There were similar questions but I am having trouble inferring a solution from them.
Assume all integrals are $\pm\infty$
I am supposed to prove that $$F^{-1}[Y_1(w)Y_{2(w)}] =\int y_1(t-\tau)y_2(\tau)d\tau$$
I have made some substitutions starting from the definition of the inverse Fourier transform and have gotten to here:
$$F^{-1}[Y_1(w)Y_{2}(w)] = \frac{1}{2 \pi} \int [\int y_1(t)e^{-i\omega t}dt ] [\int y_2(\tau)e^{-i\omega \tau}d\tau] e^{i\omega t'}d\omega$$
I then use the delta function projection property to project out t' and end up with the end result except there is an extra 2$\pi$
where does the 2$\pi$ go from that step to the ending step?
$$F^{-1}[Y_1(w)Y_{2}(w)]=\frac{1}{2 \pi} \int \int \delta (t-t'-\tau) y_1(t)dt y_2(\tau)d\tau$$
how do I get rid if the 1/2$\pi$?