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Integral equality

  1. Aug 13, 2005 #1
    Integral equality....

    let be a and b real numbers..and let be the integral...

    [tex]\int_{-\infty}^{\infty}dxf(x)x^{a}\int_{-\infty}^{\infty}g(x+y)y^{ib}=0 [/tex] so if this is zero also will be its conjugate:

    [tex]\int_{-\infty}^{\infty}dxf(x)x^{a}\int_{-\infty}^{\infty}g(x+y)y^{-ib}=0 [/tex] now let,s suppose we would have that (1-a,-b) is also a zero so:

    [tex]\int_{-\infty}^{\infty}dxf(x)x^{1-a}\int_{-\infty}^{\infty}g(x+y)y^{-ib}=0 [/tex] then my conclusion is that 1-a=a a=1/2 and there is no other solution.. :biggrin: :zzz: :biggrin: :biggrin: :biggrin: :biggrin:
     
  2. jcsd
  3. Aug 16, 2005 #2

    shmoe

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    You've *somehow* decided that if your integrals are equal then x must have the same exponent? I don't even know where to begin with what's wrong here, but by your logic shouldn't the same be true for y then, so b=0? Please, tell me this is a joke.

    Have you been reading some of the horrible "proofs" of the Riemann Hypothesis you find online? This looks awfully familiar.
     
  4. Aug 16, 2005 #3
    Are you integrating over y in the second integral? x? A nonexistent variable?
     
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