1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    VietDao29

    User Avatar
    Homework Helper

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Well defined intergrals
  1. Well defined (Replies: 12)

Loading...