The introduction to the Laplace transform chapter in my DE textbook says (I'm paraphrasing some of this):

In general, integral transforms address the question: How much is a given function $\displaystyle y(t)$ "like" a particular standard function?

For example, if $\displaystyle y(t)$ represents a radio signal and we want to compare it to $\displaystyle sin \omega t$ with frequency $\displaystyle \frace \omega {2 \pi}$, just integrate:

$\displaystyle \int_{-N}^N y(t)\; sin (\omega t)\; dt$.

If $\displaystyle y(t)$ goes up when $\displaystyle sin(\omega t)$ goes up, and goes down when $\displaystyle sin(\omega t)$ goes down, then the integral will be large, because $\displaystyle y(t)$ is very much "like" $\displaystyle sin(\omega t)$. So far, this makes perfect sense.

Then, it says you can compare $\displaystyle y(t)$ to $\displaystyle e^{-zt}$, where $\displaystyle z=s+i \omega$ is Complex, using:

$\displaystyle \int_{-N}^N y(t)\; e^{-zt}\; dt$

Which can be split into

$\displaystyle \int_{-N}^N y(t)\; e^{-st}\; dt\,\, $ and $\displaystyle \mathbf{\int_{-N}^N y(t)\; e^{-i \omega t}\; dt }$.

The last (bold) of those integrals is called the Fourier Transform (which is what I really need to learn). Then the book says:

The Fourier Transform's value at a particular value of $\displaystyle \omega$ is a measure of the extent to which $\displaystyle y(t)$ oscillates with a frequency of $\displaystyle \frac \omega {2 \pi}$.Can someone explain how that integral compares $\displaystyle y(t)$ to an oscillation of $\displaystyle \frac \omega {2 \pi}$?

I am aware of Euler's equation:

$\displaystyle e^{it}=cos(t)+i \cdot sin(t)$

If I apply it to just the Fourier integral, I get:

$\displaystyle \int_{-N}^N y(t)\; e^{-i \omega t}\; dt\, = \int_{-N}^N y(t)\; \cos (\omega t)\; dt + i \int_{-N}^N y(t)\; \sin (\omega t)\; dt$.

Unless the $\displaystyle \sin( \omega t)$ is always zero, I don't see how this can make that comparison.

Thanks for your explanations.

Jeff