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Integration help please

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to integrate [itex]\int \frac{exp(-(1/4)x^2)}{(1+1/2(x^2))^2}\,dx[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I've tried all the substitutions I can think of and I'm not getting anywhere. I know that the answer cannot be calculated fully (contains an erf term), but I can't even get this out.

    Any advice would be great, thank you :)
     
  2. jcsd
  3. Mar 22, 2013 #2

    SammyS

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    Hello Tonks93. Welcome to PF !

    The substitution u = x/2 will clean this up a little.

    Just to be clear, is the integral you're trying to solve
    [itex]\displaystyle \int \frac{exp(-(x^2/4))}{(1+(x^2/2))^2}\,dx \ ?[/itex]​
     
  4. Mar 22, 2013 #3
    Thanks for your reply - I'll try what you said now! And yes that's the right integral - sorry it wasn't totally clear.
     
  5. Mar 23, 2013 #4

    haruspex

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    u=x/√2 looks marginally better, but it's still a long way short of anything very useful.
    I can't see how to make any real progress with this as an indefinite integral, but there are methods that might help if it's for a specific range like -∞ to +∞.
     
  6. Mar 23, 2013 #5
    Thanks for your reply. I think the range should be 0 to +∞. Would this help at all?
     
  7. Mar 23, 2013 #6

    SammyS

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    As you mentioned in the Original Post, the indefinite integral contains a term with the error function.

    It looks as though you will need to use integration by parts, so I suggest using the following substitution.

    Let t = x/2 . (Using t = x/√2 is OK too.)

    Your integral becomes: [itex]\displaystyle \ \int \frac{exp(-(t^2))}{(1+2t^2)^2}\,dt \ .[/itex]

    Then do integration by parts using [itex]\displaystyle \ u=\frac{exp(-(t^2))}{t}\ [/itex] and [itex]\displaystyle \ dv=\frac{t}{(1+2t^2)^2}\,dt\ .[/itex]
     
  8. Mar 24, 2013 #7
    Thanks for your reply. I got the answer out with the substitution you said, so thankyou so much! :)
     
  9. Mar 24, 2013 #8

    SammyS

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    That was a crazy pair to use for integration by parts, wasn't it ?
     
  10. Mar 25, 2013 #9
    Haha well the substitution seems fairly logical now you've pointed it out, but it's not one I'd have thought of myself! Thanks again :)
     
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