# Calculate the Fourier transform of a product of three functions

1. Oct 18, 2009

### parton

I have a problem understanding the following:

I should calculate the Fourier transform of a product of three functions:

$$\mathcal{F} \left[ f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) \right] = \int dx_{1} dx_{2} f(x_{1}) g(x_{2}) h(x_{1} + x_{2}) e^{i p x_{1} + i q x_{2}}$$

okay, and this goes over into a convolution:

$$= \dfrac{1}{2 \pi} \hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q) = \dfrac{1}{2 \pi} \int dk \hat{f}(p-k) \hat{g}(q-k) \hat{h}(k)$$.

I know how to calculate the convolution between two functions, but here we have three and I don't undestand how to get to the last line here. Could somebody explain that to me, please?

Last edited: Oct 18, 2009
2. Oct 18, 2009

### HallsofIvy

Staff Emeritus
Re: Convolution

Pretty much the same way you extend the product of two numbers to three or the sum of two numbers to three. Writing f #g for the convolution of two functions, the convolution of three can be written (f#g)#h= f#(g#h).

3. Oct 18, 2009

### parton

Re: Convolution

$$\hat{g}(q) \ast \hat{h}(p+q) = \int dk ~ \hat{g}(p+q-k) \hat{h}(k)$$
But if I go further and calculate the second convolution, i.e. $$\hat{f}(p) \ast \hat{g}(q) \ast \hat{h}(p+q)$$, I would get a second integral. But in the equation above there is just one. I don't understand that. Maybe it is trivial, but I just don't see what I have to do?