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I am wondering if there are Fourier transforms of $\min(x,a)$ and $\max(x,a)$ functions. Please forgive me if this is a dumb question, I don't normally use Fourier transforms.

I attempted to simply solve the integral of the transform:

$$\begin{array}{rcl}\mathcal{F}(\min(x,a))&=&\int_{-\infty}^axe^{-i\omega x} \; dx+\int_a^\infty ae^{-i\omega x} \; dx \\ &=&\left[e^{-i\omega x}\left(\frac{x}{-i\omega}+\frac{1}{\omega^2}\right)\right]_{-\infty}^a+\left[\frac{ae^{-i\omega x}}{-i\omega}\right]_a^\infty \end{array}$$

However, when one evaluates the first term at $-\infty$, it's infinite. Similar situation arises when taking the integral for $\max$.

Does this mean that the Fourier transform simply does not exist? Or is there a different way to obtain the expression for it? As pointed out by @anon in chat yesterday, one can write $\min$ and $\max$ as follows:

$$\begin{cases}\min(x,a)& = xH(a-x)+aH(x-a)\\ \max(x,a)& = xH(x-a)+aH(a-x) \end{cases}$$

where $H(x)=\begin{cases}1&x\geq 0\\0&x<0\end{cases}$ is the Heaviside step function. I am not sure if that helps though.

This question is related to my previous question in that an affirmative answer of this question would solve the previous one: if there is a Fourier transform for $\min$ in closed form, then one could use it to solve the convolution of interest.

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The fouriertransforms you are looking for only exist in the sense of distributions, since both functions are not globally integrable. –  Alexander Thumm Feb 28 '12 at 18:30
    
Please excuse my ignorance, but what do mean by "in the sense of distributions"? –  M.B.M. Feb 28 '12 at 18:31
    
have a look at the Wikipedia article on distributions. –  Alexander Thumm Feb 28 '12 at 18:37

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