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Evaluating Improper Integrals

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Test for convergence:
    b. [tex]\int_0^\infty{e^{-x^2}}[/tex]


    2. Relevant equations
    Any method you choose to approach the problem in order to test for convergence.


    3. The attempt at a solution
    First, I attempted to integrate the problem. However, I am not exactly sure how to approach such integration. I would appreciate some hints as to how to approach this. It would help me in continuing to solve this problem.
     
  2. jcsd
  3. Feb 12, 2009 #2
    When doing improper integrals, replace the infinity with a variable, A for example. Then proceed through the integral while taking the limit as A approaches infinity. If any part of the process yields an answer of infinity, the integral is divergent.
     
  4. Feb 12, 2009 #3
    Yes, I realize what you say, so:

    lim (b-> [tex]\infty[/tex]) [tex]\int_0^b{e^{-x^2}}[/tex]

    I realize that [tex]e^\infty[/tex] is infinity, therefore, making it divergent. However, that's wrong because I have to integrate it first. Hence my question, how would I approach integrating [tex]{e^{-x^2}}[/tex]
     
  5. Feb 12, 2009 #4

    Dick

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    That's not very helpful because there no elementary antiderivative for e^(-x^2). You want to do a comparison test with something you can integrate. For example, if x>1 which is larger, e^(-x) or e^(-x^2)?
     
  6. Feb 12, 2009 #5
    Hmm, a comparison test does seem interesting. So lets try this out.
    Let [tex]f(x)=e^{-x^2}[/tex]
    and Let[tex] g(x)=e^{-x}[/tex]
    Where [tex]0\leq{f(x)}\leq{g(x)}[/tex]

    Given the comparison test, if g(x) is convergent, we can assume that f(x) is also convergent.

    [tex]\stackrel{lim}{b\rightarrow\infty}\int_0^b{e^{-x}}[/tex]

    [tex]\stackrel{lim}{b\rightarrow\infty}[-e^{-x}]\stackrel{b}{0}[/tex]

    [tex]\stackrel{lim}{b\rightarrow\infty}[-e^{-b} + 1][/tex]

    [tex]\stackrel{lim}{b\rightarrow\infty}[-e^{-\infty} + 1][/tex]

    Errr. The answer I am getting is that g(x) is divergent. I can draw no conclusion from this.

    Ohh no wait!

    [tex]\stackrel{lim}{b\rightarrow\infty}[1/-e^{\infty} + 1][/tex]

    [tex]\stackrel{lim}{b\rightarrow\infty}[1/{\infty} + 1][/tex]

    [tex]\stackrel{lim}{b\rightarrow\infty}[0 + 1][/tex]

    CONVERGENT!

    :) Thanks Dick & W3390, for quick responses and help.
     
  7. Feb 12, 2009 #6

    Dick

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    Err. e^(-infinity) is zero. Isn't it? I.e. limit b->infinity e^(-b)=0. g(x) is NOT divergent. And f(x)<=g(x) only for x>=1. Why don't you have to worry about 0<=x<1? It's good to think about these things.
     
  8. Feb 12, 2009 #7
    Yeah, It is, I figured it out as you can see haha. However, the f(x) <= g(x) is only for x>=1, we don't worry about x=0 -> x<1 because.. hmm, I'm honestly not sure.
     
  9. Feb 13, 2009 #8

    Dick

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    You're pretty sharp. I'll give you two minutes to figure it out and then I'm off to bed. Hint: you know the integral of f(x) converges in [1,infinity) now. Why it converges over [0,1] is even easier.
     
  10. Feb 13, 2009 #9
    Ok lets just solve this real quickly.

    See if f(x) converges from 0->1.

    Using the same, if g(x) converges, we'd be using the same steps, except from 0 to 1 now. This would give you

    -1/e + 1. Given that they are all constants (its approaching a value), we know it converges.

    I know this is probably not the theoretical answer you are seeking, I just seem to be missing something.
     
  11. Feb 13, 2009 #10

    Dick

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    Nah, simpler than that. 0<=f(x)<=1 for x in [0,1]. For x>1 compare with g(x)=exp(-x). For x<1 just compare with g(x)=1. Sorry, I rushed you. I know you would have figured it out.
     
  12. Feb 13, 2009 #11
    Not a problem. Anyways, thanks for the help.
     
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