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Function to satisfy this integral for all parameters?

  1. Jan 23, 2008 #1
    I wonder:

    Is it possible to find a function f(x) which satisfies this integral for *arbitrary* parameters a,b (but maybe of known domain)?

    [tex]\int_{-\infty}^\infty e^{-x^2}f(ax+b)\mathrm{d}x=0[/tex]

    I want [tex]f(\pm\infty)=0[/tex] and I can deduce
    [tex]\int_{-\infty}^\infty x^n e^{-x^2}f^{(m)}(ax+b)\mathrm{d}x=0[/tex]

    It would have to be oscillary I guess?!
    Last edited: Jan 23, 2008
  2. jcsd
  3. Jan 24, 2008 #2
    Yes, and no. Obviously, f(x)=0 satisfies the criterion, but this is probably not what you were looking for. If you do a change of variable in the expression, you'll see that the integrand is exp(-((x-b)/a)^2)*f(x), i.e., a Gaussian times some function. However, if you let a and b be arbitrary, you're allowing the Gaussian's position and width to vary as well, and by making a arbitrarily large, you can make the Gaussian as thin as you want. The net effect is that the integral would "pick out" the value of f(b), so unless f is 0 everywhere, the integral cannot be 0 everywhere.

    This, of course, assumes that f is well-behaved. Any function which did satisfy the criterion would be nowhere continuous, as it would have to oscillate infinitely at every point.
  4. Jan 24, 2008 #3
    And I an estimate for b and a is known? Say it's in a range around some a0 and b0.
  5. Jan 25, 2008 #4


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    Since coming here I have found questions I could answer, but from the other answers it looked like I hadn't understood the question. :redface: With that reservation:

    Firstly, why f(ax + b) ? Call it g(x). If you can get that you can translate into a function of (ax + b) afterwards.

    Now [tex] e^{-x^2}[/tex] is symmetrical about x = 0. So if g is any continuous and finite function of x that is antisymmetrical i.e. g(x) = - g(-x) , and defined from - to + infinity, it will satisfy your requirement won't it? Simplest example g(x) = x.

    If this shows I have not understood the question (guess no-one would ask one that easy) someone please tell me as other times I was left suspended.:confused::smile:
    Last edited: Jan 26, 2008
  6. Jan 27, 2008 #5


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    :redface: sorry, g(x) = x doesn't satisfy your [tex]f(\pm\infty)=0[/tex] condition, but you can easily find one that does, e.g. any function symmetrical about x = 0 that does, multiplied by x.
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