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Simple integral, example or general solution correct?

  1. Feb 1, 2010 #1
    This should be very simple, but I can find a simple example that violates my general conclusion. I clearly must be doing something wrong with my integration by parts. Any suggestions would be great.

    Define a distribution such that the density;
    [tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]
    which is a function ONLY of x.

    Now, let me arbitrarily calculate
    [tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k})[/tex]

    I should be able to integrate over kx by parts;
    [tex] k_{x} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) - \int dk_{x} \frac{d k_{x}}{d k_{x}} \int dk_{x} \int dk_{y} \int dk_{z} f(\vec{x},\vec{k}) = k_{x} \int d\vec{k} f(\vec{x},\vec{k}) - \int dk_{x} \int d\vec{k} f(\vec{x},\vec{k})[/tex]

    Using the stipulation at the beginning this is simply;
    [tex]k_{x} \eta(\vec{x}) - \int dk_{x} \eta(\vec{x})[/tex]

    Since, the density, eta, is only a function of x, this is zero.

    There must be something wrong with this because I can come up with a simple example that contradicts this;

    Let
    [tex]f(\vec{x},\vec{k})=\eta(\vec{x})\delta (\vec{k}-\vec{k'})[/tex]
    It is easy to see that
    [tex]\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})[/tex]

    Now calculating the original quantity
    [tex]\int d\vec{k} k_{x} f(\vec{x},\vec{k}) = \int dk_{x} k_{x} \eta(\vec{x})\delta (k_{x}-k'_{x}) = \eta(\vec{x}) k'_{x}[/tex]

    What gives? Where have I gone wrong?
     
  2. jcsd
  3. Feb 2, 2010 #2
    Your integration by parts looks incorrect. The surface term should not contain an integral over k_x, since that is the integration variable for which you are doing the integration by parts. Also, you should get something involving the derivative of f with respect to k_x, in the next term, and there should not be two integrals over k_x there.

    Sometimes it is also good to write out the limits of the integration domains.

    Torquil
     
  4. Feb 2, 2010 #3
    Well, I'm taking u as [tex]k_x[/tex] and v' as f in the standard notation
    [tex]\int u v' = u v| - \int u' v[/tex]
    therefore I will not have a derivative of f.

    Not sure what you mean by surface term. I do realize this is an integral over three variables and I should use the divergence theorem in general. But since u is a function of one variable only, it should be seperable. The k space over which the integral exsists also shouldn't matter, but I will check again.
     
  5. Feb 2, 2010 #4

    Mute

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    Homework Helper

    You integration by parts is incorrect. To clear things up, let [itex]g(\mathbf{k}) = \int dk'_x~f(k_x',k_y,k_z)[/itex], where the k_x integral is done from some intial value (say -infinity) to an arbitrary k_x, so that the result is still a function of [itex]\mathbf{k}[/itex]. Now, let's do the integral by parts, supposing that the k integrals run from 0 to infinity.

    [tex]\int dk_x k_x \int dk_y \int dk_z~f(\mathbf{k}) = \left. k_x \int dk_y \int dk_z g(\mathbf{k}) \right|^{k_x = \infty}_{k_x = -\infty} - \int dk_x \int dk_y \int dk_z~g(\mathbf{k}).[/tex]

    Assuming the surface term (the first term on the right hand side) is zero at +/- infinity, this gives

    [tex]\int dk_x k_x \int dk_y \int dk_z~f(\mathbf{k}) = - \int dk_x \int dk_y \int dk_z~g(\mathbf{k}).[/tex]
     
    Last edited: Feb 2, 2010
  6. Feb 4, 2010 #5
    Yes of course. v is the indefinate integral of v' and not over the boundary of the initial region. I should have realize this before. Thanks!

    Then, is there anything to be done without knowing the exact form of f(k,x)?
     
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