits from 1 to infinity...

    that's x cubed over the square root of x raised to 4 minus 3....



    after replacing infinity with b and taking the limits it seems that I have to take limits of two expressions one that goes to infinity which is ( (b^4 - 3)^1\2) / 8 but the other is undefined since I have a negative 2 inside a square root. The second expression is the square root of negative two all over 8.


    Thanks for any help XD
     
  2. jcsd
  3. Oct 15, 2004 #2

    NateTG

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    Just to make sure, that's:
    [tex]\int_1^{\infty} \frac{x^3}{\sqrt{x^4-3}} dx[/tex]
    Which is divergent.

    Perhaps there was an algebra error leading up to this?
     
  4. Oct 15, 2004 #3

    arildno

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    Dearly Missed

    Not only is it divergent, but it makes no sense outside the field of complex analysis..
     
  5. Oct 15, 2004 #4
    Yeah that's it.

    I end up with two expressions and when taking the limit of the first it's infinity but the second has a negative inside the sqaure root. I guessed it's divergent since whatever mistake I did doesnt change (I think..) that the second expression is a constant anyway... I guess. :p

    Thanks dudes. ^_^
     
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