Proof of fourier transformation of multiplication of two signals

Question

I've been trying to find a proof of the following, but still I m unable to proof it, can someone help me? $$ ℱ[x(t)g(t)] = \frac{1}{2\pi} [X(\omega)*G(\omega)] $$

Answer 1

I am very thankful to Dilip-sarwate and Gilles, who took their precious time to understand my problem and guide me.
So, Now I'm going to write the correct solution to my question. Which is as follows: $$ ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}[x(t)g(t)]e^{-j\omega t}dt $$
As we know : $$ x(t) = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha $$
Therefore, $$ ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}\bigg[\frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha\bigg]g(t) e^{-j\omega t}dt $$
After rearranging, we have $$ ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)\bigg[\int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt\bigg]d\alpha $$
As we also know that : $$ G(\omega) = \int_{t=-\infty}^{\infty}g(t)e^{-j\omega t}dt $$ and replacing $\omega$ with $\omega-\alpha$, we get: $$ G(\omega-\alpha) = \int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt $$
Therefore, $$ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)G(\omega-\alpha)d\alpha$$
Finally, we get $$ ℱ[x(t)g(t)] = \frac 1{2\pi}[X(\omega)*G(\omega)] $$
[$\because$ $x_1(t)*x_2(t) = \int_{\tau=-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau$]

Answer 2

HINT:

Write it as: $$ \mathcal{F}\big\{x_1(t)x_2(t)\big\} = \frac 1{2\pi}\int_{-\infty}^{\infty}\bigg[\mathcal{F}^{-1}\big\{X_1(\alpha)\big\}x_2(t)\bigg]e^{-j\Omega t}dt $$

You correctly used the inverse Fourier transform of $X_1(\alpha)$ for a new expression of $x_1(t)$ as in $(1)$

$$ x_1(t) = \frac 1{2\pi}\int_{-\infty}^{\infty}X_1(\alpha)e^{j\alpha t}d\alpha\tag{1} $$ Don't express $x_2(t)$ in the form of the inverse Fourier transform, leave it as it is and you'll end up with an exponential to the power $-j(\Omega - \alpha)t$. The expression can be further simplified and give equation $(2)$ $$ \frac 1{2\pi}\int_{-\infty}^{\infty}X_1(\alpha)X_2(\Omega - \alpha)d\alpha\tag{2} $$