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.(adsbygoogle = window.adsbygoogle || []).push({});

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 ofx.

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 ofx, 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?

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# Simple integral, example or general solution correct?

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