Convolutions and the convolution theorem Say $f(x)=(x-1)e^{-\pi (x-1)^{2}}$ and we want to find $f*f*f*f$ where $*$ denotes convolution, how would one go about solving this?
Using the convolution theorem ? Then we would have to find the fourier transform of this function, how can this be done?
 A: Convolution theorem states that, under some general assumptions, the Fourier transform of a convolution is identical to the product of the Fourier transform of each function, i.e.,
$$
\widehat{f_1*f_2}(\xi)=\hat{f_1}(\xi)\hat{f_2}(\xi).
$$
If you have three functions convoluted together, you need to apply the convolution theorem twice, i.e.,
$$
\widehat{f_1*f_2*f_3}=\widehat{\left(f_1*f_2\right)*f_3}=\widehat{f_1*f_2}\hat{f_3}=\hat{f_1}\hat{f_2}\hat{f_3}.
$$
In general,
$$
\widehat{f_1*f_2*\cdots*f_n}=\hat{f_1}\hat{f_2}\cdots\hat{f_n}.
$$
Back to your problem. Denote $g=f*f*f*f$, and you have
$$
\hat{g}(\xi)=\left(\hat{f}(\xi)\right)^4=\left(\int_{\mathbb{R}}\left(x-1\right)e^{-\pi\left(x-1\right)^2}e^{-2\pi ix\xi}{\rm d}x\right)^4.
$$
Note that
\begin{align}
\int_{\mathbb{R}}\left(x-1\right)e^{-\pi\left(x-1\right)^2}e^{-2\pi ix\xi}{\rm d}x&=e^{-2\pi i\xi}\int_{\mathbb{R}}\left(x-1\right)e^{-\pi\left(x-1\right)^2}e^{-2\pi i\left(x-1\right)\xi}{\rm d}\left(x-1\right)\\
&=e^{-2\pi i\xi}\int_{\mathbb{R}}ye^{-\pi y^2}e^{-2\pi iy\xi}{\rm d}y\\
&=e^{-2\pi i\xi}\left(-i\xi e^{-\pi\xi^2}\right).
\end{align}
Therefore,
$$
\hat{g}(\xi)=\left[e^{-2\pi i\xi}\left(-i\xi e^{-\pi\xi^2}\right)\right]^4=\xi^4e^{-4\pi\xi^2-8\pi i\xi}.
$$
Consequently, the Fourier inverse transform of $\hat{g}$ yields
$$
g(x)=\mathcal{F}^{-1}[\hat{g}](x)=\int_{\mathbb{R}}\hat{g}(\xi)e^{2\pi ix\xi}{\rm d}\xi=\frac{1}{512\pi^2}\left[\pi^2\left(x-4\right)^4-12\pi\left(x-4\right)^2+12\right]e^{-\pi\left(x-4\right)^2/4}.
$$
The Fourier transform and inverse transform involved above could be figured out by referring to this table of important Fourier transforms.
