# 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?