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Special Integrals Hermite(2n+1,x)*Cos (bx) and e^(-x^2/2)

  1. May 31, 2015 #1
    Do some one know how to integrate the
    Hermite(2n+1,x)*Cos (bx) and e^(-x^2/2), x from 0 to infinity?
     
  2. jcsd
  3. Jun 1, 2015 #2
    Do some one know how to integrate the
    Integrate [Hermite(2n+1,x)*Cos (bx)*e^(-x^2/2), {x,0, \infinity}]?
     
  4. Jun 3, 2015 #3

    Svein

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  5. Jun 3, 2015 #4
    Thanks. However, what I want to calculate is : Integrate[Exp[-x^2/2]Cos[ b x] HermiteH[2n+1,x],{x,0,\infinity}]. I have checked some references but I cannot find the result.
     
  6. Jun 4, 2015 #5

    Svein

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    If we leave the Hermite polynoms for a moment, we can transform the rest: [itex] e^{-\frac{x^{2}}{2}}\cos(bx)[/itex] is the real part of [itex] e^{-\frac{x^{2}}{2}+ibx}[/itex]. The exponent can be further transformed: [itex] -\frac{x^{2}}{2}+ibx=-\frac{1}{2}(x^{2}-2ibx)=-\frac{1}{2}(x-ib)^{2}-\frac{1}{2}b^{2}[/itex]. Thus you end up with [itex]e^{-\frac{1}{2}(x-ib)^{2}}\cdot e^{-\frac{1}{2}b^{2}} [/itex], where the last part is constant. Now, put z = (x-ib), then dz = dx. Use that with the contents of the link I gave you and see where you end up.
     
  7. Jun 4, 2015 #6
    Thanks. However, it is not easy as what you thought. The problem is: x \in (0,\infinity). If we take z=x-i b and we have dz=dx, but the Hermite polynomial becomes
    H[2n+1, z+ib] and integral interval becomes (-i b,\infinity). Using the mathematica, it still cannot find the solutions.
     
  8. Jun 5, 2015 #7

    Svein

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    I did not say it was easy, I just gave you an idea of how to attack it.
     
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