why do we get a zero coefficient for frequencies not in the sound-wave using Fourier transform?

Question

I'm learning about Fourier transforms and watching this video.

The instructor says that if a frequency exists in a sound, we will get a non-zero coefficient using Fourier transform and a zero coefficient for the frequencies not in the sound.

Let's look at these examples:

This is our signal with a frequency of 1

This is the Fourier transform for frequency 1 (which is in the sound):

As we can see, the centre of gravity(the average of all points)is a non-zero value.

But for a frequency not in the sound (for example, 1.2), we get a symmetric shape, and the average of all points become zero.

The formula for calculating the coefficient is this: $ \huge{ \hat{g}(t) = \int {g(t) e^{-i2\pi ft}} dt} $

Where f is the frequency, we're checking, and t is time.

Why do we get a symmetric shape for the frequencies not in the sound (therefore a zero centre of gravity) and a non-symmetric shape for frequencies based on this formula?

Answer 1

A frequency $f$ being present in the signal $g(t)$ means that we can decompose $g(t)$ into a bunch of terms, one of which is (some scalar multiple of) $e^{2 \pi i f t}$.

Now the Fourier transform of this signal is $$ \int_0^1 e^{2 \pi i f t} e^{-2 \pi i f t} \, d t = \int_0^1 1 \, d t = 1. $$

On the other hand, if we try this with some frequency not present in the signal, say $f' \neq f$, we get $$ \int_0^1 e^{2 \pi i f' t} e^{-2 \pi i f t} \, d t = \int_0^1 e^{2 \pi i (f' - f) t} \, d t = \frac{e^{2 \pi i (f' - f) t}}{2 \pi i (f' - f)} \biggr\rvert_0^1 = 0. $$

Note that the integral in the opening post (and in the video) is missing the bounds of integration, which in this case are very important.