Fourier Transform of Power Laws

Question

Is there a function $f(x)$ whose Fourier Transform has the form $\tilde{f}(\omega)=ic\omega$ where $c$ is a constant, and $i$ is the imaginary number.

Answer 1

There is not a function, but the Distribution (i.e., Generalized Function), $-c\delta'(x)$, the Dirac unit doublet, has Fourier Transform

$$\int_{-\infty}^\infty (-c\delta'(x))e^{i\omega x}\,dx=i\omega c$$

Answer 2

Here is an intuitive way to arrive at the conclusion proposed by Dr. MV.

First of all, I will be taking the Fourier transform to be $\tilde f(w) = \int_{-\infty}^\infty f(x) e^{-i x w} dx$.

Now if we set $g(x) = f'(x)$ then $\tilde g(w) = (-iw) \tilde f(w)$.

Letting $\tilde f(w) = icw$ we have $\tilde g(w) = -c$. We can thus look for a function for which the Fourier transform is a constant, and then integrate to find $f$ or we can try to find $f$ directly. My following argument will work for either case.

This is a bit advanced for what you are asking, but the following argument contains many useful rules of thumb for someone working in Fourier Analysis. Our principle tool is the Paley Wiener Theorem. Essentially, if we take a function that is compactly supported, then its Fourier transform is an entire (analytic over all of $\mathbb{C}$) function.

Moreover, if $f$ is zero outside a set $[-A,A]$ then its Fourier tranform is an entire function for which $|\tilde f(w)| < C e^{A|w|}$ for some positive $C$. The nice thing is that we can take this backwards too, if we start with an entire function satisfying the above inequality, it's inverse Fourier transform is compactly supported in $[-A,A]$.

Note that $|\tilde f(w)| = |icw| = c|w| < C e^{A|w|}$ for any $A >0$ (with $C$ changing with different choices of $A$). Thus $f$ must be compactly supported in $[-A,A]$ for every $A > 0$. By making $A$ smaller and smaller we see that $f(x) = 0$ except at $x=0$.

Thus at best we have $$f(x) = \left\{\begin{array}{cc} B & x = 0\\ 0 & x \neq 0\end{array}\right.$$ where $B$ is a constant.

Now if we try to take the Fourier transform, we find that $\tilde f(w) = \int_{-\infty}^\infty 0 \cdot e^{-ixw} dx = 0$ (we can neglect the value of $f$ at zero). This is a contradiction, and thus there is no function that gives us what we want.

Now there are objects that we care about with the Fourier transform $icw$. Names are used such as distributions, measures, and generalized functions. They do not fit in the category of functions of real or complex variables.