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Integral Sin(ax)^2 Between Infinity and 0

  1. Jun 12, 2012 #1
    Need result for integral
    Sin(ax)*Sin(ax) Between Infinity and 0

    Cant find this anywhere but there is a standard result with a in it.
     
  2. jcsd
  3. Jun 12, 2012 #2

    micromass

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    That integral won't exist in general.
     
  4. Jun 12, 2012 #3
    Should do need it for a normalisation problem

    have to square the wavefunction then integrate

    wavefuction form: sin(ax)

    so need to integrate sin(ax)^2 over all space

    problem is part of infinite square well limits should between infinity and 0.
    V=0 for x< a
    v= infinity for x>a

    Maybe I'm thinking of the wrong limits.

    should probably be between a and -a ?
     
    Last edited: Jun 12, 2012
  5. Jun 12, 2012 #4
    Think about it. It's periodic and always nonnegative. Assuming a≠0, every period will have finite area. So the sum of the areas of the infinite periods ...
     
  6. Jun 12, 2012 #5
    the wave function is zero where the potential is infinite
     
  7. Jun 12, 2012 #6
    Yeah my limits are wrong because the well is bound between a and -a

    so need integral between -a and a for sin(ax)^2
     
  8. Jun 12, 2012 #7
    You mean [itex]\displaystyle \int_{-a}^{a}\sin^2(ax)dx[/itex] I presume.
    Did you try the half-angle identity and u-substitution?
     
  9. Jun 12, 2012 #8

    rbj

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    Nick, take a look at http://en.wikipedia.org/wiki/Wikipedia:Math and learn (it's very easy) a little math-symbol paste-up, like LaTeX. perhaps there is a better description somewhere.

    just remember that [itex] \sin^2(x) [/itex] has an average value of 1/2 and if you integrate any non-zero constant over anything to [itex]\infty[/itex], you will get an infinite number. and i am wondering if the limits should be from -1/a to +1/a ? or should it be a 1/a in the sin() argument?
     
  10. Jun 13, 2012 #9
    yeah thanks its done now
     
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