Can You Solve This Complex Fourier Transform Problem?

[dingo_d]

Homework Statement

Find the Fourier transform of
f(x)=\frac{1}{(x^2+a^2)^2},\ a>0, and show by direct calculation that with inverse Fourier transform you'll get the original function f(x)!

Homework Equations

Fourier transform and it's inverse:
F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{i k x}dx
f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k)e^{-ikx}dk

The Attempt at a Solution

I have put my function into the transform and I got:
F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{e^{ikx}}{(x^2+a^2)^2}dx

I have transformed it to complex integral and since k is arbitrary I have two possible paths of integration, for k>0 in the upper plane and for k<0 in the lower plane. My resudues are:

Res(f,ia)=-i\frac{e^{-ka}}{4a^3}(1+ka) and

Res(f,-ia)=i\frac{e^ka}{4a^3}(1-ka).

So for k>0 my integral (without the (2\pi)^{-1/2}) is:
I=\pi \frac{e^{-ka}}{2a^3}(1+ka)
And for k<0:
I=\pi\frac{e^{ka}}{2a^3}(1-ka)

Now the problem is: which one do I use? Mathematica gives me this:
\frac{1}{2a^3}\cdot\ \sqrt{\frac{2}{{\pi}}}e^{-ak}\left((ak+1)\theta(k)-e^{2ak}(ak-1)\theta(-k)\right)
Where \theta(x) is Heaviside Step function.

I have noticed that the solutions of those integral are similar but how do I implement the step function? Plus how would I integrate it? I know that the derivative of step function is Delta function, but I don't know what the integral of it is?

Thanks!

 

[Matterwave]

The integral of a step function is much simpler to deal with than the derivative. It's simply the area under the step!

 

[vela]

Just break the integral up into two pieces, from -\infty to 0 and from 0 to \infty. One step function will equal 0, and the other will equal 1 in each integral.

 

[dingo_d]

Just break the integral up into two pieces, from -\infty to 0 and from 0 to \infty. One step function will equal 0, and the other will equal 1 in each integral.

Oh, ok that's fine. But still how did I get the step in the first place? When I was computing the transform I got 2 solutions for k\lessgtr 0, do I need to combine them to get the step or? And If I do how to do that?

 

[vela]

Yes, the step function just let's you indicate for which part of the domain the function it multiplies should apply. The \theta(k) says the (ak+1)[/tex] part only contributes for k&amp;gt;0 because \theta(k) is zero for k&amp;lt;0 and one for k&amp;gt;0. Similarly, the \theta(-k) says the e^{2ak}(ak-1) term contributes only for k&amp;lt;0. Using the step function allows you to combine the two solutions you got for k&amp;gt;0 and k&amp;lt;0 into one expression.

 

[dingo_d]

Thanks for clarification!

 

[dingo_d]

Hi I have a question about this problem... again :(

I know why I get these results with step function, that's all fine, but now I need to find the inverse (the original function).

So I have:
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k)e^{-i kx}dk. I put my previous solution up there, step function 'kills' the bounds in the integral (for \theta(k) my integral becomes: \int_{-\infty}^\infty \theta(k)dk=\int_0^\infty, right?) So I end up with 2 integrals which could be solved easily. But I have problem with that. If I try to solve this integral:
\int_0^\infty e^{-ik(x-ia)}(1+ka)dk
I have problem with that upper bound. I have e^{i\cdot\infty}, and what should I make of that? If I try to solve this with residues, I have a difficulty, because I don't have \int_{-\infty}^\infty or even function so that I could make \int_0^\infty=\frac{1}{2}\int_{-\infty}^\infty. So how to solve it?