Simple integral, example or general solution correct?

In summary, the conversation discusses a problem with integration by parts, specifically in regards to a distribution with a density function only dependent on one variable. The individual is struggling to find a simple example that goes against their initial conclusion. Suggestions are given to correct the integration by parts and to consider the limits of integration. It is noted that the exact form of the function f(k,x) may be necessary to fully solve the problem.
  • #1
ExtravagantDreams
82
5
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?
 
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  • #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
 
  • #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.
 
  • #4
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:
  • #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)?
 

1. What is a simple integral?

A simple integral refers to the process of calculating the area under a curve. It involves finding the anti-derivative of a given function and then evaluating it within a specific range.

2. Can you provide an example of a simple integral?

One example of a simple integral is finding the area under a straight line. For instance, if the function is f(x) = 2x + 3, the anti-derivative would be F(x) = x^2 + 3x. To find the area under the curve between x = 1 and x = 3, we would evaluate F(3) - F(1), which would give us the answer of 12.

3. How do you solve a simple integral?

To solve a simple integral, you first need to find the anti-derivative of the given function. Then, you would substitute the upper and lower limits of the integral into the anti-derivative and subtract the results to find the area under the curve.

4. What is a general solution for a simple integral?

A general solution for a simple integral is an expression that can be applied to a wide range of functions. For instance, the general solution for a simple integral would be the anti-derivative of a function, which can then be used to find the area under any curve with the same function.

5. How do you know if your solution for a simple integral is correct?

You can check the correctness of your solution by differentiating it and seeing if you get back the original function. If the derivative matches the original function, then your solution is correct. Additionally, you can also use online integral calculators to verify your answer.

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