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Homework Help: Well defined intergrals

  1. May 2, 2006 #1
    Im having trouble explaining why intergrals are well defined. For instance:

    [tex]\int_{0}^{\infty} \frac{1}{(x + 16)^{\frac{5}{4}}}dx. [/tex]

    Here do i say something like:

    [tex] \mbox{The integral behaves at zero, and at } \infty, (x + 16)^{\frac{5}{4}} > x^{\frac{5}{4}} \mbox{ therefore the integral diverges.} [/tex]
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 2, 2006 #2

    arildno

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    Your sentence is, essentially meaningless:
    "The integral behaves at zero" :confused:

    Besides, your conclusion is wrong.
     
  4. May 2, 2006 #3

    AKG

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    It depends on what you mean by "well-defined". If you mean "is a real number" then look at what happens when you apply the definition of the improper integral to what you have, and show that you really do get a real number.
     
  5. May 2, 2006 #4

    arildno

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    Actually, the value of your integral equals a most special prime.
     
  6. May 2, 2006 #5

    VietDao29

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    No, it does not make sense. This is an Improper integral. As you know:
    [tex]\mathop {\int} \limits ^ {b}_{a} f(x) dx = F(b) - F(a) [/tex], so what if b is not finite, i.e b tends to infinity? We have:
    [tex]\mathop {\int} \limits ^ {\infty}_{a} f(x) dx = \lim_{b \rightarrow \infty} \mathop {\int} \limits ^ {b}_{a} f(x) dx = \lim_{b \rightarrow \infty}(F(b)) - F(a) [/tex]
    -----------
    Say you want to evaluate:
    [tex]\mathop {\int} \limits ^ {\infty}_{1} \frac{dx}{x ^ 2}[/tex]
    We have:
    [tex]\mathop {\int} \limits ^ {\infty}_{1} \frac{dx}{x ^ 2} = \left( \lim_{x \rightarrow \infty} - \frac{1}{x} \right) + \frac{1}{1} = 1[/tex].
    Can you get this? :)
     
  7. May 3, 2006 #6
    Yeah, thanks. Its just the way my silly university asks questions and explains things that confused me. Thanks for the help :smile:
     
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