- #1

Telemachus

- 835

- 30

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 [tex]f(x)=H(x)-H(x-l)[/tex] (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 [tex]\int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega[/tex]

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:

Then the Fourier integral representation:

[tex]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[/tex]

But g(x) is odd, then:

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

Help please :)

PD, I don't know why latex isn't working in some cases, so I've attached some images.

The problem says: Given [tex]f(x)=H(x)-H(x-l)[/tex] (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 [tex]\int_0^{\infty}\frac{1-cos (\omega l)}{\omega}\sin (\omega l) d\omega[/tex]

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:

Then the Fourier integral representation:

[tex]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[/tex]

But g(x) is odd, then:

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

Help please :)

PD, I don't know why latex isn't working in some cases, so I've attached some images.

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