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Any tips on improper integrals? (also, question)

  1. Oct 20, 2004 #1
    Those things scare me.

    A lot.


    They look tame and not so im-proper when you only evaluate the integral and take the limit.... then you see that section in your texts titled " Comparison of improper integrals."

    I've looked at a few google links but all problems still seem so different to me-- I can't pin down a pattern I can use. I know about the comparison test ( 0 <= f(x) <= g(x) ) for example but applying it is kinda.. hard.

    I know I'll be able to do HW but I don't know if I'll master the subject to do a problem like this for example:

    Find the value of a (to three decimal places) that makes

    [tex]\int a e^u dx = 1[/tex]


    limits are negative infinity to positive infinity...

    btw there is no "u" in the problem... I can't use this tex thing properly so here u is = -x^2/2 that's the negative of the square root of x over 2...

    Now how do I even start such a problem? My plan of attack:

    take out a outside and break the integral into two parts (with limits as negative infinity to b for the first integral and b to infinity for the second integral) and evaluate the limits... I'm guessing it converges. I'll set that value it coverges to equal to 1 and solve for an a that will somehow transform that value to 1...

    =\
     
    Last edited: Oct 20, 2004
  2. jcsd
  3. Oct 20, 2004 #2

    Galileo

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    Well, basically this is one of those standard improper integrals which you'll just hve to look up in some handbook of integrals.
    [tex]\int \limits_{-\infty}^{\infty}e^{-\lambda x^2}dx=\sqrt\frac{\pi}{\lambda}}[/tex]

    I don't think they'll ask you to show this on your own.

    It's not that hard though, once you know how.
    -First show that the integal converges. Call the value I.
    -Find I^2 and write it as a double integral.
    -Use a change of variables (should be easy to see which change).
    -Evaluate the integral.
     
  4. Oct 20, 2004 #3
    I've got another question...

    [tex]\int 1/ ( t + 1 )^2 [/tex]

    the limits are -1 and 5..

    the trouble spot is -1 but I don't know how to deal with it by comparison... I know I'll replace it with b and take the limit to b of that integral but they want me to do it with comparison and see if it coverges or diverges..


    I'm not sure what you mean by those two steps. Do you mean I^2/2 ?
     
  5. Oct 20, 2004 #4

    Galileo

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    If you picture the graph. What does your intuition say about the integral?
    If you want to do it by comparison. What function does the integrand look like?

    About the other integral. What I meant was:
    [tex]I=\int \limits_{-\infty}^{\infty}e^{-\lambda x^2}dx[/tex]
    [tex]I^2=\int \limits_{-\infty}^{\infty}e^{-\lambda x^2}dx\int \limits_{-\infty}^{\infty}e^{-\lambda y^2}dy=\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{\infty}e^{-\lambda(x^2+y^2)}dxdy[/tex]
    You can solve this double integral by substitution and then find I.
     
  6. Oct 20, 2004 #5
    Well it explodes and goes to infinity at -1...

    It looks like

    [tex] \int 1/t^2 dt[/tex]
     
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