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Integral question

  1. Nov 6, 2006 #1
    I got lost in an example in my book. Hoping someone could explain it to me.

    For what values of p is the intergral

    from 1 to infinity [tex]\int \frac {1}{x^p}dx [/tex]


    from 1 to infinity [tex]\int \frac {1}{x^p}dx [/tex]

    = lim (t -> infinity) [tex]\frac {x^-^p^+^1}{-p+1} [/tex] (from x = 1 to x = t)

    = lim (t -> infinity) [tex]\frac {1}{p-1} [\frac {1}{t^p^-^1} - 1] [/tex]

    the only thing that confuses me about this is how the t^p-1 ended up in the denominator because after the 2nd sept I get the following:

    = lim (t -> infinity) [tex]\frac {t^p^-^1}{p-1} - \frac {1}{p-1}[/tex]

    Last edited: Nov 6, 2006
  2. jcsd
  3. Nov 6, 2006 #2


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    Homework Helper

    Isn't [tex]\int \frac{1}{x^{-p}}dx=\frac{x^{1+p}}{1+p}[/tex]?
  4. Nov 6, 2006 #3
    Oh, I'm sorry the intital inegral is

    [tex]\int \frac {1}{x^p}dx [/tex]

    thanks for catching that mistake :)
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