Why is there an exponential in Fourier's defining integral?

Question

I am having a hard time relating integration with Fourier series. Basically, I just get lost where there is an exponential in the integration to convert into the frequency domain. If someone can explain it to me like I have only taken basic Calculus that would be appreciated.

A related concept would be why there is an exponential in cosine and sine (Euler).

Answer 1

The function in the frequency domain is characterized by this sum (discrete): $$\sum_{\omega=-\infty}^{+\infty} A(\omega)cos(2\pi\omega t)$$

Or with this integral (continuous): $$\int_{-\infty}^{+\infty} A(\omega)cos(2\pi\omega t)d\omega$$

Where $A(\omega)$ is the module. So, in discrete case we have a function represents like: $$f(t)=A_1cos(2\pi t)+A_2cos(4\pi t)+A_3cos(6\pi t)+...$$

With $A_1,A_2$, etc, that depend on the function represented.

A better, and more legible way to see this formula, is using the Euler's Formula:

$$e^{jt}=cost+jsint$$

where $j=\sqrt{-1}$, $e$ is the base of the natural logarithm.

So, you can see that from an exponential we can represent harmonics.

Answer 2

The taylor series for $f(x) = e^x$ also converges for complex $x$. And for real $\theta$ there's Euler's identity: $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. That's why the exponential appears in the integral - because of its relation to $\sin$ and $\cos$.