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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.

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    $\begingroup$ The variable $k'$ integrates out and you're left with a function of $k$ similar to your "dull" Fourier transform. $\endgroup$ Apr 26, 2022 at 18:50
  • $\begingroup$ Yea, but what I meant is how to rigorously derive it? If I write $$e^{\displaystyle-i\,k\,x} = e^{\displaystyle -i(k-k')\,x}\,e^{\displaystyle-i\,k'\,x}$$ for $$\hat{f}(k) = \frac{1}{\sqrt{2\,\pi}} \int_{-\infty}^{+\infty} f(x)\,e^{\displaystyle -i\,k\,x}\,\mathrm{dx}$$ then I get $$\hat{f}(k) = \frac{1}{\sqrt{2\,\pi}} \int_{-\infty}^{+\infty} f(x)\,e^{\displaystyle -i(k-k')\,x}\,e^{\displaystyle-i\,k'\,x}\,\mathrm{dx}$$ What's not really the equivalence of the last equation. There is the integral over $k'$ missing $\endgroup$
    – Leon
    Apr 27, 2022 at 22:14

1 Answer 1

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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.

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