Conceptual Problem with Convolution Theorem

[thrillhouse86]

Hi - I'm trying to work out the following convolution problem:

I have the following integral:
<br /> \int^{\infty}_{-\infty}p(x)U(x)e^{-i \omega x}dx<br />

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

The problem I having is with the very similar integral but the exponential is now positive:
<br /> \int^{\infty}_{-infty}p(x)U(x)e^{+i \omega x}dx<br />

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:

<br /> \int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega&#039;)U(\omega-\omega&#039;)d\omega&#039;e^{-i\omega x}d\omega = p(x)u(x)<br />

And:

<br /> \int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega&#039;)U(\omega-\omega&#039;)d\omega&#039;e^{+i\omega x}d\omega = p(x)u(x)<br />

My problem is this:
If I define the \int^{\infty}_{-\infty} f(x) e^{-i\omega x} integral as the Fourier Transform - then I can write the second equation as:
<br /> \mathcal{F}^{-1}[P(\omega)*U(\omega)] =p(x)u(x)<br />
And thus applying the inverse Fourier operator to both sides I get:
<br /> [P(\omega)*U(\omega)] =\mathcal{F}[p(x)u(x)]<br />

But If I set up this convention for my Fourier Transform how do I deal with the first equation:
<br /> \int^{\infty}_{-\infty} \int^{\infty}_{-\infty}P(\omega&#039;)U(\omega-\omega&#039;)d\omega&#039;e^{-i\omega x}d\omega = p(x)u(x)<br />

This isn't a Fourier Transform operation anymore - its slightly different. Is there anything I can do from here to show what:
<br /> \int^{\infty}_{-\infty} p(x)u(x)e^{i\omega x} dx<br />

is in terms of convolutions ?

 

[AlephZero]

Think about real and imaginary parts of the integral separately.

p(x) and U(x) are real functions.

e^{i\omega x} = \cos \omega x + i \sin \omega x

e^{-i\omega x} = \cos \omega x - i \sin \omega x

That's all you need to answer the question.

 

[thrillhouse86]

Thanks AlpehZero - I guess it always helps to go back to the fundamental definitions ...