- #1

- 80

- 0

I have the following integral:

[tex]

\int^{\infty}_{-\infty}p(x)U(x)e^{-i \omega x}dx

[/tex]

Where p(x) is any real function which is always positive and U(x) is the step function

Obviously this can easily be solved using the convolution theorem because I have

[tex]

\mathcal{F}[p(x)U(x)] = P(\omega)*U(\omega)

[/tex]

The problem I having is with the very similar integral but the exponential is now positive:

[tex]

\int^{\infty}_{-infty}p(x)U(x)e^{+i \omega x}dx

[/tex]

I don't know how to deal with this integral - even though I suspect I can use the convolution theorem on it.

I've tried to derive the convolution theorem for both exponentials but I get stuck at the stage:

[tex]

\int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega')U(\omega-\omega')d\omega'e^{-i\omega x}d\omega = p(x)u(x)

[/tex]

And:

[tex]

\int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega')U(\omega-\omega')d\omega'e^{+i\omega x}d\omega = p(x)u(x)

[/tex]

My problem is this:

If I define the [tex] \int^{\infty}_{-\infty} f(x) e^{-i\omega x}[/tex] integral as the Fourier Transform - then I can write the second equation as:

[tex]

\mathcal{F}^{-1}[P(\omega)*U(\omega)] =p(x)u(x)

[/tex]

And thus applying the inverse Fourier operator to both sides I get:

[tex]

[P(\omega)*U(\omega)] =\mathcal{F}[p(x)u(x)]

[/tex]

But If I set up this convention for my Fourier Transform how do I deal with the first equation:

[tex]

\int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega')U(\omega-\omega')d\omega'e^{-i\omega x}d\omega = p(x)u(x)

[/tex]

This isn't a Fourier Transform operation anymore - its slightly different. Is there anything I can do from here to show what:

[tex]

\int^{\infty}_{-\infty} p(x)u(x)e^{i\omega x} dx

[/tex]

is in terms of convolutions ?