Homework Help: Finding Fourier transform

1. Jan 12, 2010

dingo_d

1. The problem statement, all variables and given/known data
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 $$f(x)$$!

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

3. 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!

2. Jan 12, 2010

Matterwave

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

3. Jan 12, 2010

vela

Staff Emeritus
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.

4. Jan 13, 2010

dingo_d

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?

5. Jan 13, 2010

vela

Staff Emeritus
Yes, the step function just lets you indicate for which part of the domain the function it multiplies should apply. The $\theta(k)$ says the $(ak+1)[/tex] part only contributes for [itex]k>0$ because $\theta(k)$ is zero for $k<0$ and one for $k>0$. Similarly, the $\theta(-k)$ says the $e^{2ak}(ak-1)$ term contributes only for $k<0$. Using the step function allows you to combine the two solutions you got for $k>0$ and $k<0$ into one expression.

6. Jan 13, 2010

dingo_d

Thanks for clarification!

7. Jan 18, 2010

dingo_d

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(k)e^{-i kx}dk$$. 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?