# Integral in k-space

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1. Jan 19, 2016

### ognik

1. The problem statement, all variables and given/known data
Hi - this exercise appears in a section on Fourier Transforms.

Show $\int e^{ik \cdot (r - r')} \frac{d^3 k}{(2 \pi)^3 \vec{k}^2} = \frac{1}{4 \pi}|r-r'|$
Hint: use spherical coords in k-space.

2. Relevant equations
I am not sure of the coord. transform, but from the Jacobian I used $d^3k = k^2 sin \theta dk d\phi d\theta$
Let R=r-r', then $k \cdot (r - r') = kR cos \theta$

3. The attempt at a solution

$\int e^{ikR cos \theta} \frac{d^3 k}{(2 \pi)^3 \vec{k}^2} = \frac{1}{(2\pi)^3}\int_{0}^{\infty} \,dk \int_{0}^{\pi} e^{ikR cos \theta} \,d\theta \int_{0}^{2\pi} \,d\phi$

Let $cos \theta = t, \therefore dt = sin \theta d\theta, 1 \le t \le -1$

$=> \frac{1}{(2\pi)^3}\int_{0}^{\infty} \,dk \int_{1}^{-1} e^{ikR t} \,dt \int_{0}^{2\pi} \,d\phi$

$= \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{e^{ikR t}}{ikR}|^{-1}_1 \,dk$

$= \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{e^{-ikR }-e^{ikR }}{ikR}|^{-1}_1 \,dk = \frac{1}{(2\pi)^2}\int_{0}^{\infty} \frac{-2 sin(kR)}{kR} \,dk$

Could someone tell me if I have it right so far please, and/or point out any mistakes? (I would then use residues to try and solve the integral)

2. Jan 19, 2016

### blue_leaf77

I believe the RHS should be $\frac{1}{4 \pi |r-r'|}$.

Where is $\sin \theta$ in the integrand?

One mistake there about the presence of minus sign.

You have plugged in the integration limits for $t$ in the previous step, therefore you don't need to write "$|_{1}^{-1}$" again

Except for one technical mistake I have pointed out above, you are right.

3. Jan 19, 2016

### ognik

Thanks, silly mistakes copying from my paper ....but I was over-confident about using residue and would appreciate some further guidance.

So I must evaluate $\frac{1}{4 \pi^2} \int^\infty_0 - \frac{sin kR}{kR} \,dk$ Let k = z, then this has a simple pole at z = 0?

So I would need a contour that goes from 0 to ∞ along the real axis, then maybe a quarter circle in the 1st quadrant - and back to the start down the imaginary axis... but how do I include the pole? Also I know the arc will go to 0 as R tends to ∞, but what does the 3rd part of the path do? Altogether I think I need some help with what path to use please?

4. Jan 20, 2016

### blue_leaf77

There are a couple of ways to integrate $\textrm{sinc}$ function. Apart from using residue theorem, you can also use Fourier transform, or another method, to express $1/x$ in terms of integral of an appropriate exponential function. But since you seem to be insisting on using residue theorem, you can start by remembering that the function $\frac{\sin x}{x}$ is an even function. The integral over semi-infinite $x$ axis is then equal to the half of integral over the entire $x$ axis. The appropriate complex function to calculate this integral is
$$\int \frac{e^{iz}}{z} dz$$
The contour should be a semicircle on the upper half of the complex plane. The pole is obviously at $z=0$, so the contour line will exactly pass through the pole. Then make the radius of the semicircle arbitrarily to infinity because there are no other poles in the upper plane (so are in the lower half). At this point, I hope you know the simple rule for calculating the residue if the corresponding pole lies exactly on the contour line, and especially when there are no more poles than this one which lie on the line.

5. Jan 20, 2016

### ognik

I don't think so, the book seems to cover only interior and exterior poles.

For interior I have Res(f)=$\frac{1}{(n-1)!} \lim_{{z}\to{z_0}} \frac{d^{n-1}} {{dz}^{n-1}} (z-z_0)^n f(z)$, with the integral = $2\pi i$ times the res sum.

I have found suggestions that in the case of the pole on the straight line section of the contour, the residue is just halved, but it is not clear if that is applicable to all orders of poles and/or other contours, are you able to add anything to this?

(For this simple pole I get the final answer of $\frac{\pi}{R}$ for the integral which fits the expected answer)

6. Jan 20, 2016

### blue_leaf77

That's exactly what I meant in the line that you quoted. Another alternative is to directly calculate the contour integral using a contour line formed by halving a donut, taking only the upper half. The inner and outer radii should be centered on the pole. Make the outer radius goes to infinity and the inner one to zero to calculate the integral.