# Making sense of the Fourier transform of the product of two functions [closed]

The Fourier transform of the product of two functions f(x) and g(x) is given as:

$\mathcal{F}[ f(x)g(x)] = \int_{-\infty}^{+\infty} F(\omega^\prime) G(\omega - \omega^\prime) d\omega^\prime \; = \; \mbox{convolution of} \; \; F(\omega^\prime )G(\omega^\prime)$

where $F(\omega^\prime)$ and $G(\omega^\prime)$ are the Fourier transforms of $f(x)$ and $g(x)$ respectively.

Although I understand the derivation of this formula, I've got difficulty making sense of two frequency terms $\omega$ and $\omega^\prime$. I'm fine with $\omega^\prime$ but I don't know what to make of $\omega$. Should I treat it as a constant, or should I set it to zero?

I'm really interested in the Fourier transform of the square of the second derivative of a function e.g. $\mathcal{F}[ f^{\prime\prime}(x)^2 ]$. Because this problem does not involve a shift, I don't know what to make of the shift term $\omega$.

## closed as not a real question by Charles Matthews, Daniel Moskovich, Deane Yang, José Figueroa-O'Farrill, Gjergji ZaimiSep 6 '10 at 13:27

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## 1 Answer

The notation is not quite right, let F(w) and not F(w') be the Fourier transform of f(x) (resp G(w) for g(x)), so that w is the "frequency term", and w' (or better $\lambda$ to avoid confusion) be just a dummy variable for integration.

Then the convolution integral $\int^{\infty}_{-\infty}{F(\lambda)G(w-\lambda)d\lambda}$ see $\lambda$ as the variable of integration and after integration $\lambda$ will vanish and the result will be a function of w.

For the second question: If $F(w)$ is the Fourier transform of f(x), recall $\frac{d^nf(x)}{dx^n}$ ~ $(2 \pi iw)^n F(w)$.

Then $f''(x) f''(x)$ has Fourier Transform a convolution of $(2 \pi iw)^2 F(w)$ with itself, i.e $(2\pi )^4 \int^{\infty}_{-\infty}{\lambda^2 F(\lambda) (w-\lambda)^2 F(w-\lambda)d\lambda}$.