Our teacher wrote the same formula. I would like to know if the below formula is correct, and if so, how I might be able to prove it:
$$\int\limits_{-\infty}^\infty x(t)y(t) \text{d}t =1/2π\int\limits_{-\infty}^\infty X(w)Y(-w) \text{d}w$$
Note that $X(w)$ and $Y(w)$ are the Fourier transforms of $x(t)$ and $y(t)$, respectively.
$$\int\limits_{-\infty}^\infty x(t)y(t) \text{d}t =1/2π\int\limits_{-\infty}^\infty X(w)Y(-w) \text{d}w$$
$$\int\limits_{-\infty}^\infty x(t)y(t) \text{d}t=\int\limits_{-\infty}^\infty (\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty X(\omega_1)e^{i\omega_1t}d\omega_1)(\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty Y(\omega_2)e^{i\omega_2t}d\omega_2) dt$$
$=\frac{1}{2\pi}\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty X(\omega _1)Y(\omega_2)e^{it(\omega_1+\omega2)}d\omega_1d\omega_2dt$
Now the complex exponent is suggestive. Were that a minus sign, we'd be able to use a property of the Dirac Delta function to simplify the expression.
So let's let $\omega_3=-\omega_2$. Then $d\omega_3=-\omega_2$
Then the triple integral becomes:
$=\frac{1}{2\pi}\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty X(\omega _1)Y(\omega_3)e^{it(\omega_1-\omega3)}d\omega_1d\omega_3dt$
Now integrate with respect to $t$ to get a double integral:
$=\frac{1}{2\pi}\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty X(\omega _1)Y(\omega_3) \delta(\omega_1-\omega_3)d\omega_1d\omega_3$
Now integrate with respect to $\omega_1$ and let $\omega=\omega_1$
$$=\frac{1}{2\pi}\int\limits_{-\infty}^\infty X(\omega)Y(-\omega)d\omega$$