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Integral of gaussian and more

  1. Dec 4, 2012 #1
    I have an integral of the form (from -inf to inf):

    ∫exp((a+ib)x2+icx)dx

    How do I solve that?

    I have tried setting y = √(a+ib)(x+ic/√(a+ib))

    And you then get an integral of exp(-y2) times a constant. But I don't even know if this is legal since part of the exponentials are complex. Is it possible to obtain the result with my method?
     
  2. jcsd
  3. Dec 4, 2012 #2
    Have you tried completing the square on the exponent?
     
  4. Dec 4, 2012 #3
    what do you mean? Isn't that what I do by the introduction of my y?
     
  5. Dec 4, 2012 #4
    Oh okay that didnt look clear to me as I expected you to leave it in the square form directly. I can't prove it off-hand, but the gaussian integration formula holds for complex a,b and c (where the exponent is in the form -ax^2 + bx + c) as long as the real part of a is positive
     
  6. Dec 4, 2012 #5

    Ray Vickson

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    What you want is not too hard to prove if you have a bit of complex function theory. If you take
    [tex] I = \int_{-\infty}^{\infty} e^{-(a + ib)(x + iy)^2} \, dx, \: a > 0[/tex]
    you can write it in the form of a complex contour integral:
    [tex] I = \int_{\Gamma} e^{- (a + ib) z^2} \, dz,[/tex]
    where ##\Gamma## is the line from -∞ + iy to +∞ + iy. This is the limit as M → ∞ of I_M, the integral from -M + iy to +M + iy. Since the integrand is an analytic function of z, get the same result I_M if we deform the contour into a path from -M+iy to -M, then from -M to +M, then from M to M+iy. The part from -M to +M goes along the real axis, and the parts from -M+iy to -M and from M to M+iy go to zero as M → ∞, so we are left with
    [tex] I = \int_{-\infty}^{\infty} e^{-(a + ib) x^2} \, dx.[/tex]
    Now the usual method of squaring I and converting the double integral to polar coordinates applies essentially unchanged.
     
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