1. ## Fourier transform

Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
The problem says: Given $\displaystyle f(x)=H(x)-H(x-l)$ (H(x) is the Heaviside unit step function).
a) Consider the odd extension for f and find its Fourier integral representation.
b) Using the previous incise calculate the value for $\displaystyle \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega$

Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
$\displaystyle g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}$

Then the Fourier integral representation:
$\displaystyle g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_ {-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega$

But g(x) is odd, then:

$\displaystyle g(x)=\displaystyle\frac{1}{2\pi}\displaystyle\int_ {-\infty}^{\infty} 2 \displaystyle\int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{0}^{\infty}g(x) \sin(\omega x)}dx e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \displaystyle\int_{0}^{\infty} \sin(\omega x)}dx - \displaystyle\int_{-\infty}^{0} \sin(\omega x)}dx\right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \cos(\omega x)} |_0^{\infty} - \cos(\omega x)}|_{-\infty}^{0} \right ) e^{i \omega x}d\omega=\displaystyle\frac{-i}{\pi}\displaystyle\int_{-\infty}^{\infty} \left ( \frac{2}{\omega}-\displaystyle\lim_{c \to{+}\infty}{}\cos{\omega c}-\displaystyle\lim_{c \to{-}\infty}{}\cos{\omega c}}\right ) e^{i \omega x}d\omega$

And the thing is that the integral for the cosine diverges as I see it, but I'm probably doing something wrong.

2. Originally Posted by Ulysses
Hi there. I'm starting with the Fourier transforms, and I'm having some trouble with my first exercise on this topic.
The problem says: Given $\displaystyle f(x)=H(x)-H(x-l)$ (H(x) is the Heaviside unit step function).
a) Consider the odd extension for f and find its Fourier integral representation.
b) Using the previous incise calculate the value for $\displaystyle \int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega$

Well, for a) I think I should get the fourier transfor for the sign function, I don't know if this is right, but anyway I've defined the function like this:
$\displaystyle g(x)=\begin{Bmatrix}{ 0}&\mbox{ if }& x>l\\1 & \mbox{if}& 0<x<l\\-1 & \mbox{if}& -l<x<0\\0 & \mbox{if}& x<-l\end{matrix}$
Assuming that $\displaystyle l>0$ you should have:

$\displaystyle g(x)= \left\{ \begin{array}{ll}1,&0\le x<l \\ 0,&\text{otherwise}\end{array} \right.$

What you give is the odd extension (more of less).

Then the Fourier integral representation:
$\displaystyle g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x)e^{-i \omega x}dx e^{i \omega x}d\omega$

But g(x) is odd, then:

$\displaystyle g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} 2 \int_{0}^{\infty}g(x)(-i)\sin(\omega x)}dx e^{i \omega x}d\omega$
OK to this point

.................. $\displaystyle =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{\infty}g(x)\sin(\omega x)}dx e^{i \omega x}d\omega$
You have lost the 2, and this last integral becomes (putting the 2 back in) OK found it below now fixed:

.................. $\displaystyle =\frac{-i}{\pi}\int_{-\infty}^{\infty}\int_{0}^{l}\sin(\omega x)}dx e^{i \omega x}d\omega$

3. I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks

4. Originally Posted by Ulysses
I canceled it out with the 2 in the denominator, are you sure that's wrong? I've found one mistake, it's pretty silly. I was making for -infinity to zero, but I shouldn't use that part. I see now, I shouldn't take from zero to infinity, I must use l. Thanks
You are correct about the 2.

CB