# Homework Help: Fourier transform from k-space to x

1. Nov 13, 2012

### jmm5872

I have calculated a k-space function to be f(k) = $\frac{1}{2k}$

I want to fourier transform this to find f(x), I have found many different fourier transform equations...can I use this one?

f(x) = $\frac{1}{\sqrt{2π}}$$\int$$\frac{1}{2k}$e-ikxdk Limits fo integration -Infinity to Infinity

I'm also having trouble remembering my complex analysis, would I use residue theorem for this integral? I don't quite remember exactly how that goes...

∫f(z) = 2πiRes(f,z0) Correct?

The residue for f(k) is 1/2, so I am not sure excatly what the answer means? Or if I am using residue theorem correctly?

Final answer: f(x) = $\frac{1}{\sqrt{2π}}$2πi(1/2) = $\sqrt{\frac{π}{2}}$i
This doesn't make sense as an answer.

Thanks,
Jason

2. Nov 13, 2012

### Mute

The issue with your calculation is that you seem to have forgotten that the residue theorem applies when the contour you are integrating over completely encircles a pole in the complex plane. What you calculated is the integral over a contour around the point k = 0. The integral you want to calculate is

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty dk~\frac{e^{-ikx}}{k},$$
so you need a closed contour, at least one piece of which runs along the real axis while avoiding the pole at k = 0. One typically skirts around the pole with a semi-circular arc of small radius $\epsilon$, and then closes the contour with an arc of radius R, either in the upper half plane or the lower half plane. One takes $R \rightarrow \infty$ and $\epsilon \rightarrow 0$ at the end of the calculation.

There are a number of important things to note about such a contour: 1) the pole is not enclosed by it, so the full contour integral is zero; the integrals along the segments of the contour thus sum to zero. 2) Which half-plane (negative or positive) you close the large arc in will depend on whether or not x is positive or negative.

Have you computed a contour integral like this before? Give it a shot and see what you find.