How to model a generic low frequency signal?

Question

I'm trying to apply Fourier analysis to a specific problem I have. I have essentially an integral like the following

$$ \int_{\Omega} f(t) g(t) dt $$

And I'm trying to assume that $g$ is a narrow band signal (namely the Fourier transform is defined on a compact set). I want to prove that if $g$ is really narrow band then

$$ \int_{\Omega} f(t) g(t) dt \approx \left( \int_{\Omega} f(t) dt \right) \left( \int_{\Omega} g(t) dt \right) $$

To prove this, I made this assumption

$$ g(t) = g_a(t) = \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega $$

Where $a$ is a positive real parameter. When $a \to 0$ this integral turn out to be

$$ \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega \to \hat{g}(0) $$

which means

$$ g(t) = g_a(t) \to \hat{g}(0) = \int_{-\infty}^{+\infty} g(t) dt $$

and therefore

$$ \int_{\Omega} f(t) g(t) dt \approx \int_{\Omega} f(t) \hat{g}(0) dt = \hat{g}(0) \int_{\Omega} f(t) dt = \int_{-\infty}^{+\infty} g(t) dt \int_{\Omega} f(t) dt $$

which is not exactly what I want , but it's close enough. The final question is:

is the model

$$ g_a(t) = \frac{1}{a} \int_{-\infty}^{+\infty} \hat{g}(\omega)rect \left( \frac{\omega}{a} \right) e^{-j2\pi \omega t} d \omega $$

the correct model for a narrow band signal? I'm not entirely sure because of the factor $1/a$ I've introduced. But the idea why it makes sense is the following.

If the signal has narrow band, that means the variation in time is really slow, that means the narrower is the band the more the signal tend to be constant, and in theory if it would be constant then I'd be able to factorize the function $g(t)$ from the integral.

Clarification : probably in this discussion is more correct to say "low frequency".

Answer 1

Note that you can also use Parseval's theorem. Assuming that $f(t)$ and $g(t)$ are both real-valued we have:

$$\int_{-\infty}^{\infty}f(t)g(t)dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\tag{1}$$

If $g(t)$ is a narrow-band low-pass signal, i.e., if we can model $G(\omega)$ as $G(\omega)\approx G(0)\text{rect}(\omega/\Delta\omega)$, and if we can assume that $F(\omega)$ is approximately constant (equal to $F(0)$) in the narrow interval $\Delta\omega$, then we can write

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)G^*(\omega)d\omega\approx\frac{1}{2\pi}F(0)G^*(0)\Delta\omega\tag{2}$$

where $\Delta\omega$ is twice the bandwidth of $g(t)$ (because the bandwidth is defined as the support at positive frequencies). With

$$F(0)=\int_{-\infty}^{\infty}f(t)dt$$

and

$$G^*(0)=G(0)=\int_{-\infty}^{\infty}g(t)dt$$

we obtain from $(1)$ and $(2)$

$$\int_{-\infty}^{\infty}f(t)g(t)dt\approx\frac{\Delta\omega}{2\pi}\int_{-\infty}^{\infty}f(t)dt\int_{-\infty}^{\infty}g(t)dt\tag{3}$$