Fourier Transform of Product of 2 Function (inverse convolution) Having the Fourier Transform defined as $$\mathcal{F}(f(x)) = \hat{f}(k) = \frac{1}{\sqrt{2\,\pi}} \int_{-\infty}^{+\infty} f(x)\,e^{\displaystyle -i\,k\,x}\,\mathrm{dx}$$
I am asked to form the FT of the product of two function $f_1(x)\,f_2(x)$.
Writing it out is kind of dull: $$\mathcal{F}(f_1\,f_2) = \frac{1}{\sqrt{2\,\pi}} \int_{-\infty}^{+\infty} f_1(x)\,f_2(x)\,e^{\displaystyle -i\,k\,x}\,\mathrm{dx}$$
I'm sure this is not the final answer because of some manipulations that can be done.
Presumable it is expected to discover the convolution out of nowhere:
$$ \begin{align*}\mathcal{F}(f_1\,f_2) = \hat{f}_1 \star \hat f_2 = &\int_{-\infty}^{+\infty}\hat{f}_1\,(k')\,\hat{f}_2(k-k')\,\mathrm{dk'}\\\\ &\int_{-\infty}^{+\infty}\left[\,\frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{+\infty}f_1(x)\,e^{\displaystyle -i\,k'\,x}\mathrm{dx}\,\frac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{+\infty}f_2(x)\,e^{\displaystyle -i\,(k-k')\,x}\,\mathrm{dx}\,\right]\mathrm{dk'}\end{align*}$$
Actually (if right so far) the formula can be recovered simply by rewriting the first equation: $$\\e^{\displaystyle-i\,k\,x} = e^{\displaystyle -i(k-k')\,x}\,e^{\displaystyle-i\,k'\,x}$$
I just wonder how it comes one integrates over $k'$ and if it's not total nonsense at all.
 A: Assuming the Fourier transform is defined by
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^\infty f(x)\,e^{-i\,\omega\,x}\,dx\tag{1}$$
and the inverse Fourier transform is defined by
$$f(x)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)=\int\limits_{-\infty}^\infty F(\omega)\,e^{i\,x\,\omega}\,d\omega\tag{2}$$
the convolution
$$F_1(\omega)\,*\,F_2(\omega)=\int\limits_{-\infty}^\infty F_1(\omega')\,F_2(\omega-\omega')\,d\omega'=\int\limits_{-\infty}^\infty F_1(\omega')\,\left(\int\limits_{-\infty}^\infty f_2(x)\,e^{-i\,(\omega-\omega')\,x}\,dx\right)\,d\omega'\tag{3}$$
can be evaluated by interchanging the order of integration as
$$F_1(\omega)\,*\,F_2(\omega)=\int\limits_{-\infty}^\infty f_2(x)\,\left(\int\limits_{-\infty}^\infty F_1(\omega')\,e^{i\,x\,\omega'}\,d\omega'\right)\,e^{-i\,\omega\,x}\,dx$$ $$=\int\limits_{-\infty}^\infty f_2(x)\,f_1(x)\,e^{-i\,\omega\,x}\,dx\tag{4}=\mathcal{F}_x[f_1(x)\,f_2(x)](\omega)$$
so the Fourier transform of the product of two functions is the convolution of the Fourier transforms of the two functions.
