Fourier transform from k-space to x

[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

 

[Mute]

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

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.